Java regex phone number

问题:

I am supplying the following regex : ^((?:\+27|27)|0)(\d{9})$ for number of South Africa and want to only return true for numbers that start with +27 or 27 or 0. eg; [phone], 27838776654 or 0612323434.

I tried using: regexplanet.com

and regex test

But both return false no matter what I enter (even simple regex).

Anybody know what I am doing wrong?

回答1:

In Java code, you need to double the backslashes, and in a online tester, the backslashes must be single.

Also, it is not necessary to use start and end of string markers if used with String#matches as this method requires a full string match.

Here is how your Java code can look like:

String rex = "(\\+?27|0)(\\d{9})";
System.out.println("[phone]".matches(rex)); // => true
System.out.println("27838776654".matches(rex));  // => true
System.out.println("0612323434".matches(rex));   // => true

See IDEONE demo

If you use String#find, you will need the ^ and $ anchors you have in the original regex.

Note I shortened ((?:\+27|27)|0) to (\+?27|0) as ? quantifier means 1 or 0 occurrences. The extra grouping is really redundant here. Also, If you are not using captured texts, I'd suggest turning the first group into a non-capturing (i.e. (?:\+?27|0)) and remove the round brackets from \d{9}:

String rex = "(?:\\+?27|0)\\d{9}";

回答2:

This is an expression tested on http://tools.netshiftmedia.com/regexlibrary/# for numbers that start with +27 or 27 or 0 and after with length 9.

^((0)|(27)|(\+27))(\d{9})$

It's ok with your numbers :

回答3:

This should work, here is a variant. (I removed ^ and $ for this example):

((?:\+27|27|0)(?:\d{9}))