Problem in handling exception with Regex

Hello,
I want to identify via a java REGEXP the following snippet:

Code:

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catch (Exception e) {
  // ...
  throw e;
}

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I think it comes down to having a catch, followed by a throw before a closing brace is found. On or regexp: "[.&&[^}]]* catch throw" which means the word "catch", then any character except ")" 0 or n times and then throw. But it does not work. Shortly, I am trying to handle exception with the Regex. If you have any idea about it then please post back. Thanks in advance.

Hello,
I think you can try the following, to do the same. Just try it

Code:

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catch[^\}]*throw

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which means "catch" followed by any character other than ")" followed by "throw". It works better? I have tried this with my code, though not sure that it will work for you , because everybody program and the output are different.

Hello,
Perfect indeed, but we must add a .* at the end if the string contains characters after the throw:

Code:

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System.out.System.out.println(Pattern.games("catch throw [^}]* .*", "catch (Exception e) (e.printStackTrace () throw e;)));

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And this

Code:

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System.out.System.out.println(Pattern.games("catch throw [^}]* .*", "catch (Exception e) (e.printStackTrace ();) throw e;"));

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It remains for me to recognize the opposite, namely a catch that does not throw.

Hello,
A small note, do not forget escaping character ')':

Code:

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String rg = "catch [^\\)] .* * throw "

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Quote:

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It remains for me to recognize the opposite, namely a catch that does not throw.

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He'll just change your first regular expression to make the "throw" option (with the quantifier ?) And you have one regular expression :

Code:

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String rg = "catch [^ \ \)] * (throw)?.*"

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Hope this is going to help you.

Hello,
The character ')' need not be escaped, it seems. In addition to "\ \" is escaping the '\' => As for me, it seems to walk in both cases with or without your exhaust.

Code:

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He'll just change your first regular expression to make the "throw" option (with the quantifier ?) And you have one regular expression :

---

Code:

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String rg = "catch [^ \ \)] * (throw)?.*"

---

Indeed, but I recognize the words or the throw is not and only them. This is not optional. From what I read here and there, this is not possible via a regexp, it must be code, which is not possible in my case. Negation '^' can be used on the characters: "[^ abc]" means either a or b or c. But that does not work for a group of letters.

Hello,
If you look at the javadoc of the class Pattern You will notice that several expressions using braces ( '(' and ')') as special characters, it's better to take precaution to avoid.

Quote:

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In addition to "\ \" represents the exhaust '\

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Why do I need to double / quadruple backslash character ( '\')? I think this is a good option to try, just try it and post what you get the output of the program and if there are more queries then we will try to solve them.