Java Escaping Characters in RegEx

Question

I have the below RegEx to match email address

[-0-9a-zA-Z.+_]+@[-0-9a-zA-Z.+_]+\.[a-zA-Z]{2,4}

AM trying to escape the characters as below but am still getting java compilation error. Could someone please help me to escape characters in the above RegEx?

String pattern = "\\[-0-9a-zA-Z\\.\\+_\\]\\+@\\[-0-9a-zA-Z\\.\\+_\\]+\\\.\\[a-zA-Z\\]\\{2,4\\}";

Andie2302 · Accepted Answer · 2015-01-09 12:47:39Z

6

The escaped regex is:

"[-0-9a-zA-Z.+_]+@[-0-9a-zA-Z.+_]+\\.[a-zA-Z]{2,4}"

answered Jan 9, 2015 at 12:47

Andie2302

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1 Comment

Ian Roberts Over a year ago

... the point being that characters like . and + don't need to be escaped at all inside square bracket character classes as they are treated as literals by default. The only thing that needs to be escaped from Java is the backslash before the literal dot.

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