Cannot get regular expression to return true in String.matches()

Hi,
My String that I'm attempting to match a regular expression against is: value=='ORIG')
My regular expression is: value=='ORIG'\\) The double backslashes are included as a delimiter for ')' which is a regular expression special character
However, when I call the String.matches() method for this regular expression it returns false. Where am I going wrong?
Thanks.

The string doesn't contain what you think it contains, or you made a mistake in your implementation.
public class Bar {
   public static void main(final String... args) {
      final String s = "value=='ORIG')";
      System.out.println(s.matches("value=='ORIG'\\)")); // Prints "true"
}

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matchregularexpression.jpg ‏336 KB

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Which match gets used when you use OR ('|') to specify multiple possible matches in a regex, and there are multiple matches among the supplied patterns? The first one (in the order written) which matches? Or the one which matches the most characters?
To make this concrete, suppose that you want to split a String into lines, where the line delimiters are the same as the [line terminators used by Java regex|http://java.sun.com/javase/6/docs/api/java/util/regex/Pattern.html#lt] :
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If we ignore the idiotic microsoft two char \r\n sequence, then no problem; the Java code would be:
String[] lines = s.split("[\\n\\r\\u0085\\u2028\\u2029]");How do we add support for \r\n? If we try
String[] lines = s.split("[\\n\\r\\u0085\\u2028\\u2029]|\\r\\n");which pattern of the compound (OR) regex gets used if both match? The
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For instance, if the above code is called when
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because there is an empty String between the \r and following \n. On the other hand, if the rule is use the pattern which matches the most characters, then the
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System.out.println();
if (true) return;the answer that I get is
3 lines: "a" "" "b"So it seems like the first listed pattern is used, if it matches.
Therefore, to get the desired behavior, it seems like I should use
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(?:X)
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bbatman wrote:
Which match gets used when you use OR ('|') to specify multiple possible matches in a regex, and there are multiple matches among the supplied patterns? The first one (in the order written) which matches? Or the one which matches the most characters?
The longest match wins, normally. Except for alternation (or) as can be read from the innocent sentence
The Pattern engine performs traditional NFA-based matching with ordered alternation as occurs in Perl 5.
in the javadocs. More information can be found in Friedl's book, the relevant page of which google books shows at
[http://books.google.de/books?id=GX3w_18-JegC&pg=PA175&lpg=PA175&dq=regular+expression+%22ordered+alternation%22&source=bl&ots=PHqgNmlnM-&sig=OcDjANZKl0VpJY0igVxkQ3LXplg&hl=de&ei=Dcg7S43NIcSi_AbX-83EDQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CA0Q6AEwAA#v=onepage&q=&f=false|http://books.google.de/books?id=GX3w_18-JegC&pg=PA175&lpg=PA175&dq=regular+expression+%22ordered+alternation%22&source=bl&ots=PHqgNmlnM-&sig=OcDjANZKl0VpJY0igVxkQ3LXplg&hl=de&ei=Dcg7S43NIcSi_AbX-83EDQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CA0Q6AEwAA#v=onepage&q=&f=false]
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regular expression "ordered alternation"
My first hit went right into Friedl's book.
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Last edited by n0stradamus (2012-05-03 20:27:02)

Procyon wrote:
--> echo "a:i:foo r::bar t:ei:bark" | sed 's/$[art]:[ei]\{0,2\}:[^ ]*$/1/'
1 r::bar t:ei:bark
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1 1 1
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