Re: Matching parentheses with Regular Expressions

shakah wrote:
On Jul 3, 9:52 pm, Joshua Cranmer <Pidgeo...@xxxxxxxxxxxxxxx> wrote:
James wrote:
The regular expression
\\\\{2}.+\\\\Process\\(java\\).
>
> matches, but it matches too much of it:

In that case, you probably want this regex:
\\\\{2}[^\\\\]+\\\\Process\\(java\\)
--
Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth

FWIW, you could avoid a little of the backslash escape mess
by using single-char character classes, e.g.:
Pattern.compile("[\\]{2}[^\\]+[\\]Process[(]java[)]") ;
// ...outside of a Java string that'd be [\]{2}[^\]+
[\]Process[(]java[)]

You also might get rid of some of those backslashes by substituting another character, then using replace() on the string before compiling it.

final static String PATTERN = "``{2}.+``Process`(java`)";

String myRegex = PATTERN.replace("`", "\\" );
System.out.println( myRegex );

Result:

\\{2}.+\\Process\(java\)


It just makes things more readable. Using `, or %, or # in a string, then replace that character with \'s before compiling it as a regex can save your eyes.

Incidentally, I wonder if Sun could be convinced to add this themselves. Maybe add a new operator/keyword altogether. Like # introduces new keywords or operators. It's followed by the keyword or operator. This just allows Sun to make new keywords or operators, with out breaking any existing code. So #s might give us new string constatns. Let's say ' then means like a Unix shell string, where escaping is ignored.

String regex = #s'\\{2}.+\\Process\(java\)';

Would give that literal string, without the need to escape the backslashes. Easier for regex at least. Other types of flags besides ' could be introduced too. `,$,@,%,= might do the same thing, just use a different character as a string terminator, in case you want a ' to be part of the string. """ might introduce a "here-is" operator. Etc.

Just thinking out loud....


.

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