Using variables with Include() in php

Question

This code outputs paths

function genPost() {
// Gives the full path leading to this file starting at root.
// eg. /var/www/html
$path = dirname(__FILE__);

// Lists the folders and files of the chosen path.
// FYI the variable is now an array!!!
$pathContents = scandir($path);

function makePath($key) {
    $path = dirname(__FILE__);
    $folders = scandir($path);
    $two = $folders[$key];
    $three = $path . "/" . $two;
    echo $three . "<br>";
    //echo include("$three");
}
$key = 0;
// array_key_exists() returns false when the key to the array doesn't exist
while (array_key_exists($key, $pathContents)) {
    makePath($key);
    $key = $key + 1;
}
}
echo genPost();

However when I change the while to this, it outputs nothing to the browser.

while (array_key_exists($key, $pathContents)) {
    include "makePath($key)";
    $key = $key + 1;
}

My question is how do I avoid putting a directory in the include, and instead use a variable to tell the php interpreter where to pull the file from for the include function.

I'm trying to make a script for displaying posts on a blog. This is the start of an experiment to see how to dynamically manipulate the Include() function. — Christian, Apr 02 '20 at 05:58
I find your question to be Unclear. I cannot replicate whatever you are trying to describe because I don't see `getSum()` declared or called anywhere. Are you trying to `echo include`? — mickmackusa, Apr 02 '20 at 06:14
I updated the post. sorry for being unclear. Let me know if this helps. — Christian, Apr 02 '20 at 06:21
PHP does resolve variables inside double quoted strings - but not function calls. `include "makePath($key)";` needs to be `include makePath($key);` And that function needs to _return_ the path, not echo it. — CBroe, Apr 02 '20 at 06:25
This all WAAAAAAAAAY too much convolution. See the top answer in the duplicate. — mickmackusa, Apr 03 '20 at 12:42

score 1 · Answer 1 · answered Apr 02 '20 at 06:32

The issue that you are using include with string ( normally should be string that represent a path, not a function call) and this will cause wrong path error.

your makePath should be:

function makePath($key) {
    $path = dirname(__FILE__);
    $folders = scandir($path);
    $two = $folders[$key];
    $three = $path . "/" . $two;
    return $three;
}

and when you need to use it:

while (array_key_exists($key, $pathContents)) {
   include makePath($key);
   $key = $key + 1;
}

Oh my! This is it!! I didn't catch the typo in echo $three . "
"; Thank you for the advice on the return, as well!!! — Christian, Apr 02 '20 at 06:36

Using variables with Include() in php

1 Answers1