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How do I "exicute" a variable?

Posted on 2007-11-14
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Last Modified: 2009-12-16
Hello.

How do I "exicute" a variable?

$code = "echo('Hello World');"

0
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Question by:hankknight
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9 Comments
 
LVL 49

Expert Comment

by:Roonaan
ID: 20280877
hankknight,

ob_start();
// execute code

$code = ob_get_clean();


Kind regards,

-r-
0
 
LVL 17

Expert Comment

by:nplib
ID: 20280945
define execute,

cause you can't quite execute a variable,

variables store information,

you can execute a function

function myFirstFunction() {
     $var = "Hello World";
     echo $var;
}

myFirstFunction();
///this will output hello world to the screen

or

function myFirstFunction() {
     $var = "Hello World";
     return $var;
}

$var = myFirstFunction();

////This will return the variable in the function to your global variable.

Although this a simple demonstration, if there were a series of complex instructions in the function the output is all that would be returned
0
 
LVL 16

Author Comment

by:hankknight
ID: 20280980
If $code = "echo('Hello World');" then I want this to be done:
echo('Hello World')

If $code = "include('file.php');" then I want this to be done:
include('file.php');
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LVL 27

Expert Comment

by:Cornelia Yoder
ID: 20280989
$code = "echo('Hello World');"
exec($code);

http://us2.php.net/manual/en/function.exec.php
0
 
LVL 49

Accepted Solution

by:
Roonaan earned 2000 total points
ID: 20281086
you would use eval(). Exec is used to do system calls.

-r-
0
 
LVL 17

Expert Comment

by:nplib
ID: 20281100
exec() is for executing external commands that exist on the server.

Why do you want it to work that way, what are you trying to achieve?

I don't think it can work that way.
$variables are either
Integer,
String,
Boolen,
Arrays,
or Objects.

functions are executables
create functions like

function myEcho ($var) {
     echo $var;
}

then all you do is type
myEcho("hello World");
 and it will print it to the screen,

or create and include function,

function myInclude($var) {
    include($var);
}

then all you do is type,

myInclude("page.php");

and it will include it,

but these are all predefined executables in PHP,

So What are you trying to achieve and don't just say, trying to execute a variable, because you can't.
0
 
LVL 27

Expert Comment

by:Cornelia Yoder
ID: 20282641
They're exactly right, my bad, sorry ... eval() is the one you want, not exec().

http://us2.php.net/manual/en/function.eval.php

$code = "echo('Hello World');"
eval($code);
0
 
LVL 21

Expert Comment

by:nizsmo
ID: 20282822
so what about the if statement? something like this?

<?php
if($code == "echo('Hello World');")
    eval($code);

if($code == "include('file.php');")
    eval($code);
?>

Hope this helps.
0
 
LVL 27

Expert Comment

by:Cornelia Yoder
ID: 20283308
nizsmo, I don't quite understand why you want if statements.  eval() will execute whatever statement is in $code, so it shouldn't be necessary to check what that is.  Am I missing something?
0

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