preg_match: nothing to repeat / no match

Question

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

and am getting:

Warning: preg_match() [function.preg-match]:

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')) {

Answer 2

'/^\+[0-9]$/' means that begining of the line has to be plus sign folowed by a number then end of line.

'/^\+[0-9]+$/' means that begining of the line has to be plus sign folowed by a one or more numbers then end of line.