How to use regular expression to find string

hi,
who know how to get all digits from the string "Alerts 4520 ( 227550 ) ( 98 Available )" by regular expression, thanks
br, Andrew

Liu,
You can use RegEx as
d+
Whether you are using CL_ABAP_REGEX class then
report zars.
data: regex   type ref to cl_abap_regex,
      matcher type ref to cl_abap_matcher,
      match   type c length 1.
create object regex exporting pattern = 'd+'
                              ignore_case = ''.
matcher = regex->create_matcher( text = 'Test123tes456' ).
match = matcher->match( ).
write match
You can find more details regarding REGEX and POSIX examples here
http://www.regular-expressions.info/tutorial.html
a®

Similar Messages

How to write regular expression to find desired string piece *duplicate*

Hi All,
Suppose that i have following string piece:
name:ali#lastname:kemal#name:mehmet#lastname:cemalI need
ali
mehmetI use following statement
SQL> select lst, regexp_replace(lst,'(name:)(.*)(lastname)(.*)','\2',1,1) nm from (
2 select 'name:ali#lastname:kemal#name:mehmet#lastname:cemal' as lst from dual
3 );
LST NM
name:ali#lastname:kemal#name:mehmet#lastname:cemal ali#lastname:kemal#name:mehmet#
SQL> But it does not return names correctly. When i change 5th parameter(occurence) of regexp_replace built-in function(e.g. 1,2), i may get ali and mehmet respectiveley.
Any ideas about regexp?
Note : I can use PL/SQL instr/substr for this manner; but i do not want to use them. I need regexp.
Regards...
Mennan
Edited by: mennan on Jul 4, 2010 9:53 PM
thread was posted twice due to chrome refresfment. Please ignore the thread and reply to How to write regular expression to find desired string piece

The approach is to do cartesian join to a 'number' table returning number of records equal to number of names in the string.I have hardcoded 2 but you can use regexp_count to get the number of occurrences of the pattern in the string and then use level <=regexp_count(..... .
See below for the approach
with cte as(
select
'name:ali#lastname:kemal#name:mehmet#lastname:cemal' col ,level lev
from dual connect by level <=2)
select substr(regexp_substr('#'||col,'#name:\w+',1,lev),7)
from cte
/

How to use regular expression to delete a character?

Hello,
I have a query,
select partition_name from dba_tab_partitions where table_owner='xxx'and num_rows <>0 and table_name = 'xxx';
P5
P6
P7
P12
P13
P14
P17
P18
P19
P20
P24
How can I use regular expression in above SQL query to get result without letter 'P', like..
5
6
7
12
13
14
17
18
19
20
24
thank you

I find answer...
select regexp_replace(partition_name,'P','')
thanks anyway

How to use regular expressions to generate test data ?

Hi
Someone can help me on what I have to do in order to create test data with regular expressions ?
For example, I want to introduce a random telephone number (XXX-XXXX) in the phone number Form Field, I want to create the phone number using regular expressions in order to test different values in each playback of the script.
I don't want to use VB or vbscript in e-tester, I'm just trying to do this with e-load nav editor and e-load
Thanks a lot

Hi and thanks for your answer!, it's a great trick ^_^
I'm doing a research on how to improve the execution speed of the scripts in e-load, so actually I'm trying to avoid the use of databanks and VB code also.
I was expecting that maybe e-load, e-load nav editor or e-tester can automatically generate test data via Regular Expressions. Someone Knows if this is possible ?
Also can anyone tell me what the option "Automatically Generated (complex)" means ? I think that this will help me a lot
*you can find this option in e-load Nav Editor when you select a parameter in the tree view, then go to the "type" listbox in the properties pane, there you will find this option and some more options like :"Databanked variable", "Custom Dynamic Value", "Function".. etc.
Thanks again

How to use Regular expression

Hi,
I have a file that contains this format (separated by ;(semicolon) ):
user id;user name;email address;password;integer;list of integer(separated by ,(comma))
below is the example data :
abc;Abc;[email protected];password1;1;1,2
def;Def;[email protected];password;2;1,2,3
ghi;Ghi;[email protected];password;2;1
my question is how to verify the valid input for each row using regular expression..? TQ

@Op. Doing a correct validation of e-mailaddresses
is very hard using regular expressions (doingbasic
validation is however easy)
http://www.regular-expressions.info/email.html
I like the RFC 822 compliant regexp :)

Unable To Use Regular Expression To Find Function Names

Hi,
I am trying to create a regular expression to find function and procedure names without the static designation and the parameter list. Sample source document:
static function test
static function test(i,j)
function test
function test(i,j)
static procedure test
static procedure test(i,j)
procedure test
procedure test(i,j)
For each of the above samples, I would like only the word "test" to be found.
Thanks

I suggest starting with this expression:
^\s*(static\s+)?(function|procedure)\s+(?<NAME>\w+)
Programmatically, the name can be extracted from the group called “NAME”.
The expression can be improved.

Using regular expressions to find and replace code.

Hi! Semi-newbie coder here.
I'm trying to strip out code from multiple pages, I've tried regular expressions but I'm struggling to understand them. I also need to do it across a LOT of pages, so I need an automated way of doing it.
The best way I can explain is with an analogy:
I want to delete any string of characters that start with c, ends with t and includes anything inbetween, so it would pick up "cat, cut, chat, coconut, can do it" whatever appears in the middle of those.
Except, instead of c and t, I want it to find strings of code starting with <div class="advert" and ending with Vote<br> while picking up everything in between, (including spaces, code, comments, etc.). Then, deletes that whole string including the starting and ending.Is there a regular expression I could use in dreamweaver that could do this? Is there a way to do this at all?

Let me begin by saying I'm a complete idiot with DW's Reg Ex. I use Search Specific Tag whenever possible. See screenshot below.
Try this on your Current Document to see if it works. Then make a back-up copy of site before attempting it on Entire Local Site as you cannot "Undo" this process.
Good luck,
Nancy O.

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 = { };
}

How to use regular expression replace for this special characters?

hi,
I need to replace the below string, but i couldnt able to do if we use the special charaters '+', '$' . can anyone suggest a way to do this?
select REGEXP_REPLACE('jan + feb 2008','jan + feb 2008', 'feb',1,0,'i') from dual
anwers should be :- feb

you should use escape character \.
the regular expression will look like as follows:
select REGEXP_REPLACE('jan + feb 2008','jan \+ feb 2008', 'feb',1,0,'i') from dual
hope this is what you needed.
cheers,
Davide

How to use regular expression in editor?

I tried to add some text to all line in selected text in editor
ie .
x:= x+1 ; change to x:=x+1; log_something;
y:= y-1; -> y:= y-1 ; log_something;
and so on.
Then
1. select some text in editor
2.pressed Ctrl-R
3.in "text to search for" I put "^$.*$$"
4.in "replace with" I put "\1 log_something;
5.check "Regular expressions"
6. check "Selected text only"
I got "The search text "^$.*$$" was not found."
What am I doing wrong?

Thanks. But now editor wants to replace whole text, not only selected by me, regardless of <<check "Selected text only" >>
Update: if I pick "scope: all" instead of "prompted" it shows popup "text not found" but replaces text as requested.
Message was edited by: user4879976

How to write regular expression to find desired string piece

Hi All,
Suppose that i have following string piece:
name:ali#lastname:kemal#name:mehmet#lastname:cemalI need
ali
mehmetI use following statement
SQL> select lst, regexp_replace(lst,'(name:)(.*)(lastname)(.*)','\2',1,1) nm from (
2    select 'name:ali#lastname:kemal#name:mehmet#lastname:cemal' as lst from dual
3 );
LST                                                NM
name:ali#lastname:kemal#name:mehmet#lastname:cemal ali#lastname:kemal#name:mehmet#
SQL> But it does not return names correctly. When i change 5th parameter(occurence) of regexp_replace built-in function(e.g. 1,2), i may get ali and mehmet respectiveley.
Any ideas about regexp?
Note : I can use PL/SQL instr/substr for this manner; but i do not want to use them. I need regexp.
Regards...
Mennan

Hi, Mennan,
You can nest REGEXP_SUBSTR withing REGEXP_REPLACE to get the n-th occurrence, like this:
SELECT     lst
,      REGEXP_REPLACE ( REGEXP_SUBSTR ( lst
                                  , 'name:[^#]*#lastname'
                           , 1
                           , n
                 , 'name:(.*)#lastname'
                 , '\1'
                 )      AS nm If the pattern occurs fewer than n times, the expression above returns NULL.

How to use regular expression using pattern and match concept for this scenario?

Hi Guys,
I have a string "We have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle"
I need to replace the numbers based on the below condition.
if more then 5, replace with many
if less then 5, replace with a few
if it is 1, replace with "only one"
below is my code, I am missing the equating part to replace the numbers could any one of you please help me out fixing this.
private static String REGEX="(\\d+)";
private static String INPUT="We have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle";
//String pattern= "(.*)(\\d+)(.*)";
private static String REPLACE = "replace with many";
public static void main(String[] args) {
// Create a Pattern object
       Pattern r = Pattern.compile(REGEX);
    // Now create matcher object.
       Matcher m = r.matcher(INPUT);
       //replace the value 7 by the replace string
//How to equate the (\\d+) greater than a number and use it the below code.
              INPUT = m.replaceAll(REPLACE);
       //Print the final Result;
        System.out.println(INPUT);
Thanks and Regards,

Hi,
Try the following which makes use of "appendReplacement" instead with the "start" and "end" methods to locate and check the searched "regExp" string before dynamically setting the "replace" string:
String regExp = "\\d+";
String input = "We have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle";
String replace;
Pattern p = Pattern.compile(regExp);
// get a matcher object
Matcher m = p.matcher(input);
StringBuffer sb = new StringBuffer();
while (m.find()) {
   Integer x = Integer.valueOf(input.substring(m.start(), m.end()));
   replace = (x >= 5) ? "many" : (x == 1) ? "only one" : "few";
   m.appendReplacement(sb, replace);
m.appendTail(sb);
System.out.println(sb.toString());
HTH.
Regards,
Rajen
P.S: Please mark the post as answered/helpful if it resolves your issue for the benefit of all community members.

How to use regular expressions

Hey ,
I found my self getting troubled with using regex in java.
I know that in order to use regex, i need to import two classes
import java.util.regex.Matcher;
import java.util.regex.Pattern;
but I become entangled with the implementation.
I need to check if inserted string contains a number inside, so i build a pattern of regex. and compiled it. then i used matcher method. but from here all methods i use, i'm getting only the first digit of the number. ---> "sdasda25" i'm getting (after using group() ) only the digit 2.
is there any method that i can pass over all chars in the string (by loop) and to check if that particular char isMatch for my pattern (That's the way to implement it in C#, and i thought that here it will be the same...).
Tanx

both of you were right, I didnt write a '+' at the end of my pattern
thank you both

How to use regular expression in el

${fn:replace(str,'\r\n','')}
${fn:replace(str,'\\r\\n','')}
both can not lead to the result of:
str.replaceAll("\r\n","");

My guess is a whole other level of escapes:
${fn:replace(str,"\\\\r\\\\n","")}
But I thought that would be the solution for any regex...

Find/Replace Using Regular Expressions

Can someone help me with this...I am using Regular expressions to
FIND:
http.*lid=([^&"]*)[^"]*
REPLACE:
$set(\1,ID_id,code)$
So that in the following it will change this:
a href="http://www.test.com/shc/s/home_10153_12605?lid=Search" rilt="Search"
To this:
a href="$set(Search,ID_id,code)$" rilt="Search
Those expressions work in Notepad++ but when i use dreamweaver it just replaces the http... with "$set(\1,ID_id,code)$" and doesnt reference the "search"
Any help?
Thanks

Let me begin by saying I'm a complete idiot with DW's Reg Ex. I use Search Specific Tag whenever possible. See screenshot below.
Try this on your Current Document to see if it works. Then make a back-up copy of site before attempting it on Entire Local Site as you cannot "Undo" this process.
Good luck,
Nancy O.

How to use regular expression to find string

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