How to get the second parameter out of URL (PHP)

Question

Can Someone help me.

I have a link like this:

http://www.somesite.com/somepage.php?&validation=3545&validation=1431

but i want to get the 1431 from the second validation as a variable. I tried to use this but it didn't work:

$Validationcode2 = $_GET['validation=3545&validation'];

But i think it doesn't works that way.

Parameter names should be unique. Do you have control over URL? — chris85
– chris85, Commented Dec 21, 2015 at 22:07
Do you mean this: http://www.somesite.com/somepage.php?validation=3545&validation=1431 (Removed extra ampersand) — 3ocene
– 3ocene, Commented Dec 21, 2015 at 22:09
Possible duplicate of How to get multiple parameters with same name from a URL in PHP — chris85
– chris85, Commented Dec 21, 2015 at 22:10
no, each validation should be unique (validation1=3545&validation2=1431) — Chaibi Alaa
– Chaibi Alaa, Commented Dec 21, 2015 at 22:10

Chris Trudeau · Accepted Answer · 2015-12-21 22:08:49Z

1

You'll want the GET parameters to be unique, so don't have two validation variables, assuming that was a mistake:

If you know the name you can reference it like:

$second_param = $_GET["validation"];

If you always want the second variable regardless of name you could do it like this:

$key_array = array_keys($_GET);
$second_param = $_GET[$key_array[1]];

answered Dec 21, 2015 at 22:08

Chris Trudeau

1,4373 gold badges16 silver badges20 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

3ocene · Accepted Answer · 2015-12-21 22:14:54Z

1

Use $_SERVER['QUERY_STRING'] to get the entire query string. Then you can apply a regex (try /[?&]validation=3545&validation=(\d*)/) to get the value you're looking for.

Regex demo: http://www.phpliveregex.com/p/e1L

$_SERVER documentation: http://php.net/manual/en/reserved.variables.server.php

edited Dec 21, 2015 at 22:14

answered Dec 21, 2015 at 22:07

3ocene

2,2311 gold badge19 silver badges31 bronze badges

Comments

jbrahy · Accepted Answer · 2015-12-22 00:01:43Z

0

It's definitely better to use the validation[] construct but the funny thing is that it actually works out to print the second parameter of &validation=1431

$url = "http://www.somesite.com/somepage.php?&validation=3545&validation=1431";
$parsed_url = parse_url($url);
$query = $parsed_url['query'];
parse_str($query,$parts);

print_r(array('parts'=>$parts));

this prints out

Array(
    [parts] => Array(
        [validation] => 1431
    )
)

answered Dec 22, 2015 at 0:01

jbrahy

4,4252 gold badges46 silver badges58 bronze badges

Collectives™ on Stack Overflow

How to get the second parameter out of URL (PHP)

3 Answers 3

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related