PHP include does not work

Question

I'm working with two files - a.php which contains the class Background, and b.php which includes a.php . *In a.php , out of the class scope , there's a echo statement " background check" .

When I load b.php , I can see the output "background check" , but when I try to create a background object the next warning message appears :

Fatal error: Class 'Background' not found in ....

Here's a code sample from b.php :

<?php
        include ('http://localhost/wT/sf/a.php');
        $url2="http://www.google.com";
        $b = new Background($url2);
?>

It is if the URL returns only source code. It's bad practice, but it can be done. — WWW
– WWW, Commented Jul 24, 2012 at 13:37

WWW · Accepted Answer · 2012-07-24 13:36:13Z

4

If the output of http://localhost/wT/sf/a.php (in a web browser) isn't PHP source code, then your include will not work. You probably (in your case) have to include the file from the filesystem, not through an HTTP address.

answered Jul 24, 2012 at 13:36

WWW

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3 Comments

Itamar Over a year ago

The include function returns the next error : failed to open stream: No such file or directory

cHao Over a year ago

Might want to make sure the document root is listed in include_path. Then, just include stuff relative to it.

WWW Over a year ago

Or just use the file's absolute path to include it.

Naftali · Accepted Answer · 2012-07-24 13:36:25Z

3

Use relative routes for the includes, not a URL.

In other words, the file that you are including has to be on the same server (or able to be accessed on the same server) as the file that you are including it in.

answered Jul 24, 2012 at 13:36

Naftali

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2 Comments

Itamar Over a year ago

The include works though , becuase I can see the output of the echo statement from a.php ...the problem is that I can not create the object I've created in a.php

Ron Over a year ago

Loading php-executable content by url may work, but it is wrong by concept

Kalpesh · Accepted Answer · 2012-07-24 13:38:18Z

1

Try something like this to make this work:

<?php
        include ('/var/www/wT/sf/a.php'); //your filesystem location to a.php
        $url2="http://www.google.com";
        $b = new Background($url2);
?>

answered Jul 24, 2012 at 13:38

Kalpesh

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3 Comments

Itamar Over a year ago

Include using the filesystem location does not work : here's the file location : wT/sf/a.php

Kalpesh Over a year ago

please post content of a.php to better help you

Kalpesh Over a year ago

the file location can't be wT/sf/a.php, there should be something liek /var/www/.. what is your server's root directory?

frangly · Accepted Answer · 2012-07-24 13:41:31Z

0

In include method you should use relative file path

Files are included based on the file path given or, if none is given, the include_path specified. If the file isn't found in the include_path, include will finally check in the calling script's own directory and the current working directory before failing. The include construct will emit a warning if it cannot find a file

answered Jul 24, 2012 at 13:41

frangly

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Comments

Naftali · Accepted Answer · 2012-07-24 13:51:08Z

0

You are using web-url instead of use system path like

<?php
        include ($_SERVER['document_root'].'/a.php');
        $url2="http://www.google.com";
        $b = new Background($url2);
?>

edited Jul 24, 2012 at 13:51

Naftali

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answered Jul 24, 2012 at 13:38

Viral Solani

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1 Comment

Ron Over a year ago

better include dirname(__FILE__).'/a.php';

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