Solved

variable named variable ($$)

Posted on 2004-09-14
8
210 Views
Last Modified: 2006-11-17
here is what i am trying to do, which i thought i have done before but it does not seem to work.

$Foo= 'Bar';

$$Foo = '1';

if ($Bar == '1'){
//DO SOMETHING
} elseif ($OtherName == '1'){
//DO SOMETHING ELSE
}

basically i need to be able to set the name of a variable, then later on give it a value and test for the value. i will need the variables name to be able to be set to anything i want, because i will have a bunch of different if's at the end.

thank you
0
Comment
Question by:qwertq
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8 Comments
 
LVL 14

Expert Comment

by:dfu23
ID: 12055917
why not use an associative array/hash
0
 

Author Comment

by:qwertq
ID: 12055951
could you be more specific?
0
 
LVL 33

Expert Comment

by:snoyes_jw
ID: 12056420
Your code seems to function as written.  Try
echo $Bar;
to see if it got set correctly.
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LVL 14

Accepted Solution

by:
dfu23 earned 50 total points
ID: 12057068
http://us2.php.net/types.array

$Foo = array("Bar" => "1", "OtherName" => 1);

if ($Foo["Bar"] == "1") {
    // Do Something
}
elseif ($Foo["OtherName"] == "1") {
    // Do Something Else
}
0
 
LVL 8

Expert Comment

by:hendridm
ID: 12057559
Your code seems to work for me in 4.3.8:
<?
$foo = "bar";
$$foo = 1;

echo $bar; //Displays '1'
?>

Not sure which PHP you're using, but maybe this would work for you:
<?
$foo = "bar";
eval("\$$foo = 1;");

echo $bar; //Displays '1'
?>
0
 

Expert Comment

by:mary_BE
ID: 12082169
You might wanna try this :
${$Foo} = '1';
0

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