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calling php file fron another php file while passing arguments

Posted by: admin July 12, 2020 Leave a comment

Questions:

i need to call a php inside another php file and pass some arguments also. how can i do this??
i tried

include("http://.../myfile.php?file=$name");
  • but gives access denied. i read like v must not set allow_url_open to OFF.

if i write like

$cmd = "/.../myfile.php?file=".$name";
$out =exec($cmd. " 2>&1");
echo $out;
  • gives error as /…/myfiles.php?file=hello: no such file or directory.

how can i solve this???

How to&Answers:

You don’t have to pass anything in to your included files, your variables from the calling document will be available by default;

File1.php

<?php

$variable = "Woot!";
include "File2.php"; //if in the same folder

File2.php

<?php
echo $variable;

Answer:

the location in your code is incorrect:

$cmd = "/.../myfile.php?file=".$name;

Answer:

you can include file over http only if the allow_url_fopen is set to TRUE, and the same parameter allow you to pass variables to the files…

Answer:

You should not include files via HTTP connection – that’s almost always a serious security problem.

If you must do this, you have to set allow_url_include and allow_url_fopen to ON but neither is a recommended procedure.