Problem forming regular expression

Hi,
I want to use regular expression to match a repeating pattern until it meets a particular character. Eg.
The string I am trying to match is a function definition.
function abc()
This is the pattern I am using to match the above string.
Pattern p =
Pattern.compile("function abc.*\r*\n*\\{(\r*\n*.*)*(\r\n)*\\}", Pattern.MULTILINE);
"function abc" takes care of the method signature
" .*\r*\n*\\{" takes care of all spaces/characters/newline till it encounters the opnening brace
"(\r*\n*.*)*" this iswhere the problem lies. This matches all the individual new lines in the method body. But I dont want it to match the closing brace "}" . It is doing that currently. How can I avoid that? So that the next part of the pattern i.e. "\\}" actually matches the closing brace?

DrLaszloJamf wrote:
Isn't the basic difference between regular grammars and context free grammars? The latter can generate matched parens, the former can't?I guess you mean by "regular grammar" a "regular language" (forgive my ignorance if they are the same)? But yes, AFAIK that is a difference. However, there are regex engines that can cope with these recursive matches (an undefined number of back references). Perl is one of them.
It is an often heard complaint about implementations of regular expression engines: they have grown too big for what they were created for.

Similar Messages

Urgent!!! Problem in regular expression for matching braces

Hi,
For the example below, can I write a regular expression to store getting key, value pairs.
example: ((abc def) (ghi jkl) (a ((b c) (d e))) (mno pqr) (a ((abc def))))
in the above example
abc is key & def is value
ghi is key & jkl is value
a is key & ((b c) (d e)) is value
and so on.
can anybody pls help me in resolving this problem using regular expressions...
Thanks in advance

"((key1 value1) (key2 value2) (key3 ((key4 value4)
(key5 value5))) (key6 value6) (key7 ((key8 value8)
(key9 value9))))"
I want to write a regular expression in java to parse
the above string and store the result in hash table
as below
key1 value1
key2 value2
key3 ((key4 value4) (key5 value5))
key4 value4
key5 value5
key6 value6
key7 ((key8 value8) (key9 value9))
key8 value8
key9 value9
please let me know, if it is not possible with
regular expressions the effective way of solving itYes, it is possible with a recursive regular expression.
Unfortunately Java does not provide a recursive regular expression construct.
$_ = "((key1 value1) (key2 value2) (key3 ((key4 value4) (key5 value5))) (key6 value6) (key7 ((key8 value8) (key9 value9))))";
my $paren;
   $paren = qr/
           [^()]+ # Not parens
         |
           (??{ $paren }) # Another balanced group (not interpolated yet)
    /x;
my $r = qr/^(.*?)$(\w+?) (\w+?|(??{$paren}))$\s*(.*?)$/;
while ($_) {
     match()
# operates on $_
sub match {
     my @v;
     @v = m/$r/;
     if (defined $v[3]) {
          $_ = $v[2];
          while (/\(/) {
               match();
          print "\"",$v[1],"\" \"",$v[2],"\"";
          $_ = $v[0].$v[3];
     else { $_ = ""; }
C:\usr\schodtt\src\java\forum\n00b\regex>perl recurse.pl
"key1" "value1"
"key2" "value2"
"key4" "value4"
"key5" "value5"
"key3" "((key4 value4) (key5 value5))"
"key6" "value6"
"key8" "value8"
"key9" "value9"
"key7" "((key8 value8) (key9 value9))"
C:\usr\schodtt\src\java\forum\n00b\regex>

Problem with Regular Expression

Hi There!!
I have a problem with regular expression. I want to validate that one word and second word are same. For that I have written a regex
Pattern p=Pattern.compile("([a-z][a-zA-Z]*)\\s\1");
Matcher m=p.matcher("nikhil nikhil");
boolean t=m.matches();
if (t)
          System.out.println("There is a match");
     else
          System.out.println("There is no match");
The result I am getting is always "There is no match
Your timely help will be much appreciated.
Regards

Ram wrote:
ErasP wrote:
You are missing a backward slash in the regex
Pattern p = Pattern.compile("([a-z][a-zA-Z]*)\\s\\1");
But this will fail in this case.
Matcher m = p.matcher("Nikhil Nikhil");It is the reason for that *[a-z]*.The OP had [a-z][a-zA-Z]* in his code, so presumably he know what that means and wants that String not to match.

Problem on regular expression

Hello everyone,
I am working on a project that require me to seperate the following examples:
+0,+0.0000+E1,-4.33+E1,-4.222+E1,-6.33+E2,-6.55+E2
What I need is the final four results seperately:
-4.33+E1,
-4.222+E1,
-6.33_E2,
-6.55+E2.
I am totally fresh to regular expression. Any help is well appreciated!
Thanks,
+Kunsheng
Solved!
Go to Solution.

Take a look at the code below. It doesn't make much use of RegEx, but I think it is simpler.
Message Edited by jmcbee on 03-05-2009 12:42 PM
Message Edited by jmcbee on 03-05-2009 12:43 PM
CLA, CLED, CTD,CPI, LabVIEW Champion
Platinum Alliance Partner
Senior Engineer
Using LV 2013, 2012
Don't forget Kudos for Good Answers, and Mark a solution if your problem is solved.
Attachments:
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Problems with regular expressions.

Hi all.
I want to substitute within an xml file a string with the following structure:
Referencer="//@Model/@OwnedElement.0/@OwnedElement.1/@OwnedElement.0/@OwnedElement.1/@OwnedElement.33 ..."
where the dots mean that the string within the quotes may be repeated (separated by spaces) many times.
I've tried to match such expression with java regex, with no luck. I thought running
line.contains("Referencer=\"[^\"\\r\\n]*\"");
would return true, but it didn't.
Could anybody point me the right way?.
Thank you all in advance.

Method String.contains() does not take a regular expression - you need to use method String.matches() or class Pattern.
This looks to be a good problem to solve using http://elliotth.blogspot.com/2004/07/java-implementation-of-rubys-gsub.html .

Problem with Regular Expressions

Hi Everyone:
I'm having a problem finding the easiest way to retrieve the replacement text that has been edited to insert back-references from previous matches. For instance,
Lets say I want to use the below regular expression
foo_bar([1-9])_fun
on the search text
foo_bar1_fun
foo_bar2_fun
foo_bar3_fun
and then the replacement text would be
foobar_fun$1
so in the end I would get
foobar_fun1
foobar_fun2
foobar_fun3
What I would like to do is be able to extract the replacement text that has been modified with the back reference matches after I use the Matcher.appendReplacement(stringbuffer, string) method.
So to clarify further, after I find the first match and use the appendReplacement Method, I would like to extract just the replacement that was made to the text, in this case foobar_fun1
Thanks for any help!

Alright, thanks for the reply. I'll try and make this a little more clear
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class LeClassName {
   public static void main(String[] args) {
      String input = "12341234foo_bar1_fun24342345foo_bar2_fun3522545foo_bar3_fun3423432";
      Pattern pattern = Pattern.compile("foo_bar([1-9])_fun");
      StringBuffer sb = new StringBuffer();
      Matcher matcher = pattern.matcher(input);
      while (matcher.find()) {
         matcher.appendReplacement(sb, "foobar_fun$1");
         //after the first pass through, I would like to extract, foobar_fun1
         // but what is in sb is 12341234foobar_fun1
      System.out.println(sb.toString());
}I did find a solution myself after a bit of thinking but if anyone can come up with a better solution do tell and I'll award that person the answer. Thanks again!
My Solution:
private String fixBackReplacements() throws IndexOutOfBoundsException {
            String currentMatch = this.findMatch.group();
            Pattern temporaryPattern = Pattern.compile(findText);
            Matcher temporaryMatcher = temporaryPattern.matcher(currentMatch);
            return temporaryMatcher.replaceFirst(replaceText);
}

Problem with Regular Expression: XML comment parsing

I have create pattern to parse xml comments:
It works fine but only problem that I have is that it parse only last occurrence :(

sabre150 wrote:
MrJbee wrote:
I have create pattern to parse xml comments:
It works fineI'm surprised it works fine since the ".*" is greedy and will gobble as much as it can and you will end up matching from the start of the first comment to the end of the last comment.
but only problem that I have is that it parse only last occurrence :(Make the ".*" reluctant using ".*?" or, best of all, use an XML parser.
Edited by: sabre150 on Jan 11, 2010 4:00 PMFirst at thanks for reply.
Second it is not so greedy like it look. Because of that:
(?!.*(<!--))I can`t use xml parser since It is only part of my JEditor with styles. I use reg exp to highlight comments, strings, chars and so on....
By the way I try to replace .* => .*? and it is still no results. I mean only last is selected. Maybe you have an example on how to match structure if the do not
have specified substrings.
Something like this:
hello[^(guten tag)]*bye // I know that this is wrongEdited by: MrJbee on Jan 11, 2010 1:22 PM

Problem with regular expression validation

hi,
   how can i validate folder structure
   conditions
     no two forward slashes(/) in a sequence no special characters
    " /w" and   " '/' "
   e:/tomcat/ss/ss.text
       if(valStr.matches("((['/'])[a-zA-Z0-9_]*)")){
                System.out.println("matches");
         else
                System.out.println("notmatches");
this is no correct .
please give me right solution
with regards
ss

hI,
/appmg/dd/
is the paths for solaris( starts with /)
i have to test if it is valid path name or not
with regards
ss

Regular Expressions in CS5.5 - something is wrong

Hello Everybody,
Please correct me, but I think, I found a serious problem with regular Expressions in Indesign CS5.5 (and possibly in other apps from CS5.5).
Let's start with simple example:
var range = "a-a,a,a-a,a";
var regEx = /(a+-a+|a+)(,(a+-a+|a+))*/;
alert( "Match:" +regEx.test(range)+"\nLeftContext: "+RegExp.leftContext+"\nRightContext: "+RegExp.rightContext );
What I expected was true match and the left and the right context should be empty. In Indesign CS3 that is correct BUT NOT in CS5.5.
In CS 5.5 it seems that the only first "a-a" is matched and the rest is return as the rightContext - looks like big change (if not parsing error in RegExp engine).
Please correct me if I am wrong.
The second example - how to freeze ID CS5.5:
var range = "a-a,a,a-a,a";
var regEx = /(a+-a+|a+)(,(a+-a+|a+)){8,}/;
alert( "Match:" +regEx.test(range)+"\nLeftContext: "+RegExp.leftContext+"\nRightContext: "+RegExp.rightContext );
As you can see it differs only with the {8,} part instead of *
Run it in CS5.5 and you will see that the ID hangs (in CS3 of course it runs flawlessly}.
The third example - how to freeze ID 5.5 in one line (I posted it earlier in Photoshop forum because similiar problem was called earlier):
alert((/(n|s)? /gmi).test('s') );
As you can guess - it freezes the CS5.5 (CS3 passes the test).
Please correct me if I am doing something wrong or it's the problem of Adobe.
Best regards,
Daniel Brylak

Hi Daniel,
Thanks for sharing. Really annoying indeed.
Just to complete your diagnosis, what you describe about CS.5 is the same in CS5, while CS4 behaves as CS3.
var range = "aaaaa";
var regEx = /(a+-a+|a+)(,(a+-a+|a+))*/;
alert([
    "Match:" +regEx.test(range),
    "LeftContext: "+RegExp.leftContext.toSource(),        // => CS3/4: EMPTY -- CS5+: EMPTY
    "RightContext: "+RegExp.rightContext.toSource()        // => CS3/4: EMPTY -- CS5+: ",a,a-a,a"
    ].join('\r'));
So there is a serious implementation problem of the RegExp object from ExtendScript CS5.
I don't think it's related to the greedy modes. By default, JS RegExp quantifiers are greedy, and /a*/ still entirely captures "aaaaaa" in CS5+.
By the way, you can make any quantifier non-greedy by adding ? after the quantifier, e.g.: /a*?/, /a+?/, etc.
I guess that Adobe ExtendScript has a generic issue in updating the RegExp.lastIndex property in certain contexts—see http://forums.adobe.com/message/3719879#3719879 —which could explain several bugs such as the Negative Class bug —see http://forums.adobe.com/message/3510078#3510078 — or the problems you are mentioning today.
@+
Marc

Regular Expressions and comments

Hello,
I've got a problem with regular expressions. I Want to find special words like "todo" in java comments, but wasn't able to manage that satisfactory. I hope someone of you can give me an advice! :-)
example of a comment:
* comment text. todo: what is still todo.
First of all, I tried this:
/\*(.*?)todo(.*?)\*/
but this version seems to ignore the borders of the comment and shows me also parts of the code.
Then I tried this:
/\*([^/]*?)todo(.*?)\*/
this version returns good results, but doesn't notice html formatting like
* <code> .... </code> text. todo: text
So my idea now is to check whether a star precedes the slash, but I don't really know how to combine it.
Or is there another simplier solution? Thanks for your hints!
Greetz Jan

Thanks for your answer, but when I take this expression the programm seems to hang-up. An operation that usually finishs in 3 secs didn't even in 10 minutes. :-( Do you have any idea what could be wrong?
btw. can I take this one:
pattern = Pattern.compile("/\\*(?:[^\\*]+|\\*(?!/))*todo.*?\\*/", Pattern.DOTALL | Pattern.CASE_INSENSITIVE);
matcher = pattern.matcher(text);
text = matcher.replaceAll("");or do I have to take a this instead of replaceAll():
while (matcher.find()) {
text = text.substring(0, matcher.start(1)) + text.substring(matcher.end(1), text.length());
matcher = pattern.matcher(text);
}any hints?
Greetz Jan

Rplacing space with &nbsb; in html using regular expressions

Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What number does this represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

sorry i missed some information in above,The modified information was in red color
Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADIN G="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What&nbsb;number&nbsb;does&nbsb;this&nbsb;represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

Regular expressions and characters as č, �, ť ...

Hi. I have this problem with regular expressions. The next piece of code is work only with string without diacritics. If I use as pattern character with diacritic the matcher is not find any occurence of a character. The code:
String pat = "č"; // "a"
String text = "naj čalamada"
Pattern pattern = Pattern.compile(pat, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) { }If I uncomment character 'a' its occurence will be founded but occurence of 'č' is not found. :(
Can you tell me what is wrong ??
If I use flag Pattern.CASE_INSENSITIVE | Pattern.CANON_EQ nothing will be changed.
Thanks for all replays.

The file encoding is same as system encoding and it is utf-8.
How can I specify encoding as an inout parameter when I compile it?
Piece of code or concrete funcions help me more...

Regular expressions: the shorter the string, the longer the time to analyse

Hello there !
I've got a problem with regular expressions...
I analyse those two strings with the same regular expression, searching for the "message" word.
The first of them, which is much longer than the other one, is instantly analysed... but the second one takes about to one second !
Do you see the problem ? (I do not, as you can guess...).
I tried to put the same special characters in the two ones... but it still works slowly for the second one.
This is instantly analysed:
"Coucou----- Origin.a.l. .message -----From: clems= To: Cl�ment S=ent: Wednes=day, April 30, 2003 1:38 PMSubject: Fw: mail de test----- Original Message -----From: clemsTo: Cl�ment Sent: Wednesday, April 30, 2003 1:37 PMSubject: mail de testcoucou, ceci est un mail de t.e.s.t... !!!Clems.".matches("((.*\\W)*\\Q" + "message" + "\\E(\\W.*)*)")
This takes about to one second:
"This is a b message ubodyub.brHere s a href= http:www.tano.fr.fm the second linea.".matches("((.*\\W)*\\Q" + crit�reActuel.getLeMot(i).toLowerCase() + "\\E(\\W.*)*)")
Pleaaase, help ;)
Clement

Well in fact everything seems to work... there was another problem...
My regular expression seems to be wrong and the code too...
I apologize... sorry for the waste of time.
See you later :) for my next wrong problem ! :))
Clement

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

Evaluate Regular expression complexity

Hi all
I've a problem on regular expression usage in my application.
I'm using a regular expression to identify objects and fetch them to be served depending on an input string with has to be matched.
each object has a property representing a regular expression to be matched to be candidate for fetching.
my program receive an external input string, then loops on the full objects collection identifying which are the object whose regular expression match the input stream.
doing an example:
obj1) key = "J*SDK"
obj2) key = "Ja*6*"
obj3) key = "JEE*"
if the input string is "Java 6.0 SDK" obj1 and obj2 are cadidate, while obj3 is discarded.
up to now everithing is fine, now here is my question:
i want only one object as output and I want the one best matching my input string.
this means that
-> obj1 is matching 4 chatacter ans has only one wildchar
-> obj2 is matching 3 characters and has two wildcard
so obj2 is discarded since it's regular expression is more complex than the obj1 one
my problem is HOW to evaulate correctly such complexity for each candidate object to be able to choose my best object.
is there some formal rule / api for this?
I'd like to match all wildcards into the regex, but doing this "by hand" would surely result in some bug due to some missing case, so a "third party" API or a grammar rule would be useful.
hoping for you help.
regards
Michele Sacchetti

ok, after days and days of research i came up to this solution:
1) I used this (http://www.brics.dk/automaton) package for regular expression which let me access the internal state automa data
2) use the getShortestExample() method to retrieve the shortest string matching the given regExp
3) evaluate the Levenshtein Distance between the given string and the one to be matched
PROs:
1) the regexp logic is fully handled by the same state machine which take cares of pattern matching in the first phase
2) the library provide me a non-regexp string to be used with comparison (e.g Levenshtein Distance evaluation)
CONs:
1) the methods getShortestExample is unaware of string to be matched, so if i use "aab|aaa" to match "aab" the method gets the first shortest sort alfabetically, that is "aaa", so I get a LD of 1 even if it should be 0, but it's quite a good deal for my application.
@endasil : your solution rely on grouping, and is based on a pre-parsing done manually so it basically went back to the "manual" parsing i wanted to avoid
Another way I'd like to give a try but had to give up was to use ANTLR (www.antlr.org) to create a parser for regular expression and then evaluating the resulting "tree size" of the parser, but wasn't able to find a formal description of RegExp grammar on the net.
do you have any suggestion or comment on my solution (or other to give a try? )

Problem forming regular expression

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