PHP If Else issue - always return blank

Question

Stupid question time.

This if statement always shows nothing.

$new = 'test = new user id';

if (stripos($new, 'test') !== FALSE) {
    $chk = explode("=",$new);
    $chk = $chk[1];
    echo $chk;
} else {
    $chk = '';
}

echo $chk;

When I echo $chk within the if else it shows the correct value, but the final echo is always blank.

Why?

Shouldn't $chk = "new user id" if the string contains that?

Also is there a neater more succinct way of writing this?

Thanks

Seems to perform fine for me: 3v4l.org/blkhW

Jim
– Jim

2014-03-28 10:26:11 +00:00
Commented Mar 28, 2014 at 10:26 — Jim
– Jim, Commented Mar 28, 2014 at 10:26
For me its getting fine

GautamD31
– GautamD31

2014-03-28 10:26:14 +00:00
Commented Mar 28, 2014 at 10:26 — GautamD31
– GautamD31, Commented Mar 28, 2014 at 10:26
Could you provide the whole code?

blue
– blue

2014-03-28 10:29:49 +00:00
Commented Mar 28, 2014 at 10:29 — blue
– blue, Commented Mar 28, 2014 at 10:29

Ananth · Accepted Answer · 2014-03-28 10:29:44Z

1

use this code. it works fine.

$new = 'test = new user id';

if (stripos($new, 'test') === FALSE) {
        $chk = '';

    } else {
          $chk = explode("=",$new);
        $chk = $chk[1];


    }

    echo $chk;

answered Mar 28, 2014 at 10:29

Ananth

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Comments

Elixir Techne · Accepted Answer · 2014-03-28 10:27:26Z

0

Initialize $chk before your condition statement.

$new = 'test = new user id';
$chk = '';
if (stripos($new, 'test') !== FALSE) {
        $chk = explode("=",$new);
        $chk = $chk[1];
        echo $chk;
    } else {
        $chk = '';
    }

    echo $chk;

answered Mar 28, 2014 at 10:27

Elixir Techne

1,85615 silver badges21 bronze badges

3 Comments

Rikesh Over a year ago

And how this relate's to the question asked by OP's ?

N.B. Over a year ago

Why do you think this answers the question at all? What does it fix? It initializes a variable, which isn't necessary at all.

blue Over a year ago

That is always a good idea, but it won't help here simply because he is changing $chk in both blocks...

Ramesh Dahiya · Accepted Answer · 2014-03-28 10:44:03Z

0

try this:

<?php
    $new = 'test = new user id';
    $start = strpos($new,'=')?strpos($new,'='):strlen($new);
    echo substr($new,$start +1);
?>

answered Mar 28, 2014 at 10:44

Ramesh Dahiya

1839 bronze badges

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