!!!!Variable Swap!!!!(not what is sounds like)

[IznSe7en]

Question, got a script here, you dont really need to know much about it because im focusing on something independent here. BTW thanks in advance to whoever replies.

Ok: ive got a script that posts to itself and sends an ID number. then

$variable = '$array[' . $fileid . ']';
so
$variable = '$array[1]';

Can i swap that so that the one on the right is the variable, i cant figure out how to get the name on the right to become a variable itself? Help?


-Thanks,
Izzy

[mrsmiley]

You need to use the $$ in your code. An example would be something like this.

$foo = "bar";
$bar = "testing";

echo $$foo would output the text "testing" to the screen. The $$ means to use the text in that variable as the name of another variable. I think the manual called it variable variables.

In your case (if I've interpreted it correctly), try this ...

<?php
$test[1] = 'cheese';
$var = '$test[1]';
$$var = 'testing';
echo $$var;
?>

It will output the text 'testing' rather than 'cheese'.

Is this what you are looking for?

[IznSe7en]

<?php
$test[1] = 'cheese';
$var = '$test[1]';
$$var = 'testing';
echo $$var;
?>

Thanks for the help, yet it doesnt really help. Look at the above code.
The second and third line are irrelevant. $$var will echo cheese because it is defined as a variable, after the first $ im assuming php just sees at as another variable. Thanks for your help though, i am going to look up variable variables, sounds familier.

[IznSe7en]

All i need to do is have a certain number that is submitted from the previous form via post. And have it pertain to an area in the array.

For example:

$somevar = $_POST['somevar'];

echo $somevar; //returns 4

echo $myarray[4];

Just gotta figure that one out, sigh....

[jumpfroggy]

Seems like you're looking for

echo $myarray[ $somevar ];

Assuming $somevar is 4, this will be equivalent to $myarray[ 4 ]. Is this what you were looking for?