PHP Variables.

August 2nd, 2005, 11:01 am

I'm in an odd situation. I have a seperate php script included into the main INDEX php file.

This has a list of variables with values assigned to them. When i submit a form a variable will need to be made to that value thats been posted. So then when it forwards to another website it forwards to the variable thats identical to.

e.g)

variables.php

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<?
$var1 = "google.com";
$var2 = "yahoo.com":
?>

1. <?
2. $var1 = "google.com";
3. $var2 = "yahoo.com":
4. ?>

index.php

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include "variables.php";
if(isset($_POST['submit'])) {


$name = $_POST['name'];
//lets say $name is "var1"


$server = "$".$name;

header("Location: $server");
exit();

}

1. include "variables.php";
2. if(isset($_POST['submit'])) {
5. $name = $_POST['name'];
6. //lets say $name is "var1"
9. $server = "$".$name;
11. header("Location: $server");
12. exit();
14. }

Instead of the script forwarding to 'google.com' it tries to go to '$var1'.


What am i missing? Something im not doing right?

August 2nd, 2005, 11:11 am

You're setting $server to the string $var1. You need to use:

PHP Code: [ Download ] [ Select ]

$server = $$name;

instead

August 2nd, 2005, 11:29 am

I get:

Quote:

Parse error: parse error, unexpected T_VARIABLE in /home/~~/public_html/~~/index.php on line 27

... with that.

August 2nd, 2005, 11:41 am

Sorry, but you were right. It worked.

August 2nd, 2005, 11:47 am

Sorry for doing three posts in a row.

What if the situation was that i need to make a variable like:

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$nameis_var1 = "google.com";
$nameis_var2 = "yahoo.com";

1. $nameis_var1 = "google.com";
2. $nameis_var2 = "yahoo.com";

and it was like:

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$name = $_POST['name'];
//lets say $name is "var1"


$server = $nameis_$name;

1. $name = $_POST['name'];
2. //lets say $name is "var1"
5. $server = $nameis_$name;

$server variable would not work at all. I get an error as said in third post in this topic.