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???
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Answer
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;