hankknight
asked on
How do I "exicute" a variable?
Hello.
How do I "exicute" a variable?
$code = "echo('Hello World');"
How do I "exicute" a variable?
$code = "echo('Hello World');"
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
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
ASKER
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');
echo('Hello World')
If $code = "include('file.php');" then I want this to be done:
include('file.php');
ASKER CERTIFIED SOLUTION
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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.
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.
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);
http://us2.php.net/manual/en/function.eval.php
$code = "echo('Hello World');"
eval($code);
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.
<?php
if($code == "echo('Hello World');")
eval($code);
if($code == "include('file.php');")
eval($code);
?>
Hope this helps.
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?
ob_start();
// execute code
$code = ob_get_clean();
Kind regards,
-r-