PHP using a variable in a variable name

Question

I have two preexisting variables:

$side (which is either F or B)
and
$plate (which is a 3 digit number)

I am trying to build two arrays called

$sidecountF
and $sidecountB

using the following code:

$sidecount{$side}[$plate] = 1;

assuming $side is F and $plate is 200, I'm hoping that the end result is that

$sidecountF[200] = 1;

I am, at the beginning, declaring sidecountF and sidecountB as arrays with

$sidecountF = array();
$sidecountB = array();

So now I'm stumped.

What exactly are you asking? Your code appears to be correct. — Ben
– Ben, Commented Aug 19, 2010 at 14:39

Artefacto · Accepted Answer · 2010-08-19 14:42:08Z

11

${"sidecount$side"} = array();

But you're better off using arrays:

$sidecount = array("F" = array(), "B" => array());
$sidecount[$side][$plate] = /* ... */

answered Aug 19, 2010 at 14:42

Artefacto

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1 Comment

Agreed with array using! Gives much more flexibility in the end... At least it always happens with my scripts.

RobertPitt · Accepted Answer · 2010-08-19 14:47:45Z

0

$_blank = array(
    'sidecount' . $side => array()
);

extract($_blank);

this would be another way of doing it, it also is not bound to creating 1 variable with ${""} you can create several variables at once.

answered Aug 19, 2010 at 14:47

RobertPitt

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