Regular expression equivalent to s.indexOf("\t")==-1

hi,
is there a way to define a regex such that:
String s = "..."; // any string s
boolean a = s.indexOf("\t")!=-1;
boolean b = s.matches("<SOME EXPRESSION>");
System.out.println(a==b);always outputs true?
thanks,
asjf

boolean a = s.indexOf("\t")==-1;
boolean b = s.matches("<SOME EXPRESSION>");
"(?!.*\t).*"
"(?!.*\\t).*"Both should do the job.
Regards

Similar Messages

Functionality equivalent to Regular Expressions

Hi,
We have a old, very badly designed database(which by the way is the culprit of all the problems).
There is a varchar2 column with values like -
C500
500FCG
50-100
MX
ADW
Y
10-5
500
B1
C500
I am trying to retrieve only digits with no leading or trailing alphabets. But i still want numbers like 50-100.
I know in 10g this could be easily done with regular expressions. But we are still using oracle 9i.
Is it possible within the sql statement or is there any other way to do it(like using some perl programs calling from plsql).
Any suggestions greatly appreciated.
Thanks,
Siri

Sorry Mohana for not explaining my req clearly.
Actually i was able to get the results by changing your query a little.
Here is the query -
select NVL(replace(translate(cov_limit, 'ABCDEFGHIJKLMNOPQRSTUVWXYZ','*'),' '), cov_limit) as After, cov_limit as Before
from test;
SQL> /
BEFORE
C500
500FCG
50-100
MX
ADW
Y
10-5
500
B1
C500
1000
AMD
CAD
MAD
WAT
FCG
CG
AD
DW
A
A500
AFTER
500
500
50-100
MX
Y
10-5
500
1
500
1000
FCG
CG
DW
*500
Only thing i am not able to understand is why the alphabet 'A' is not getting replaced with blank.
Thanks & Regards,
Sree

Query help in regular expression

Hi all,
SELECT * FROM emp11
WHERE INSTR(ENAME,'A',1,2) >0;
Please let me know the equivalent query using regular expressions.
i have tried this after going through oracle regular expressions documentation.
SELECT * FROM emp11
WHERE regexp_LIKE(ename,'A{2}')
Any help in this regard would be highly appreciated .
Thanks,
P Prakash

please go here
Introduction to regular expressions ...
Thanks,
P Prakash

Help regarding regular expression

HI All ,
Please see the following string
String s = "IF ((NOT NUM4 IS ALPHABETIC ) AND NUM3 IS ALPHABETIC-UPPER AND (NUM5 IS GREATER OR EQUAL TO 3) AND (NUM5 IS NOT GREATER THAN 3) AND (NUM3 GREATER THAN 46) AND (NUM5 GREATER THAN NUM3) OR NUM3 LESS THAN 78) .";
My problem is: i want to capture the part of this line which contains "ALPHABETIC ,ALPHABETIC-UPPER for ex :NOT NUM4 IS ALPHABETIC , NUM3 IS ALPHABETIC-UPPER.from that I have to capture the word num4 , num3 which are in these phrases only ;from the whole string whereever it exists along with the phrase,Can any one help me out by suggesting something.num4 and num3 are variable names

I suspect you're right, Sabre, but I can't resist...
import java.util.regex.*;
* A rewriter does a global substitution in the strings passed to its
* 'rewrite' method. It uses the pattern supplied to its constructor, and is
* like 'String.replaceAll' except for the fact that its replacement strings
* are generated by invoking a method you write, rather than from another
* string. This class is supposed to be equivalent to Ruby's 'gsub' when given
* a block. This is the nicest syntax I've managed to come up with in Java so
* far. It's not too bad, and might actually be preferable if you want to do
* the same rewriting to a number of strings in the same method or class. See
* the example 'main' for a sample of how to use this class.
* @author Elliott Hughes
public abstract class Rewriter
private Pattern pattern;
private Matcher matcher;
   * Constructs a rewriter using the given regular expression; the syntax is
   * the same as for 'Pattern.compile'.
public Rewriter(String regularExpression)
    this.pattern = Pattern.compile(regularExpression);
   * Returns the input subsequence captured by the given group during the
   * previous match operation.
public String group(int i)
    return matcher.group(i);
   * Overridden to compute a replacement for each match. Use the method
   * 'group' to access the captured groups.
public abstract String replacement();
   * Returns the result of rewriting 'original' by invoking the method
   * 'replacement' for each match of the regular expression supplied to the
   * constructor.
public String rewrite(CharSequence original)
    this.matcher = pattern.matcher(original);
    StringBuffer result = new StringBuffer(original.length());
    while (matcher.find())
      matcher.appendReplacement(result, "");
      result.append(replacement());
    matcher.appendTail(result);
    return result.toString();
public static void main(String[] args)
    String s = "IF ((NOT NUM4 IS ALPHABETIC ) " +
                "AND NUM3 IS ALPHABETIC-UPPER " +
                "AND (NUM5 IS GREATER OR EQUAL TO 3) " +
                "AND (NUM5 IS NOT GREATER THAN 3) " +
                "AND (NUM3 GREATER THAN 46) " +
                "AND NUM645 IS ALPHABETIC " +
                "AND (NUM5 GREATER THAN NUM3) " +
                "OR NUM3 LESS THAN 78 " +
                "AND NUM34 IS ALPHABETIC-UPPER " +
                "AND NUM92 IS ALPHABETIC-LOWER " +
                "AND NUM0987 IS ALPHABETIC-LOWER) .";
    String result =
      new Rewriter("(NUM\\d+) +IS +(ALPHABETIC(?:-(?:UPPER|LOWER))?)")
        public String replacement()
          String type = group(2);
          if (type.endsWith("UPPER"))
            return "Character.isUpper(" + group(1) + ")";
          else if (type.endsWith("LOWER"))
            return "Character.isLower(" + group(1) + ")";
          else
            return "Character.isLetter(" + group(1) + ")";
      }.rewrite(s);
    System.out.println(result);
}

Introduction to regular expressions ... last part.

Continued from Introduction to regular expressions ... continued., here's the third and final part of my introduction to regular expressions. As always, if you find mistakes or have examples that you think could be solved through regular expressions, please post them.
Having fun with regular expressions - Part 3
In some cases, I may have to search for different values in the same column. If the searched values are fixed, I can use the logical OR operator or the IN clause, like in this example (using my brute force data generator from part 2):
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE data IN ('abc', 'xyz', '012');There are of course some workarounds as presented in this asktom thread but for a quick solution, there's of course an alternative approach available. Remember the "|" pipe symbol as OR operator inside regular expressions? Take a look at this:
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(abc|xyz|012)$')
;I can even use strings composed of values like 'abc, xyz , 012' by simply using another regular expression to replace "," and spaces with the "|" pipe symbol. After reading part 1 and 2 that shouldn't be too hard, right? Here's my "thinking in regular expression": Replace every "," and 0 or more leading/trailing spaces.
Ready to try your own solution?
Does it look like this?
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(' || REGEXP_REPLACE('abc, xyz , 012', ' *, *', '|') || ')$')
;If I wouldn't use the "^" and "$" metacharacter, this SELECT would search for any occurence inside the data column, which could be useful if I wanted to combine LIKE and IN clause. Take a look at this example where I'm looking for 'abc%', 'xyz%' or '012%' and adding a case insensitive match parameter to it:
SELECT data
FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE REGEXP_LIKE(data, '^(abc|xyz|012)', 'i')
; An equivalent non regular expression solution would have to look like this, not mentioning other options with adding an extra "," and using the INSTR function:
SELECT data
FROM (SELECT data, LOWER(DATA) search
          FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE search LIKE 'abc%'
    OR search LIKE 'xyz%'
    OR search LIKE '012%'
SELECT data
FROM (SELECT data, SUBSTR(LOWER(DATA), 1, 3) search
          FROM TABLE(regex_utils.gen_data('abcxyz012', 4))
WHERE search IN ('abc', 'xyz', '012')
; I'll leave it to your imagination how a complete non regular example with 'abc, xyz , 012' as search condition would look like.
As mentioned in the first part, regular expressions are not very good at formatting, except for some selected examples, such as phone numbers, which in my demonstration, have different formats. Using regular expressions, I can change them to a uniform representation:
WITH t AS (SELECT '123-4567' phone
             FROM dual
            UNION
           SELECT '01 345678'
             FROM dual
            UNION
           SELECT '7 87 8787'
             FROM dual
SELECT t.phone, REGEXP_REPLACE(REGEXP_REPLACE(phone, '[^0-9]'), '(.{3})(.*)', '(\1)-\2')
FROM t
;First, all non digit characters are beeing filtered, afterwards the remaining string is put into a "(xxx)-xxxx" format, but not cutting off any phone numbers that have more than 7 digits. Using such a conversion could also be used to check the validity of entered data, and updating the value with a uniform format afterwards.
Thinking about it, why not use regular expressions to check other values about their formats? How about an IP4 address? I'll do this step by step, using 127.0.0.1 as the final test case.
First I want to make sure, that each of the 4 parts of an IP address remains in the range between 0-255. Regular expressions are good at string matching but they don't allow any numeric comparisons. What valid strings do I have to take into consideration?
Single digit values: 0-9
Double digit values: 00-99
Triple digit values: 000-199, 200-255 (this one will be the trickiest part)
So far, I will have to use the "|" pipe operator to match all of the allowed combinations. I'll use my brute force generator to check if my solution works for a single value:
SELECT data
FROM TABLE(regex_utils.gen_data('0123456789', 3))
WHERE REGEXP_LIKE(data, '^(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$')
; More than 255 records? Leading zeros are allowed, but checking on all the records, there's no value above 255. First step accomplished. The second part is to make sure, that there are 4 such values, delimited by a "." dot. So I have to check for 0-255 plus a dot 3 times and then check for another 0-255 value. Doesn't sound to complicated, does it?
Using first my brute force generator, I'll check if I've missed any possible combination:
SELECT data
FROM TABLE(regex_utils.gen_data('03.', 15))
WHERE REGEXP_LIKE(data,
                   '^((25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$'
; Looks good to me. Let's check on some sample data:
WITH t AS (SELECT '127.0.0.1' ip
             FROM dual
            UNION
           SELECT '256.128.64.32'
             FROM dual
SELECT t.ip
FROM t WHERE REGEXP_LIKE(t.ip,
                   '^((25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})\.){3}(25[0-5]|2[0-4][0-9]|[01]?[0-9]{1,2})$'
; No surprises here. I can take this example a bit further and try to format valid addresses to a uniform representation, as shown in the phone number example. My goal is to display every ip address in the "xxx.xxx.xxx.xxx" format, using leading zeros for 2 and 1 digit values.
Regular expressions don't have any format models like for example the TO_CHAR function, so how could this be achieved? Thinking in regular expressions, I first have to find a way to make sure, that each single number is at least three digits wide. Using my example, this could look like this:
WITH t AS (SELECT '127.0.0.1' ip
             FROM dual
SELECT t.ip, REGEXP_REPLACE(t.ip, '([0-9]+)(\.?)', '00\1\2')
FROM t
; Look at this: leading zeros. However, that first value "00127" doesn't look to good, does it? If you thought about using a second regular expression function to remove any excess zeros, you're absolutely right. Just take the past examples and think in regular expressions. Did you come up with something like this?
WITH t AS (SELECT '127.0.0.1' ip
             FROM dual
SELECT t.ip, REGEXP_REPLACE(REGEXP_REPLACE(t.ip, '([0-9]+)(\.?)', '00\1\2'),
                            '[0-9]*([0-9]{3})(\.?)', '\1\2'
FROM t
; Think about the possibilities: Now you can sort a table with unformatted IP addresses, if that is a requirement in your application or you find other values where you can use that "trick".
Since I'm on checking INET (internet) type of values, let's do some more, for example an e-mail address. I'll keep it simple and will only check on the
"[email protected]", "[email protected]" and "[email protected]" format, where x represents an alphanumeric character. If you want, you can look up the corresponding RFC definition and try to build your own regular expression for that one.
Now back to this one: At least one alphanumeric character followed by an "@" at sign which is followed by at least one alphanumeric character followed by a "." dot and exactly 3 more alphanumeric characters or 2 more characters followed by a "." dot and another 2 characters. This should be an easy one, right? Use some sample e-mail addresses and my brute force generator, you should be able to verify your solution.
Here's mine:
SELECT data
FROM TABLE(regex_utils.gen_data('a1@.', 9))
WHERE REGEXP_LIKE(data, '^[[:alnum:]]+@[[:alnum:]]+(\.[[:alnum:]]{3,4}|(\.[[:alnum:]]{2}){2})$', 'i'); Checking on valid domains, in my opinion, should be done in a second function, to keep the checks by itself simple, but that's probably a discussion about readability and taste.
How about checking a valid URL? I can reuse some parts of the e-mail example and only have to decide what type of URLs I want, for example "http://", "https://" and "ftp://", any subdomain and a "/" after the domain. Using the case insensitive match parameter, this shouldn't take too long, and I can use this thread's URL as a test value. But take a minute to figure that one out for yourself.
Does it look like this?
WITH t AS (SELECT 'Introduction to regular expressions ... last part. URL
             FROM dual
            UNION
           SELECT 'http://x/'
             FROM dual
SELECT t.URL
FROM t
WHERE REGEXP_LIKE(t.URL, '^(https*|ftp)://(.+\.)*[[:alnum:]]+(\.[[:alnum:]]{3,4}|(\.[[:alnum:]]{2}){2})/', 'i')
Update: Improvements in 10g2
All of you, who are using 10g2 or XE (which includes some of 10g2 features) may want to take a look at several improvements in this version. First of all, there are new, perl influenced meta characters.
Rewriting my example from the first lesson, the WHERE clause would look like this:
WHERE NOT REGEXP_LIKE(t.col1, '^\d+$')Or my example with searching decimal numbers:
'^(\.\d+|\d+(\.\d*)?)$'Saves some space, doesn't it? However, this will only work in 10g2 and future releases.
Some of those meta characters even include non matching lists, for example "\S" is equivalent to "[^ ]", so my example in the second part could be changed to:
SELECT NVL(LENGTH(REGEXP_REPLACE('Having fun with regular expressions', '\S')), 0)
FROM dual
;Other meta characters support search patterns in strings with newline characters. Just take a look at the link I've included.
Another interesting meta character is "?" non-greedy. In 10g2, "?" not only means 0 or 1 occurrence, it means also the first occurrence. Let me illustrate with a simple example:
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^.* +')
FROM dual
;This is old style, "greedy" search pattern, returning everything until the last space.
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^.* +?')
FROM dual
;In 10g2, you'd get only "Having " because of the non-greedy search operation. Simulating that behavior in 10g1, I'd have to change the pattern to this:
SELECT REGEXP_SUBSTR('Having fun with regular expressions', '^[^ ]+ +')
FROM dual
;Another new option is the "x" match parameter. It's purpose is to ignore whitespaces in the searched string. This would prove useful in ignoring trailing/leading spaces for example. Checking on unsigned integers with leading/trailing spaces would look like this:
SELECT REGEXP_SUBSTR(' 123 ', '^[0-9]+$', 1, 1, 'x')
FROM dual
;However, I've to be careful. "x" would also allow " 1 2 3 " to qualify as valid string.
I hope you enjoyed reading this introduction and hope you'll have some fun with using regular expressions.
C.
Fixed some typos ...
Message was edited by:
cd
Included 10g2 features
Message was edited by:
cd

Can I write this condition with only one reg expr in Oracle (regexp_substr in my example)?I meant to use only regexp_substr in select clause and without regexp_like in where clause.
but for better understanding what I'd like to get
next example:
a have strings of two blocks separated by space.
in the first block 5 symbols of [01] in the second block 3 symbols of [01].
In the first block it is optional to meet one (!), in the second block it is optional to meet one (>).
The idea is to find such strings with only one reg expr using regexp_substr in the select clause, so if the string does not satisfy requirments should be passed out null in the result set.
with t as (select '10(!)010 10(>)1' num from dual union all
select '1112(!)0 111' from dual union all --incorrect because of '2'
select '(!)10010 011' from dual union all
select '10010(!) 101' from dual union all
select '10010 100(>)' from dual union all
select '13001 110' from dual union all -- incorrect because of '3'
select '100!01 100' from dual union all --incorrect because of ! without (!)
select '100(!)1(!)1 101' from dual union all -- incorrect because of two occurencies of (!)
select '1001(!)10 101' from dual union all --incorrect because of length of block1=6
select '1001(!)10 1011' from dual union all) --incorrect because of length of block2=4
select '10110 1(>)11(>)0' from dual union all)--incorrect because of two occurencies of (>)
select '1001(>)1 11(!)0' from dual)--incorrect because (!) and (>) are met not in their blocks
--end of test data

Using Regular Expressions to Find Quoted Text

I have run into a couple problems with the following code.
1) Slash-Star and Slash-Slash commented text must be ignored.
2) It does not detect backslashed quotes, or if that backslash is backslashed.
Can this be accomplished with Regular Expressions, or should I implement this using if/indexOf logic?
Thank You in advance,
Brian
    * Finds position of next quoted string in a line
    * of source code.
    * If no strings exist, then a Pointer position of
    * (0,0) is returned.
    * @param startPos position to start search from
    * @param argText the line of text to search
    * @returns next string position
   public Pointer getQuotedStringPosition(int startPos, String aString) {
      String argText = new String( aString );
      Pattern p = Pattern.compile("[\"][^\"]+[\"]");
      Matcher m = p.matcher( argText.substring(startPos); );
      if( m.find() )
         return new Pointer( m.start() + startPos, m.end() + startPos );
      else
         return new Pointer( 0, 0 ); // indicates nothing was found
   }

YATArchivist was right about the regular expressions.
I think I've got it but somebody test it if you want. Let me know what you find.
I've included a barebones Position class as well...
import java.util.regex.*;
import java.io.*;
import java.util.*;
@author Joshua A. Logan, Jr.
public class RegexTest
   private static final String SLASH_SLASH = "(//.*)";
   private static final String SLASH_STAR =
                           "(/\\*(?:[^\\*]|(?:\\*(?!/)))+(\\*/)?)";
   private static final Pattern COMMENT_PATTERN =
                     Pattern.compile( SLASH_SLASH + "|" + SLASH_STAR );
   private static final Pattern QUOTED_STRING_PATTERN =
                  Pattern.compile( "\" ( (?:(\\\\.) | [^\\\"])*+ )     \"",
                                   Pattern.COMMENTS );
   // Breaking the above regular expression down, you'd have:
   //   " ( (?: (\\ .) | [^\\ "] ) *+ )   "
   //   ^          ^     ^     ^       ^      ^
   //   |          |     |     |       |      |
   //   1          2     3     4       5      6
   // which matches:
   // 1) The starting quote...
   // Followed by something that is either:
   // 2) some escaped sequence ( e.g. _\n_ or even _\"_ ),
   // 3)                ...or...
   // 4) a character that is neither a _\_ nor a _"_ .
   // 5) Keep searching this as much as possible, w/o giving up
   //                    any found text at the end.
   //        Note: the text found would be in group(1)
   // 6) Finally, find the ending quote!!
   public static Position [] getQuotedStringPosition( final String text )
      Matcher cm = COMMENT_PATTERN.matcher( text ),
              qm = QUOTED_STRING_PATTERN.matcher( text );
      final int len = text.length();
      int startPos = 0;
      List positions = new ArrayList();
      while ( startPos < len )
         if ( cm.find(startPos) )
            int commStart = cm.start(),
                commEnd   = cm.end();
            // are we starting @ a comment?
            if ( commStart == startPos )
               startPos = commEnd;
            else if ( qm.find(startPos) )
               // Search for unescaped strings in here.
               int stringStart = qm.start(1),
                   stringEnd   = qm.end(1);
               // Is the quote start after comment start?
               if ( stringStart > commStart )
                  startPos = commEnd; // restart search after comment end...
               else if ( (stringEnd > commEnd) ||
                         (stringEnd < commStart) )
                  // In this case, the "comment" is actually part of
                  // the quoted string. We found a match.
                  positions.add( new Position(text, qm.group(1),
                                              stringStart,
                                              stringEnd) );
                  int quoteEnd = qm.end();
                  startPos = quoteEnd;
               else
                  throw new IllegalStateException( "illegal case" );
            else
               startPos = commEnd;
         else
            // no comments were found. Search for unescaped strings.
            int quoteEnd = len;
            if ( qm.find( startPos ) ) {
               quoteEnd = qm.end();
               positions.add( new Position(text,
                                           qm.group(1),
                                           qm.start(1),
                                           qm.end(1)) );
            startPos = quoteEnd;
      return positions.isEmpty() ? Position.EMPTY_ARRAY
                                 : (Position[])positions.toArray(
                                          Position.EMPTY_ARRAY);
   public static void main( String [] args )
      try
         BufferedReader br = new BufferedReader(
                  new InputStreamReader(System.in) );
         String input = null;
         final String prompt = "\nText (q to quit): ";
         System.out.print( prompt );
         while ( (input = br.readLine()) != null )
            if ( input.equals("q") ) return;
            Position [] matches = getQuotedStringPosition( input );
            // What does it do?
            for ( int i = 0, max = matches.length; i < max; i++ )
               System.out.println( "-->" + matches[i] );
            System.out.print( prompt );
      catch ( Exception e )
         System.out.println ( "Exception caught: " + e.getMessage () );
class Position
   public Position( String target,
                    String match,
                    int start,
                    int end )
      this.target = target;
      this.match = match;
      this.start = start;
      this.end = end;
   public String toString()
      return "match==" + match + ",{" + start + "," + end + "}";
   final String target;
   final int start;
   final int end;
   final String match;
   public static final Position [] EMPTY_ARRAY = { };
}

Regular Expressions to Omit Characters

I am new to Java. I am trying to use regular expressions to omit a pattern. The construct [^abc] can be used to omit a or b or c, but how can I omit the string abc from a regular expression? Many thanks.

Why not use the regular expression "abc" and see if it doesn't match? Or even use this:
boolean matches = (targetString.indexOf("abc") == -1); // == -1 means "not found", so matches become true if abc is not in targetString
if (matches) System.out.println("Yay, no abc in " + targetString + "!");
else System.out.println("Eek, get that abc out of my sight!");

Searching in reverse with regular expressions

I have recently constructed a "Find" dialog for my editor and would like to support the use of regular expressions (for people who know more about them than I do.) I have radio buttons that allow the user to search down (from the current caret position to the end of the document) or up (from the current caret position to the beginning of the document.)
I have it working fine with literal text via String.indexOf() and String.lastIndexOf(), but I'm not sure how to implement an "upwards" search using regular expressions. It seems that java.util.regex only provides for searching from the current caret position downwards. Any suggestions?

two not-very-good-for-big-document suggestions
1) search up by searching down (from the beginning) to the last match before the position searched from
2) when the user asks for a search find all matches, number them and keep the index in the doc where they were found, then look up the largest index less than the carets index
you might try constructing succesively larger strings that end at the current caret position, and matching them one by one. If you're near the bottom, and theres no matches, this is also likely to be sloow..
asjf

Regular expressions: serious design flaw?

I wondered why sometimes, the replaceAll() method works in my code and sometimes it throws a java.util.regex.PatternSyntaxException. I wrote a little test case to show the problem
private void regularExpressionTest() {
        String escapers = "\\([{^$|)?*+."; //characters to escape to avoid PatternSyntaxException
        String text = "The quick brown fox jumps over the lazy dog.";
        System.out.println("Original: "+text);
        String regExp = "o";
        String word = "HELLO";
        String result = text.replaceAll(regExp, word);
        System.out.println("Replace all '"+regExp+"' with "+word+": "+result);
        text = "The quick brown {animal} jumps over the lazy dog.";
        System.out.println("Original: "+text);
        regExp = "{animal}";
//        regExp = escapeChars(regExp, escapers); //escape possible regular expression characters
        word = "catterpillar";
        result = text.replaceAll(regExp, word);
        System.out.println("Replace all '"+regExp+"' with "+word+": "+result);
}//regularExpressionTest()The output is:
Original: The quick brown fox jumps over the lazy dog.
Replace all 'o' with HELLO: The quick brHELLOwn fHELLOx jumps HELLOver the lazy dHELLOg.
Original: The quick brown {animal} jumps over the lazy dog.
java.util.regex.PatternSyntaxException: Illegal repetition
{animal}
     at java.util.regex.Pattern.error(Pattern.java:1528)
     at java.util.regex.Pattern.closure(Pattern.java:2545)
     at java.util.regex.Pattern.sequence(Pattern.java:1656)
     at java.util.regex.Pattern.expr(Pattern.java:1545)
     at java.util.regex.Pattern.compile(Pattern.java:1279)
     at java.util.regex.Pattern.<init>(Pattern.java:1035)
     at java.util.regex.Pattern.compile(Pattern.java:779)
     at java.lang.String.replaceAll(String.java:1663)
     at de.icomps.prototypes.Test.regularExpressionTest(Test.java:57)
     at de.icomps.prototypes.Test.main(Test.java:34)
Exception in thread "main" As the first replacement works as espected, the second throws an Exception. Possible because '{' is a special character for the regular expression API. In case I know the regular expression, i could escape these special characters. But in my generic server app, the strings are all parameters, so the only way for replaceAll() to work is, to escape all possible special characters in the regular expression.
1. Is there a complete list of all special characters for regular expressions that need to be escaped?
2. Is there a similar replaceAll() method without regular expressions? So that all occurences of a substring can be replaced by another string? (So far, I wrote my own method but this is of course more time consuming for massive string operations.)

1. The complete list of specially-recognized characters are listed in the Java 1.4.* API. (Of course, new ones may eventually be added as the regex package matures).
2. It is time consuming to program in general. You should have written your own utility method that goes through a String using indexOf and building up the String in a StringBuffer, and apparently you did. Now you have the utility method...you no longer need to write that method again.
3. Or you could have written some kind of method that automatically escapes the specially-recognized characters...

How to make a regular expression case insensitive?

Hi,
I am using SAP UI5 .
I am validating the user input using a regular expression to make sure user dose not enter a keyword which is reserved.
var regexAppID = /^(?=^(?:(?!(^Admin$|^AdminData$|^Push$|^smp_cloud$|^resource$|^test-resources$|^resources$|^Scheduler$|^odata$|^applications$|^Connections$|^public|^Public$|[\.]{2,}|^[\.])).)*$)(?=^[a-zA-Z]+[a-zA-Z0-9_\.]*$).*$/;
textField.mBindingInfos.value.type.oConstraints.search = regexAppID;
The regex blocks the user from entering the keywords Admin,AdminData etc.
It works fine if the user enters the value as it is given in regex (ie case sensitive example "Admin").
But when user enters "admin" or "aDmin" etc the regex is not catching it and my server crashes as it bypasses these keywords.

Can't you use this expression?
^/(?!ignoreme$)(?!ignoreme2$)[a-z0-9]+$
Then you will have only a match when the word is something else then "ignoreme" or "ignorme2".
Why don't you just check it in an IF statement? if(input==="ignoreme" or ... ) ? It's both static.. or you could put all options in an array and use the statement indexOf to find if it is in the array.
JavaScript Array indexOf() Method
Kind regards,
Wouter

Urgent help regarding Java regular expressions.

hello everyone,
I am trying to parse a html file which contains
dyn.Img("http://www.boston.com/news/nation/articles/2007/06/21/bill_clinton_takes_bigger_campaign_role&h=306&w=410&sz=13&hl=en&start=43","","QFo9lqKeMR7uzM:","http://cache.boston.com/resize/bonzai-fba/AP_Photo/2007/06/21/1182410228_1931/410w.jpg","125","93","\x3cb\x3eBill Clinton\x3c/b\x3e takes bigger campaign \x3cb\x3e...\x3c/b\x3e","","","410 x 306 - 13k","jpg","www.boston.com","","","http://tbn0.google.com/images","1")
the given above function many times. I have to fetch the whole functions into an array. So i have to write a regular expression which recognises the whole above string.
Can anyone please help me.
Thank you,
chaitanya

well if this is all you want
http://cache.boston.com/resize/bonzai-fba/AP_Photo/2007/06/21/1182410228_1931/410w.jpg
You can always substring it like chuck said
***BUT all the images would have to be .jpg for this to work***
back = we.indexOf(".jpg");
int x = 0;
while (back < web.lastIndexOf(".jpg"))
                back = web.indexOf("http",back+1);
                picture[x] = web.substring(front, back);
                x++;
                front = back;
              }       Might not be the best code but it worked with a website i had to parse
Message was edited by:
mark07

Regular Expressions + using them in Vector analysis

Hello,
I would be very glad for any hint concerning this problem:
Consider this ilustrational code sample
import java.util.regex.*;
Vector sequence = new Vector();
        sequence.add("-");
        sequence.add("-");
        sequence.add("A");
        sequence.add("-");
        sequence.add("B");I want to find the first occurence of any alphabetical character ...I thought I can do it like this:
int index = sequence.indexOf("\\w");or from the other point of view like this:
int index = sequence.indexOf("[^-]");Unfortunately none of these are working. Do you know why and how to fix this? There are supposed to be only alphabetical characters and "-" character...
thanks a lot
adam

mAdam wrote:
well, the code I posted above should suit only as an illustration of my problem.
ActualIy I have a bunch of those sequences in an Arraylist...it can be from 5 - 100 sequences(vectors) and I need to use some methods which are available only for Collections (frequency(), insertElementAt()...etc.). I am not a proffesional programmer so I do hope that I am using it correctly instead of "simple" String[][] BunchOfSequences = {seq1,seq2,...}I am not aware of any frequency() or insertElementAt() methods in one of Java's collection classes.
Is it possible to use regular expressions to find a first occurence of a "a-z"(or "A-Z"..it doesnt matter) in a vector then? Or I just need to find it in a "for loop"?
thx
adamRegexes only work on Strings, not on collections holding Strings. So yes, you will need to use a for statement (or similar).
List<String> list = ...
for(String s : list) {
if(s.matches("\\w")) {
}

How to fetch substring using regular expression

Hi,
I am new to using regular expression and would like to know some basic details of how to use them in Java.
I have a String example= "http://www.google.com/foobar.html#*q*=database&aq=f&aqi=g10&fp=c9fe100d9e542c1e" and would like to get the value of "q" parameter (in bold) using regular expression in java.
For the same example, when we tried using javascript:
match = example.match("/^http:\/\/(?:(?!mail\.)[^\.]+?\.)?google\.[^\?#]+(?:.*[\?#&](?:as_q|q)=([^&]+))?/i}");
document.write('
' + match);
We are getting the output as: http://www.google.com/foobar.html#q=database,*database* where the bold text is the value of "q" parameter.
In Java we are trying to get the value of the q parameter separately or atleast resembles the output given by JavaScript. Please help me resolving this issue.
Regards
Praveen

BalusC wrote:
Regex is a cumbersome solution for fixed patterns like URL's. String#substring() in combination with String#indexOf would most likely already suffice.I usually agree, although, in this case, finding the exact parameter might be difficult without a small regex, perhaps:
"\\wq=\\s*"in conjunction with Pattern/Matcher, used similarly to an indexOf() to find the start of the parameter value.
Winston

? on Regular Expressions

I am trying to use a regular expression to spot an empty string but it isn't working within Oracle ... regexp_like(:P13_ADDRESS,'^$')
Shouldn't this expression work '^$' ???????

Justin,
a few Oracle facts regarding the handling of null values:
1) the empty string '' is equivalent to NULL
2) an operation (using oracle built-in operators / functions) on a NULL value will always result in a null value
consider the following example for illustration purposes:
set serveroutput on
DECLARE
   l_str   VARCHAR2 (100);
BEGIN
   l_str := '';
   IF l_str IS NULL THEN
      DBMS_OUTPUT.put_line ('empty string is null');
   END IF;
   IF TO_NUMBER ('') IS NULL THEN
      DBMS_OUTPUT.put_line ('conversions on null will yield null as output');
   END IF;
END;
That's what it seems to be doing, as my example showed. I don't quite agree with this,
NULL is a RDBMS concept, it should match a empty string (^$), but Oracle's regexp
implementation doesn't seem to be doing this, so just NVL it.Well, regarding the logic behind (2), that all operations on a null value will result in a null value, it does indeed make sense that the regular expression applied to null returns null.
And in a boolean context (like in the if-clause) null will be equivalent to false. This can be tricky at times!
Thus I would use the following code (illustrated by a stored function) to check for an empty string
CREATE OR REPLACE FUNCTION is_empty (p_str IN VARCHAR2)
   RETURN BOOLEAN IS
   l_str   VARCHAR2 (2000) := TRIM (p_str); -- remove leading and trailing spaces
BEGIN
   IF l_str IS NULL THEN
      RETURN TRUE;
   ELSE
      RETURN FALSE;
   END IF;
END;
/~Dietmar.

Match Regular Expression does not match what Match Pattern does

I have read through a lot of posts about how Match Pattern does not match what Match Regular Expression will due to not processing some characters.
However, I found a problem with the other way. A simple Reg-Ex that works in Match Pattern but not Match Regular Expression.
What I have here is just an example. I want to use Match Regular Expression so I can specify some sub-matches.
The reg-ex is for: one or more non-numeric characters, a space, one or more numeric characters. At the start of the string.
How can I get this working in Match Regular Expression? I am working in LabVIEW 2010f2 32 bit. Here is the code snippet and the results:
Rob
Solved!
Go to Solution.

Robert Cole wrote:
I think I prefer the ~ for negation since ^ is also used for beginning of the string. But we work with what we have.
Let me offer you a tip and perhaps defend the honor of the regex a little bit. One of my favorite features of regexes is the ability to specify character classes (and their negation). One of the reasons I have to think about the ~ versus ^ is that I rarely use ^ in a regex alternative.
Some examples:
[0-9] = \d (digit)
[^0-9] = \D (not a digit)
The equivalent regex for your case is: \D+ \d+

Regular expression equivalent to s.indexOf("\t")==-1

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