Java Regular Expression how to split special characters

Question

In java Regex i want to replace all the special characters to escape sequence.how can i do. Example ::

//special chars ex "dd[u]i.* " to "dd//[u//]i//.//*"

You might want to note that escaping is done using `\ ` (backslash) and not `/`. — devnull, Mar 05 '14 at 08:26
[This answer](http://stackoverflow.com/a/13696653/2235132) might help. — devnull, Mar 05 '14 at 08:32

score 1 · Answer 1 · answered Mar 05 '14 at 08:48

To escape all special regexp control characters this method can be used:

Matcher.quoteReplacement(String s)

It returns a regular expression that matches exact s.

This comes from the javadoc:

Returns a literal replacement String for the specified String. This method produces a String that will work as a literal replacement s in the appendReplacement method of the Matcher class. The String produced will match the sequence of characters in s treated as a literal sequence. Slashes ('\') and dollar signs ('$') will be given no special meaning.

score 0 · Answer 2 · answered Mar 05 '14 at 08:43

0

To split or treat special characters in java pattern as normal one. You have to backshlas it. \\.; \\* it might be treat now as '.' and '*'.

answered Mar 05 '14 at 08:43

RMachnik

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Java Regular Expression how to split special characters

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