$x=include path?

$x="include('home/adulu/public_html/123.php');";

echo "$x";(i'd like it print 123.php scripts)

but it just print text ->"include('home/adulu/public_html/123.php');"

please help. thanks

 

why of course.. it should print what is in your variable $x which is the string "include('home/adulu/public_html/123.php');" set your $x to '123.php' and you'll get what you want... what really are you trying to achieve here anyway?

 

<?php 
echo str_replace('/','', $_SERVER['REQUEST_URI']); // 123.php
?>

PHP:

 

example:123.php's content is <?php $y=(1+2); echo "$y"; ?>

I hope i can
$x="include('home/adulu/public_html/123.php');";
echo "$x";
rusult print is 3

but actually is print include('home/adulu/public_html/123.php'); this is not my wish.

 

Yes.. What do you want to achieve here...? If you want to execute PHP code inside a string, use "eval" function ..
eval($x);

 

my achieve
because i'd like to determine include or not include
$a=yes;
if($a=="yes")
{
$x="include('home/adulu/public_html/123.php');";(this is wrong code that is asume)
}
else
{
$x="nothing";
}
------------content------i am only can modify above code area
echo "$x";
(I hope this [echo "$x";] can include 123.php file)

 

<?php

ob_start();
include('fileYouWantToStore.php');
$a = ob_get_contents();
ob_end_clean();

echo $a;

PHP:

 

Thanks so much, this is what i wish.

 

This is less code to do what you wanted:
( file_get_Contents returns as a string, returns as FALSE on failure )

<?php

$file = file_get_contents('filepath/file.php');
echo $file;

PHP:

 

This wouldn't execute the code, just saves it.