PHP dynamic variable

Question

I'm working in PHP. I have a dynamic number of arrays called 'placeholder0, placeholder1, placeholder2....' and I want to json_encode them

$encode = json_encode(array('tabs' => $tabs, 'placeholder0' => $placeholder0, 'placeholder1' => $placeholder1, 'placeholder2' => $placeholder2 ..... ));

The number of values inside the 'tabs' array is the same as the number of 'placeholder' arrays.

How can I do this?

Mihai Matei · Accepted Answer · 2016-03-29 17:26:22Z

1

$result = array('tabs' => $tabs);

for ($i = 0; $i < count($tabs); $i++) {
    $result['placeholder' . $i] = ${'placeholder' . $i};
}

file_put_contents($jsonLocation, json_encode($result));

edited Mar 29, 2016 at 17:26

answered Mar 29, 2016 at 16:47

Mihai Matei

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4 Comments

Mihai Matei Over a year ago

:)) sorry.. post it and I will delete mine :))

Hanky Panky Over a year ago

Nope this answer is great!

Mark Titorenkov Over a year ago

when I try file_put_contents($jsonLocation, var_dump(json_encode($result)) ); it only comes up on the screean and doesn't get outputed in the file

Mihai Matei Over a year ago

I've updated the answer.. var_dump is just a debugging function which helps you to see the type and value of a variable..

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