- Original Message -
From: "Shawn McKenzie" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, July 14, 2003 1:51 PM
Subject: Re: [PHP] Re: Eval var from query
> Thanks Kevin! That works great. It outputs: hi my name is Shawn
>
> Now if I want
Got it!
eval( '$newdata = "'.$data.'";');
Thanks!
Shawn
"Shawn McKenzie" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Thanks Kevin! That works great. It outputs: hi my name is Shawn
>
> Now if I want to assign $data to another var, let's say $newdata and have
it
> eval the $n
Thanks Kevin! That works great. It outputs: hi my name is Shawn
Now if I want to assign $data to another var, let's say $newdata and have it
eval the $name var inside of that. How would that work?
Meaning I want to $newdata = hi my name is Shawn
Thanks!
Shawn
"Kevin Stone" <[EMAIL PROTECTED
;";';
$name = 'Shawn';
eval($code); // prints "hi may name is Shawn".
Hope that makes it more clear.
- Kevin
- Original Message -
From: "Kevin Stone" <[EMAIL PROTECTED]>
To: "PHP-GENERAL" <[EMAIL PROTECTED]>
Sent: Monday, J
The string you send to eval() must be valid PHP code. So try this..
eval( 'echo "'.$data.'";');
- Kevin
- Original Message -
From: "Shawn McKenzie" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, July 14, 2003 1:15 PM
Subject: [PHP] Re: Eval var from query
> eval($data)
>
>
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