PHP - If variable is not set, get post variable

Question

I am trying to pass through a variable to a new name if it is set, or get a POST variable if the first variable is not set.

This is my code but it is not working:

if (isset($randombg)) {
    $background = $randombg;    
} else {  
   $background = $_POST['bg']) 
};

What is wrong with it, or how can I fix this?

Please, could you substitute `is not working` with what really happens? See you some messages or typos? — sensorario, Nov 19 '16 at 09:29

score 1 · Accepted Answer · answered Nov 19 '16 at 09:25

You have typos in your code. There is a ) in $background = $_POST['bg']) and a semicolon

you this should work:

$defaultbg = '#FFFFFF';

if (isset($randombg)) {
    $background = $randombg;    
} elseif( isset( $_POST['bg'] ) ) {  
   $background = $_POST['bg'];
   }else{
       $background = $defaultbg;
       }

score 1 · Answer 2 · edited May 23 '17 at 11:59

[Feature] PHP 7.0.x +

If you are using PHP 7.0.x+ you can use following syntax:

$background = $randombg ?? $_POST['bg'];

In this case $background will get value from first variable which is set and is not null.

You can do even something like this:

$background = $randombg ?? $_POST['bg'] ?? 'No background given'; 
// if $randombg and $_POST['bg'] will be not set or null $background will become 'No background given'

More about that feature you can read in PHP RFC: Null Coalesce Operator and PHP: New Features

About your code

You have syntax in your code. More informations below:

if (isset($randombg)) {
    $background = $randombg;    
} else {  
   $background = $_POST['bg']) // you have syntax here, delete )
}

You can also use short syntax:

$background = (isset($randombg)) ? $randombg : $_POST['bg'];

And it works as follows:

$background = (condition) ? 'when true' : 'when false';

More about it you can read here (Shorthand if/else)

Nice, didn't know about the shorthand for 7.x. I was trying to avoid using `isset()`. — yaakov, Sep 09 '18 at 04:09

score 0 · Answer 3 · answered Nov 19 '16 at 09:26

0

Mauybe $randombg is null or empty (that does not exactly means that is not set).

Try this:

if ($randombg) {
    $background = $randombg;    
} else {  
    $background = $_POST['bg'];
}

Or a shorten syntax:

$background = null == $randombg
    ? $_POST['bg']
    : $randombg;

answered Nov 19 '16 at 09:26

sensorario

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Please, don't do the second one. It's just too huge of a huge source of "why doesn't this code work" – Qrchack Nov 19 '16 at 09:40
1

Why not this sytax? It is way much better to read and much shorter if you write it in one line. – Karol Gasienica Nov 19 '16 at 09:57
Sometimes code does not work because developers cant read the error. – sensorario Nov 19 '16 at 10:01

PHP - If variable is not set, get post variable

3 Answers3