Dynamically setting a variable in PHP?

Question

So i want to deo something like this and not sure how

    for($s=0; $s < 5; $s++ ){
      $pre_config_query = "select * from preconfig where code = '{$industry_string}_{$s}_{$class_string}'";
      $pre_config_station = mysql_query($pre_config_query);
      $it_exists = mysql_num_rows($pre_config_station);
      if($it_exists>0){
        $pre_config = mysql_fetch_assoc($pre_config_station);
        $pre{$s} = $pre_config['id'];

I want the end product to have these 5 variables named

    print $pre1;
    print $pre2;
    print $pre3;
    print $pre4;
    print $pre5;

That have the $pre_config['id'] if present....any ideas

Why don't you just use an array or an associative array? Infact, it would probably make easier whatever you're planning. — user350814
– user350814, Commented Nov 6, 2010 at 15:56

Lekensteyn · Accepted Answer · 2010-11-06 15:52:55Z

2

You can use variable variables to accomplish that.

First, define a variable with the desired name:

$varname = "pre$s";

Second, assign a value to it:

$$varname = $pre_config['id'];

That's all!

answered Nov 6, 2010 at 15:52

Lekensteyn

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Mark · Accepted Answer · 2010-11-06 16:03:23Z

1

this works but I'm not sure I'm answering your question.

<?php
for($s=1; $s < 6; $s++ ){
    $it_exists=1;
      if($it_exists > 0){
        $pre_config = array('id'=>rand(10,99));
        ${"pre".$s} = $pre_config['id'];
      }
}
echo $pre1."<br/>";
echo $pre2."<br/>";
echo $pre3."<br/>";
echo $pre4."<br/>";
echo $pre5."<br/>";

?>

answered Nov 6, 2010 at 16:03

Mark

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