Regular expressions anyone?

Hi all,
Where can i get a package to use regular expressions in java?
Dars

Where can i get a package to use regular expressions
in java?Look at jregex library:
http://jregex.sourceforge.net
Demo applet:
http://jregex.sourceforge.net/demoapp.html

Similar Messages

Help on regular expression

hi all
i need to validate telephone number , and i tried to use the following expression :
"[\\d][\\d][\\d]-[\\d][\\d][\\d][\\d]" or
"[0-9][0-9][0-9]-[0-9][0-9][0-9][0-9]" they did not seem to work. can anyone provide some help on this? thanks in advance.

Hi,
Can anyone provide me the regular expression for the following?
1. String Should contain 7 letters (can include '*') followed by 5 alphanumeric characters i.e.5 numbers/letters
For this one, I've written reg expn = ([a-zA-Z]*[*]*[a-zA-Z]*){7}\\w{5}
But, somehow this is not working...can anyone pleaseeeee explain wots wrong with this one or give me alternate solution???????
2. String should contain 7 numbers or 7 digits consisting of 1 letter & 6 numbers in varying order, no spaces.
For this one, reg expn = (\\d{7})|((\\d*[a-zA-Z]\\d*){7})
But this is also not working...... Please help!!!!!
3. String should contain 9 numbers or 8 alphanumeric characters i.e. 8 numbers/letters or 13 numbers
This one looks really simple, but still (\\d{9})|(\\w{8})|(\\d{13}) is not working.....Please helpppp
I'll appreciate if any Regular Expresion expert can resolve my problem!!! Also, I'm using gnu-regexp-1.0.8.jar as we r not using JDK 1.4!!!! We've to use only this library, no other choice...so, please suggest!!!!!!

Use regular expressions to extract .llb filename from path

I am trying to be clever (always a dangerous thing) and use a regular expression to extract the name of a library from a filepath converted to a string. Whilst I appreciate there are other ways to do this, regex would appear to be a very powerful neat way, should I be able to get it to work.
i.e if I have a string of the type, C:\applications\versions\library.llb\toplevel.vi, I want to be able to extract library.llb from the string, given that it will be of variable length, may include numbers & spaces and may be within a file hierarchy of variable depth. In other words, I want to extract the portion of the string between the last \ that ends with .llb
The best I have managed so far, is \\+.*llb which returned everything minus the drive letter and the toplevel.vi
Can anyone help me achieve this, or am i better using an alternative method (e.g filepath to array string, search for .llb)
Thanks
Matt
Solved!
Go to Solution.

Hi Matt,
attached you'll find two other options.
Mike
Message Edited by MikeS81 on 04-13-2010 01:30 PM
Attachments:
Path.PNG ‏7 KB

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");
}

Regular expressions and limiting matched input

Hi everyone :) I am trying to put together a regular expression that matches strings that contain elements of the form;
{<some text>}
However, each piece of text may contain multiple embedded instances of this pattern. I want to ensure that I am always getting the first (or outermost) instance.
So, if I had;
{OneStart}{TwoStart}{TwoEnd}{OneEnd}
I want to make sure that I get 'One' first and 'Two' second. So I have to place a stipulation in my regular expression to match this pattern only if it has not located the patten previously.
At the moment, I have this -
([^\\{]*?)(\\{TagStart\\})(.*?)(\\{TagEnd\\})(.*)
What I think that I have to do is modify the first capture group '([^\\{]*?)' which at the moment only does not match if a preceding '{' is found to match only if a preceding '{<text>}' sequence is not found.
Anyone got any idea how to do this?
Thanks in advance.
Ben

Doesn't it work anyway, since you're using greedy operators? If not, won't it work if you remove the .* at the end and use find() rather than matches()? And finally, what's the (.*?) supposed to match? Looks to me like that should be .*

Regular expressions and back references

Just wanted to know if anyone else noticed that.
In the javadoc of java.util.regex.Pattern in the "Back references" section it says that you need to use \n to match capturing group but it does not work. To match a capturing group one need to use a "$" sign which is not standard for this type of operation.
For example, the following code should work according to the API and most other regular expression engines:
Pattern.compile("([A-Z])").matcher("ThisIsATestString").replaceAll(" \1");
But to make this work you need to use:
Pattern.compile("([A-Z])").matcher("ThisIsATestString").replaceAll(" $1");
So, is this just a doc bug or am I missing something?
Someone have any idea why Sun choose to use the "$" sign instead of the regular "\" sign??
TIA,
Shaul

The doc you're referring to is talking about using back-refereneces within the regex, not in the replacement string. For instance, if you wanted to find all instances of things like "foo-foo" or "bar-bar", you would use a Pattern like Pattern p = Pattern.compile("([a-z]+)-\\1");For the most part, they've made the syntax the same a Perl's regexes, and that's why they use $n instead of \n in the replacement string. The replacement string is described in the Matcher javadoc.

Regular expressions in Rename Wizard Utility

Hi guys,
Does anyone know how to use regular expressions (and it possible) in OBIEE's internal "Rename Wizard" utility. For example, I have the following columns:
% Change Variance ColumnName1
% Change Variance ColumnName2
% Change Variance ColumnName3
and I want them to be
Variance ColumnName1 % Change
Variance ColumnName2 % Change
Variance ColumnName3 % Change
Is there another way to do this except than by hand? Your help is appreciated.

user447618,
Regular Expressions were first introduced in SQL and PL/SQL as of Oracle Database 10g. What version of the database are you running?
Sergio

Regular Expressions - Logical AND

I know this isn't Java, and it's not really an algorithm, but I can't figure this out. I hope amongst this bright group of developers someone can help me.
I am searching for a regular expressions that will match a series of words.
Example:
Given the words: "ship book"
What regular expression could be used to find both the word "ship" and the "book"?
I have found one expression that will do it ... ship.*book|book.*ship
But that expression doesn't scale. Does anyone know of a better way?
Thanks,
BacMan

Hi,
How about something like:
public class Regexp1
   private static final Pattern all = Pattern.compile(
         "^(\\s*\\b(monkey|turnip|ship|book)\\b\\s*)*$" );
   private static final Pattern p = Pattern.compile(
         "\\s*\\b(monkey|turnip|ship|book)\\b\\s*" );
   public static void main( String[] argv )
      find( "ship book turnip monkey" );
      find( "monkey giraffe mango" );
      find( "ship shipship ship" );
   public static void find( String text )
      System.out.println( "Text: " + text );
      Matcher m = p.matcher( text );
      System.out.println( "Matches all: " + all.matcher( text ).matches() );
      while ( m.find() )
         System.out.println( "Matching word: '" + m.group(1) + "'" );
   }which will produce this when run:
Text: ship book turnip monkey
Matches all: true
Matching word: 'ship'
Matching word: 'book'
Matching word: 'turnip'
Matching word: 'monkey'
Text: monkey giraffe mango
Matches all: false
Matching word: 'monkey'
Text: ship shipship ship
Matches all: false
Matching word: 'ship'
Matching word: 'ship'Pattern all tests to see if all the words are present.
Pattern p finds each matching word and ignores others.
Ol.

Regular Expressions, please help.

Hello everyone.
Can I get a Java Regular Expression to match with a word of the following language...
Start --> Expression;
Expression --> [0-9]+;
Expression --> Expression * Expression;
So the regexp should match with words like:
4;
4664;
4 * 763;
5 * 4534 * 23534;
04 * 002 * 1 * 10 * ...
I would be very happy, if anyone could help.

I dont think that I need to learn anything more.
I am sure it is not possible to make, what I want.
I want to build a compiler.
I just finished the abstract syntax of my language. Now I need a possibility to compile the concrete syntax of my language to the abstract one.
But I think, it is not possible with regular expressions.
Cause I need possibility to match a syntax of type chomsky 2.
I think regular expressions only match chomsky 3 languages.
But the "Backtracking"-mechanism of Java RegExp could do this.
I am not sure with this.
If you have any ideas please post.

Regular Expression Assistance

Using a C350 for Email Encryption Only
I have a regular expression that searches for a number format that could match a medical record number, among other numbers. Getting many false positives due to a matching number being found in many places in the html tags of the email.
Anyone have any good ideas on how to eliminate this? I have quite a few negative lookbehind and negative lookaheads in there already to look for dashes, backslashes, slashes, etc., but still can't weed them all out.
Thanks!

Can you paste in some examples of what you're trying to match against?
Using a C350 for Email Encryption Only
I have a regular expression that searches for a number format that could match a medical record number, among other numbers. Getting many false positives due to a matching number being found in many places in the html tags of the email.
Anyone have any good ideas on how to eliminate this? I have quite a few negative lookbehind and negative lookaheads in there already to look for dashes, backslashes, slashes, etc., but still can't weed them all out.
Thanks!

Encoding in Regular Expressions

Can anyone tell me how to use Encoding in Regular Expressions to read Chinese Character..........................thanks
Sarang.

You need to read
http://java.sun.com/docs/books/tutorial/extra/regex/index.html
and then you can use the \uxxxx to define the characters you need. A very very simple example is
           Pattern p = Pattern.compile("[\u0030-\u0070]+");
            Matcher m = p.matcher("A\u0055BCDE");
            if (m.matches())
                System.out.println(m.group());
            }

Using regular expressions in java

Does anyone of you know a good source or a tutorial for using regular expressions in java.
I just want to look at some examples....
Thanks

thanks a lot... i have one more query
Boundary matchers
^      The beginning of a line
$      The end of a line
\b      A word boundary
\B      A non-word boundary
\A      The beginning of the input
\G      The end of the previous match
\Z      The end of the input but for the final terminator, if any
\z      The end of the input
if i want to use the $ for comparing with string(text) then how can i use it.
Eg if it is $120 i got a hit
but if its other than that if should not hit.

Regular expressions and string matching

Hi everyone,
Here is my problem, I have a partially decrypted piece string which would appear something like.
Partially deycrpted:     the?anage??esideshe?e
Plain text:          themanagerresideshere
So you can see that there are a few letter missing from the decryped text. What I am trying to do it insert spaces into the string so that I get:
                    The ?anage? ?esides he?e
I have a method which splits up the string in substrings of varying lengths and then compares the substring with a word from a dictionary (implemented as an arraylist) and then inserts a space.
The problem is that my function does not find the words in the dictionary because my string is only partially decryped.
     Eg:     ?anage? is not stored in the dictionary, but the word �manager� is.
So my question is, is there a way to build a regular expression which would match the partially decrypted text with a word from a dictionary (ie - ?anage? is recognised and �manager� from the dictionary).

I wrote the following method in order to test the matching using . in my regular expression.
public void getWords(int y)
int x = 0;
for(y=y; y < buff.length(); y++){
String strToCompare = buff.substring(x,y); //where buff holds the partially decrypted text
x++;
Pattern p = Pattern.compile(strToCompare);
for(int z = 0; z < dict.size(); z++){
String str = (String) dict.get(z); //where dict hold all the words in the dictionary
Matcher m = p.matcher(str);
if(m.matches()){
System.out.println(str);
System.out.println(strToCompare);
// System.out.println(buff);
If I run the method where my parameter = 12, I am given the following output.
aestheticism
aestheti.is.
demographics
de.o.ra.....
Which suggests that the method is working correctly.
However, after running for a short time, the method cuts and gives me the error:
PatternSyntaxException:
Null(in java.util.regex.Pattern).
Does anyone know why this would occur?

Regular Expression Fails String replaceAll

I am trying to use regular expressions to replace double backslashes in a string with a single backslash character. I am using version 1.4.2 SDK. Upon invoking the replaceAll method I get a stack trace. Does this look like a bug to anyone?
String s = "text\\\\"; //this results in a string value of 'text\\'
String regex = "\\\\{2}"; //this will match 2 backslash characters
String backslash = "\\";
s.replaceAll(regex,backslash); java.lang.StringIndexOutOfBoundsException: String index out of range: 1
     at java.lang.String.charAt(String.java:444)
     at java.util.regex.Matcher.appendReplacement(Matcher.java:551)
     at java.util.regex.Matcher.replaceAll(Matcher.java:661)
     at java.lang.String.replaceAll(String.java:1663)
     at com.msdw.fid.fitradelinx.am.client.CommandReader.read(CommandReader.java:55)
     at com.msdw.fid.fitradelinx.am.client.CommandReader.main(CommandReader.java:81)
Exception in thread "main"

Skinning the cat -
public class Fred12
    public static void main(String[] args)
            String s = "text\\\\"; //this results in a string value of 'text\\'
            String regex = "[\\\\]{2}"; //this will match 2 backslash characters
            String backslash = "\\\\";
            System.out.println(s.replaceAll(regex,backslash));
            String s = "text\\\\"; //this results in a string value of 'text\\'
            String regex = "(\\\\){2}"; //this will match 2 backslash characters
            String backslash = "\\\\";
            System.out.println(s.replaceAll(regex,backslash));
            String s = "text\\\\"; //this results in a string value of 'text\\'
            String regex = "(?:\\\\){2}"; //this will match 2 backslash characters
            String backslash = "\\\\";
            System.out.println(s.replaceAll(regex,backslash));
            String s = "text\\\\"; //this results in a string value of 'text\\'
            String regex = "(?:\\\\)+"; //this will match 2 or more backslash characters
            String backslash = "\\\\";
            System.out.println(s.replaceAll(regex,backslash));
            String s = "text\\\\"; //this results in a string value of 'text\\'
            String regex = "\\\\\\\\"; //this will match 2 backslash characters
            String backslash = "\\\\";
            System.out.println(s.replaceAll(regex,backslash));
}

Regular Expressions in 1.3.1

I'm in a project that I need to use regular expressions. Does anyone knows a library that I can use?
I can only use JDK 1.3.1_01.
Thanks.

There's a third party regex library for 1.3 available
from jakarta or someplace. You should be able to find
it with google.Thanks for the advice......I search for the library that you said and I also found JRegex (http://jregex.sourceforge.net/).
I will try both libraries and see which one is better...
thanks..

Regular expressions anyone?

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