Question

I am using this: if(!preg_match('/^+[0-9]$/', '+1234567'))

and am getting:

Warning: preg_match()
[function.preg-match]: Compilation
failed: nothing to repeat at offset 1

any ideas why?

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update: Now using this: if(!preg_match('/^\+[0-9]$/', '+1234567'))

and am getting no match.

any ideas why?

Answer 1

+ is a special character that indicates 1 or more of the previous character, and by not escaping it you are applying it to the caret. escape it with \ and it will match a literal plus sign.

if(!preg_match('/^\+[0-9]$/', '+1234567'))

EDIT:

The reason why it didn’t match is because you specified 1 digit from 0-9 and the end of the string with $. You need to make it a variable amount of digits.

if(!preg_match('/^\+[0-9]+$/', '+1234567')) {

Shorter version:

if(!preg_match('/^\+\d+$/', '+1234567')) {