Regular expression does not work properly - php

Question

elseif(!eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$", $user_email))

Even i tried with preg_match in this too i am getting

Warning:

preg_match() [function.preg-match]: No ending delimiter '^' found in C:\wamp\www\Saravanan\jQuery\jQuery - Samples\sign_and_login_using_jquery_ajax_and_php\sign_and_login_using_jquery_ajax_and_php\vpb_save_details.php on line 29
Call Stack

I believe the error message is pretty clear. As is [the manual](http://php.net/manual/en/function.preg-match.php). — PeeHaa, Apr 09 '13 at 07:24
Error messages try hard not to lie. And you're using an [obsolete function](http://php.net/eregi) — Touki, Apr 09 '13 at 07:24
And even if you didn't get that error, that still wouldn't be the way to [validate an emailaddress](http://stackoverflow.com/questions/12026842/how-to-validate-an-email-address-in-php/12026863#12026863) — PeeHaa, Apr 09 '13 at 07:26
POSIX regex are deprecated in PHP 5.. so don't use ereg,eregi etc.. also if ur using preg_match.. enclosed it in delimmeter "/regex/" as in comment by @elclanrs .. — Dinesh, Apr 09 '13 at 07:31

score 2 · Answer 1 · answered Apr 09 '13 at 07:25

2

Use this one:

if(!filter_var($user_email,FILTER_VALIDATE_EMAIL))

answered Apr 09 '13 at 07:25

Rashad

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+1 I wonder why people don't use this more often. – Anirudh Ramanathan Apr 09 '13 at 07:27
Thanks Rashad Aghayev, now its working grate! Chears!!!!! – user2219660 Apr 09 '13 at 08:24

score 1 · Answer 2 · answered Apr 09 '13 at 07:27

1

and if the var comes from POST:

if (!filter_input(INPUT_POST, 'email', FILTER_VALIDATE_EMAIL))

answered Apr 09 '13 at 07:27

MGE

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It should be mentioned that there's a slight difference between failing the filter and the variable not being posted though. – Ja͢ck Apr 09 '13 at 07:29

Regular expression does not work properly - php

2 Answers2