PHP How can I pass values from one array to another?

Question

I'm trying to pass some of the values from theOptions array and drop them into a new array called $theDefaults.

$theOptions = array(

    'item1' => array('title'=>'Title 1','attribute'=>'Attribute 1','thing'=>'Thing 1'),
    'item2' => array('title'=>'Title 2','attribute'=>'Attribute 2','thing'=>'Thing 2'),
    'item3' => array('title'=>'Title 3','attribute'=>'Attribute 3','thing'=>'Thing 3')

);

So, $theDefaults array should look like this:

$theDefaults = array(

    'Title 1' => 'Attribute 1',
    'Title 2' => 'Attribute 2',
    'Title 3' => 'Attribute 3'

);

However, I cannot figure out how to do this. Have tried this but it is clearly not quite working.

$theDefaults = array();

foreach($theOptions as $k=>$v) {
    array_push($theDefaults, $v['title'], $v['attribute']); 
}

but when I run this...

foreach($theDefaults as $k=>$v) {
    echo $k .' :'.$v;
}

It returns this. 0 :Title 11 :Attribute 12 :Title 23 :Attribute 24 :Title 35 :Attribute 3

Looks to be soooo close, but why are the numbers in the array?

Explosion Pills · Accepted Answer · 2013-03-29 04:22:26Z

6

It's even simpler than that:

$theDefaults = array();
foreach($theOptions as $v) {
    $theDefaults[$v['title']] = $v['attribute']; 
}

answered Mar 29, 2013 at 4:22

Explosion Pills

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3 Comments

castis Over a year ago

Noting that the issue is that array_push pushes all arguments after the first to the end of the array.

Hanky Panky Over a year ago

Oooops, beat me to it by few seconds. :)

Starfs Over a year ago

wow. that was fast. Thanks! it worked. apparently i can declare your answer correct in 8 minutes.

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