CODEWITHSUNDEEP
javaregex
Logan
Logan

Reputation: 2515

Regular Expression allowing special characters

I am using the following pattern to match my String:

[a-zA-Z0-9]*

Even when I am passing the String *$#, its getting matched by the regular expression. Could someone explain what am I doing wrong or why this is happening?

Upvotes: 0

Views: 1788

Answers (2)

Ryan Stewart
Ryan Stewart

Reputation: 128919

[a-zA-Z0-9]* means 0 or more of any of those characters. If you're using Matcher.find(), it'll find that anywhere/everywhere because it can match anywhere in a string.

Upvotes: 2

Anirudha
Anirudha

Reputation: 32817

You should use ^(start of the string) and $(end of the string).

So,the regex would be 

^[a-zA-Z0-9]*$

[a-zA-Z0-9]* would match anywhere in the string if you use find method..Using ^ and $ would match the entire input from start till end

if you use matches method you don't need to have ^,$ as it tries to match the entire string

Upvotes: 3

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