Can somebody please show me how to do a Java regex that takes in a string and returns a string with all characters removed BUT a-z and 0-9?
I.e. given a string a%4aj231*9.+ it will return a4aj2319
thanks.
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3 Answers
\d is digit, \p{L} is a-z and A-Z.
str.replaceAll("[^\\d\\p{L}]", "");
3 Comments
vbn
thanks for the quick reply..I just realized, what if I want to preserve spaces as well?
lins314159
Just add any characters you don't want to replace within the square brackets.
Sean
\p{L} also matches a ton of other Unicode characters, such as Δ, ね, and 傻. This expression will leave them all intact.
str = str.replaceAll("[^a-z0-9]+", "");
If you also meant to include uppercase characters, then you could use
str = str.replaceAll("[^A-Za-z0-9]+", "");
or the slightly leeter
str = str.replaceAll("[_\\W]+", "");
Comments
If you want a-z and 0-9 but not A-Z then
str.replaceAll("[^\\p{Lower}\\p{Digit}]", "");
1 Comment
lins314159
I didn't go with \w because that includes underscores as well.
