Execute php file from another php

Question

I want to call a PHP file that starts like

<?php function connection () {    //Statements } 

I call from the PHP like this:

<?php exec ('/opt/lampp/htdocs/stuff/name.php'); ?> 

I get:

line1-> cannot open ?: No such file line 3 //Connection: not found line 4 Syntax errror: "(" 

Why doesn't this correctly execute the name.php file?

Answer 1

It's trying to run it as a shell script, which interprets your <?php token as bash, which is a syntax error. Just use include() or one of its friends:

For example, in a.php put:

<?php print "one"; include 'b.php'; print "three"; ?> 

In b.php put:

<?php print "two"; ?> 

Prints:

eric@dev ~ $ php a.php onetwothree 

Answer 2

exec is shelling to the operating system, and unless the OS has some special way of knowing how to execute a file, then it's going to default to treating it as a shell script or similar. In this case, it has no idea how to run your php file. If this script absolutely has to be executed from a shell, then either execute php passing the filename as a parameter, e.g

exec ('/usr/local/bin/php -f /opt/lampp/htdocs/.../name.php)') ; 

or use the punct at the top of your php script

#!/usr/local/bin/php <?php ... ?>