Concatenate strings for variable identifier in PHP

Question

I know this is not the proper way to do this, however I am trying to put a quick fix on a form that was done by another developer. Basically I want to add an incremental number to a variable inside a while statement:

$count = 1;
while ($r = mysql_fetch_array($query)) {
    $variable . $count = $r['somefield'];
        $count++
}

So that makes the variables:

$variable1 $variable2 $variable3 ....etc

Basically any time you need variable variables, you're doing something wrong. Use arrays. That's what they're for. — Mike
– Mike, Commented Jun 11, 2012 at 21:16
Thanks Mike. As I stated.. I know it is the completely wrong way to do it. I just needed a quick fix for a form that another developer had done with some 400 fields and NO ARRAYS!! For this case, variable variables saved me some 4 or 5 hours from having to rewrite the entire structure of the form. — Jeff Thomas
– Jeff Thomas, Commented Jun 11, 2012 at 21:20

nothrow · Accepted Answer · 2012-06-11 21:14:57Z

4

$varname = 'variable' . $count;
$$varname = $r['somefield'];

http://www.php.net/manual/en/language.variables.variable.php

answered Jun 11, 2012 at 21:14

nothrow

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2 Comments

Jeff Thomas Over a year ago

This solved my dilemma! Thank You! I hate it when I open up a script that was written years ago by an inexperienced developer.

nothrow Over a year ago

@Prince, I can really feel your pain. :-)

Niet the Dark Absol · Accepted Answer · 2012-06-11 21:16:29Z

1

You'd be better off with an array...

$variable[] = $r['somefield'];

You can use variable variables, however it is probably not a good idea, especially for a trivial case like this one.

answered Jun 11, 2012 at 21:16

Niet the Dark Absol

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