Regular expression java

folks
I have a xml file that contains a pattern such as
<!CDATA[irc]]>
I need help to convert this into
<irc>
Any help with java will do
thanks

String s = "<!CDATA[irc]]>";
s = s.replaceAll("!CDATA\\[|\\]\\]", "");
System.out.println(s); // <irc>

Similar Messages

Regular Expressions (java.util.regex)

I am developing using a product that must
use java 1.2.2_05a but I want to use regular
expressions, does anybody know where I can
get of the package java.util.regex without
having to download the whole java 1.4 release.
Or does someone know of an alternative that
I can use ?

There is another regex pack for java available from Apache Foundation Project. You can try it.
Take a look at http://jakarta.apache.org/

Doubt in Regular Expressions : java.util.regex

I want to identify words starting with capital letters in a sentence and I want to replace the matched word with "#" added in front of it.... For example, if my input sentence is
"Christopher Williams asked Mike Muller a question"
my output should be,
"#Christopher #Williams asked #Mike #Muller a question"
How do I do that using java.util.regex ?
In perl we can can use *"back references"* in *"replacement string"* . Perl replacement accepts back references whereas java replacement method accepts only strings....
Please help me.....

Your replacement is swallowing the space before the uppercase character, and won't match at the beginning of the line.
Also, it's unnecesarily verbose. String has a replaceAll method (that calls the same methods of Pattern and Matcher under the covers)sentence = s.replaceAll("(^| )([A-Z])", "$1#$2");Disclaimer: I'm no prome, sabre or u/a :-) That can probably be simplified.
db

Regular Expressions and XML Schemas

The standard regular expression syntax in JDK1.4 doesn't support some of the constructs used by pattern facets in XML schemas.
For example the "Name" type is specified by the pattern "\i\c*" but the java.util.regex regular expression handler throws an exception when it encounters this with the following message:
Illegal/unsupported escape squence near index 1
Is anyone aware of any regular expression java libraries or tools that will cope with the XML syntax?

Thanks, but perhaps I should clarify. I realise that \i\c* is invalid syntax in the java regexp handler. My point is that in XML Schema regular expressions \i is meaningful - in fact it means
"the set of initial name characters, those matched by Letter | '_' | ':' "
I was wondering if anyone knew of a regular expression library that understood and correctly interpreted that.

Matching substrings between square brackets using regular expressions

Hello,
I am new at Java and have a problem with regular expressions. Let me describe the issue in 3 steps:
1.- I have an english sentence. Some words of the sentence stand between square brackets, for example "I [eat] and [sleep]"
2- I would like to match strings that are in square brackets using regular expressions (java.util.regex.*;) and here is the code I have written for the task
+Pattern findStringinSquareBrackets = Pattern.compile("\\[.*\\]");+
+     Matcher matcherOfWordInSquareBrackets = findStringinSquareBrackets.matcher("I [eat] and [sleep]");+
+//Iteration in the string+
+          while ( matcherOfWordInSquareBrackets.find() )+
+{+
+          System.out.println("Patter found! :"+ outputField.getText().substring(matcherOfWordInSquareBrackets.start(), matcherOfWordInSquareBrackets.end())+"");     +
+          }+
3- the result I have after running the code described in 2 is the following: *Patter found!: [eat] and [sleep]*
That is to say that not only words between square brackets are found but also the substring "and". And this is not what I want.
What I would like to have as a result is:
*Patter found!: [eat]*
*Patter found!: [sleep]*
That is to say I want to match only the words between the square brackets and nothing else.
Does somebody know how to do this? Any help would be great.
Best regards,
Abou

You can find the words by looping through the sentence and then return the substring within the indexes.
int start=0;
int end=0;
for(int i=0; i<string.length(); i++)
   if(string.substring(i,i+1).equals("[");
start=i;
if(start!=0)
if(string.substring(i,i+1).equlas("]");
end=i;
return string.substring(start,end+1);
}something like that. This code will only find the firt word however. I do not know much about regex so I cannot help anymore.
Edited by: elasolova on Jun 16, 2009 6:45 AM
Edited by: elasolova on Jun 16, 2009 6:46 AM

Regular Expression or Common Validator Utility Class

Hi all,
Just to check whether is there a common validator utility class that make use of Regular Expression for input and output validation in Web Dynpro?
Thank a lot
Regards

Hi
U can make use of following classes for validating a regular expression.
java.util.regex.Pattern
java.util.regex.Match
Regards
Surender Dahiya

Logical AND in Java Regular Expressions

I'm trying to implement logical AND using Java Regular Expressions.
I couldn't figure out how to do it after reading Java docs and textbooks. I can do something like "abc.*def", which means that I'm looking for strings which have "abc", then anything, then "def", but it is not "pure" logical AND - I will not find "def.*abc" this way.
Any ideas, how to do it ?
Baken

First off, looks like you're really talking about an "OR", not an "AND" - you want it to match abc.*def OR def.*abc right? If you tried to match abc.*def AND def.*abc nothing would ever match that, as no string can begin with both "abc" and "def", just like no numeric value can be both 2 and 5.
Anyway, maybe regex isn't the right tool for this job. Can you not simply programmatically match it yourself using String methods? You want it to match if the string "starts with" abc and "ends with" def, or vice-versa. Just write some simple code.

Java – Regular Expressions – Finding any non digit byte in a multiple byte

Hello,
I’m new to JAVA and Regular Expressions; I’m trying to write a regular expression that will find any records that contain a non digit byte in a multiple byte field.
I thought the following was the correct expression but it is only finding records that contain “all” non digit bytes.
\D{1,}
\D = Non Digit
{1,} = at least 1 or more
Below is my sample data. I would like the regular expression to find all of the records that are not all numeric. However when I use the regular expression \D{1,} it is only finding the 2 records that all bytes are non digits. (i.e. “ “ and “A “)
“ 111229”
“2 111229”
“20091229”
“200912c9”
“201#1229”
“20101229”
“20110229”
“20111*29”
“20111029”
“20111229”
“20B11229”
“A “
“A0111229”
Please note I have also tried \D{1,}+ and \D{1,}? And they also do not return my desired results
Any assistance someone can provide would be greatly appreciated.

You don't show the code you are using but I surmise you are using String.matches() which requires that the whole target must match the regular expression not just part of it. Instead you should create a Pattern and then a Matcher and use the Matcher.find() method. Check the Javadoc for Pattern and Matcher and look at the Java regex tutorial - http://docs.oracle.com/javase/tutorial/essential/regex/ .
P.S. You can re-use the Pattern object - you don't have to create it every time you need one.
P.P.S. Java regular expressions work with characters not bytes and characters are not not not bytes.

Java Regular Expressions and Pattern

I have a file that i first want to get all the lines that match a given pattern. Then from these lines that match i want to extract two values.
Example line for the pattern to match
INFO | jvm 1 | 2006/11/07 15:14:09 | INFO | Tue Nov 07 15:14:09 CET 2006 | XLDB PPS Data Dumper: MESSAGE:- 406 Processing .. '[ /opt/nexus/horizon/raw_data/network/pp_CE01S4H_sta_20050703T015717_SYDP3001_546.bdf ]'
So all the lines that are like these i want to extract two variables
2006/11/07 15:14:09
and
/opt/nexus/horizon/raw_data/network/pp_CE01S4H_sta_20050703T015717_SYDP3001_546.bdf
so i can store these variables in a database.
Can someone help me with writing the pattern to match and the regular express to extract? Also if anyone else has a better way of doing this i am all ears and i have a lot of log files to go through.

import java.util.regex.*;
class Main
public static void main(String[] args)
    String txt="INFO | jvm 1 | 2006/11/07 15:14:09 | INFO | Tue Nov 07 15:14:09 CET 2006 | XLDB PPS Data Dumper: MESSAGE:- 406 Processing .. '[ /opt/nexus/horizon/raw_data/network/pp_CE01S4H_sta_20050703T015717_SYDP3001_546.bdf ]'";
    String re1=".*?";     // Non-greedy match on filler
    String re2="((?:2|1)\\d{3}(?:-|\\/)(?:(?:0[1-9])|(?:1[0-2]))(?:-|\\/)(?:(?:0[1-9])|(?:[1-2][0-9])|(?:3[0-1]))(?:T|\\s)(?:(?:[0-1][0-9])|(?:2[0-3])):(?:[0-5][0-9]):(?:[0-5][0-9]))";     // Time Stamp 1
    String re3=".*?";     // Non-greedy match on filler
    String re4="((?:\\/[\\w\\.]+)+)";     // Unix Path 1
    Pattern p = Pattern.compile(re1+re2+re3+re4,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
    Matcher m = p.matcher(txt);
    if (m.find())
        String timestamp1=m.group(1);
        String unixpath1=m.group(2);
        System.out.print("("+timestamp1.toString()+")"+"("+unixpath1.toString()+")"+"\n");
}

Java Regular Expressions in J2EE

Does anybody know when Java Regular Expressions will be available in J2EE. They are currently in the latest release of J2SE in the java.util.regex package.

They are in the Standard Edition, so it does not make sense that they will also be in Enterprise Edition some day. You need to have the standard JRE installed before you can use the J2EE classes anyway.
If you want to use the regular expressions, install version 1.4 (beta) of the J2SE and use the current version of J2EE on top of that.
Jesper

Java Built-in regular expressions versus Jakarta

Are there any advantages to using the Jakarta regular expression package, over the built-in Java regular expressions?
I wasn't able to find much information on the Jakarta regexp package, except for the Javadoc and I didn't find that very informative.
Thanks!
Jeff

Well, the String.replaceAll(String, String) method uses the Java regexp internally, and I'm not sure there's a way to change that. So that at least is one place that will use it. I don't know for sure, but Jakarta's ORO is supposed to be fully compatible with Perl 5, and also supports other regexp types as well, so if you need that aspect of it, then go with Jakarta. I heard some of Java's are a little limited in some capabilities. I can't find any particular pages that refer to any comparisons of them at to compare runtime performance.

Regular expressions with boolean connectives (AND, OR, NOT) in Java?

I'd like to use regular expression patterns that are made up of simple regex patterns connected via AND, OR, or NOT operators, in order to do some keyword-style pattern matching.
A pattern could look like this:
(.*Is there.*) && (.*library.*) && !((.*badword.*) || (^$))
Is there any Java regex library that allows these operators?
I know that in principle these operators should be available, since Regular languages are closed under union, intersection, and complement.

AND is implicit,
xy -- means x AND yThat's not what I need, though, since this is just
concatenation of a regex.
Thus, /xy/ would not match the string "a y a x",
because y precedes x.So it has to contain both x and y, but they could be
in any order?
You can't do that easily or generally.
"x.*y|y.*x" wouldll work here, but obviously
it will get ugly factorially fast as you add more
terms.You got that right: AND means the regex operands can appear in any order.
That's why I'm looking for some regex library that does all this ugly work for me. Again, from a theoretical point of view, it IS possible to express the described semantics of AND with regular expressions, although they will get rather obfuscated.
Unless somebody has done something similar in java (e.g., for C++, there's Ragel: http://www.cs.queensu.ca/~thurston/ragel/) , I will probably use some finite-state-machine libraries and compile the complex regex's into automata (which can be minimized using well-defined operations on FSMs).
>
You'd probably just be better off doing multiple
calls to matches() or whatever. Yes, that's another possibility, do the boolean operators in Java itself.
Of course, if you
really are just looking for literals, then you can
just use str.contains(a) && !str.contains(b) &&
(str.contains(c) || str.contains(d)). You don't
seem to need regex--at least not from your example.OK, bad example, I do have "real" regexp's in there :)

Help with java regular expressions

Hi all ,
i am going to match a patternstring against an input string and print the result here is my code:
     import java.util.regex.*;
     import java.util.*;
     public class Main {
          private static final String CASE_INSENSITIVE = null;
          public static void main(String[] args)
          CharSequence inputStr = "i have 5 years FMCG saLEs exp on java/j2ee and i worked on java and j2ee and 2 projects on telecom java j2ee domain with your with saLEs maNAger experience of java j2ee and c# having very good on c++ exposure in JAVA"
         String patternStr = "\"java j2ee\" and \"c#\"";
          StringTokenizer st = new StringTokenizer(patternStr,"\",OR");
         Matcher matcher=null;
          while(st.hasMoreTokens()){
               String s=st.nextToken();
               Pattern pattern = Pattern.compile(s,Pattern.CASE_INSENSITIVE);
           matcher = pattern.matcher(inputStr);
           while (matcher.find()) {
              String result = matcher.group();
             if(!result.equalsIgnoreCase(" "))
                         System.out.println("result:"+result);
     when i compile this code i am getting the expected result...ie
result:java j2ee
result:java j2ee
result: and
result: and
result: and
result: and
result: and
result: and
result:c#
but when i replace String patternStr = "\"java j2ee\" and \"c#\""; with
String patternStr = "\"java j2ee\" and \"c++\""; i am just getting c in the result instead of c++ ie i am getting result :
result:java j2ee
result:java j2ee
result: and
result: and
result: and
result: and
result: and
result: and
result:C
result:c
result:c
result:c
result:c
result:c
result:c
In the last lines i should get result:c++ instead of result: c
Any ideas please
Thanks

In the last lines i should get result:c++ instead of result: cThe regular expression parser considers the plus sign '+' a special
character; it means: one or more times the previous regular expression.
So 'c++' means one or more 'c's on or more times. Obviously you don't
want that, you want a literal '+' plus sign. You can do that by prepending
the '+' with a backslash '\'. Unfortunately, the javac compiler considers
a backslash a special character and therefore you have to 'escape'
the backslash also, by adding another backslash. The result looks
like this:"c\\+\\+"kind regards,
Jos

Problems with java regular expressions

Hi everybody,
Could someone please help me sort out an issue with Java regular expressions? I have been using regular expressions in Python for years and I cannot figure out how to do what I am trying to do in Java.
For example, I have this code in java:
import java.util.regex.*;
String text = "abc";
          Pattern p = Pattern.compile("(a)b(c)");
          Matcher m = p.matcher(text);
if (m.matches())
               int count = m.groupCount();
               System.out.println("Groups found " + String.valueOf(count) );
               for (int i = 0; i < count; i++)
                    System.out.println("group " + String.valueOf(i) + " " + m.group(i));
My expectation is that group 0 would capture "abc", group 1 - "a" and group 2 - "c". Yet, I I get this:
Groups found 2
group 0 abc
group 1 a
I have tried other patterns and input text but the issue remains the same: no matter what, I cannot capture any paranthesized expression found in the pattern except for the first one. I tried the same example with Jakarta Regexp 1.5 and that works without any problems, I get what I expect.
I am using Java 1.5.0 on Mac OS X 10.4.
Thank to all who can help.

paulcw wrote:
If the group count is X, then there are X plus one groups to go through: 0 for the whole match, then 1 through X for the individual groups.It does seem confusing that the designers chose to exclude the zero-group from group count, but the documentation is clear.
Matcher.groupCount():
Group zero denotes the entire pattern by convention. It is not included in this count.

SQL Injection and Java Regular Expression: How to match words?

Dear friends,
I am handling sql injection attack to our application with java regular expression. I used it to match that if there are malicious characters or key words injected into the parameter value.
The denied characters and key words can be " ' ", " ; ", "insert", "delete" and so on. The expression I write is String pattern_str="('|;|insert|delete)+".
I know it is not correct. It could not be used to only match the whole word insert or delete. Each character in the two words can be matched and it is not what I want. Do you have any idea to only match the whole word?
Thanks,
Ricky
Edited by: Ricky Ru on 28/04/2011 02:29

Avoid dynamic sql, avoid string concatenation and use bind variables and the risk is negligible.

Regular expression java

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