How to check whether a string contains at least one alphabet in java?

Question

I want such a validation that My String must be contains at least one alphabet.

I am using the following:

String s = "111a11";
boolean flag = s.matches("%[a-zA-Z]%");

flag gives me false even though a is in my string s

Don't use %. That is for SQL LIKE, not regexp. Use `.*` instead So just `s.matches(".*[a-zA-Z].*");` — ppeterka, Jan 11 '13 at 12:27

Bhesh Gurung · Answer 1 · 2014-09-20T16:15:00.743

98

You can use .*[a-zA-Z]+.* with String.matches() method.

boolean atleastOneAlpha = s.matches(".*[a-zA-Z]+.*");

edited Sep 20 '14 at 16:15

answered Jan 11 '13 at 12:52

Bhesh Gurung

can we check the min and max number of characters in this regex? – Mohammad Faisal Apr 06 '16 at 13:05
4

You post a question for it. – Bhesh Gurung Apr 06 '16 at 13:48
Man I know this is several years old but thank you guys so much! Been looking all over for the proper regex expression and this is exactly what I needed. I'm going to brush up on regex expressions but man I was struggling. Thanks again! – wiregh0st Mar 06 '20 at 05:22

score 24 · Accepted Answer · edited Jun 06 '14 at 18:46

24

The regular expression you want is [a-zA-Z], but you need to use the find() method.

This page will let you test regular expressions against input.

and here you have a Java Regular Expressions tutorial.

edited Jun 06 '14 at 18:46

Kevin Panko

answered Jan 11 '13 at 12:30

Luciano

3

If you consider other languages, you could use `boolean m = str.matches(".*[\\p{L}]+.*]")` – dragos2 Dec 20 '13 at 10:16

2 Answers2