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Home/ Questions/Q 1089009
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T23:09:20+00:00 2026-05-16T23:09:20+00:00

Trying to create a variable with unique name for each $item . To prevent

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Trying to create a variable with unique name for each $item.

To prevent error “Only variables can be passed by reference”.


If there are 5 items in array $items, we should get 5 unique variables:

$item_name_1;
$item_name_2;
$item_name_3;
$item_name_4;
$item_name_5;

All of them should be empty.

What is a true solution for this?

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  1. Editorial Team
    Editorial Team
    2026-05-16T23:09:21+00:00Added an answer on May 16, 2026 at 11:09 pm

    You can dynamically create variable names doing the following:

    $item_name_{$count} = $whatever;
    

    I must warn you though, this is absolutely bad style and I’ve never seen a good reason to use this. In almost every use case an array would be the better solution.

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