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Pass a variable from the source file to an included file in PHP

Answer 1

+5 A:

If you mean to use include() use include() not readfile() which is for outputting the actual contents of a file... readfile() won't parse PHP code at all, so there is no way to pass variables to it.

The reason you are seeing PHP code being executed is that your web server is actually making a second HTTP request (to itself?), and running a second process to execute your PHP. This second process will not have the same session, nor will it have access to any cookies, or other variables from the first process.

You could pass some variables along using the GET parameters, but if the file you want to include is on the same server, you should just use include(). If the file you want to include is on another server, you should stop and think about what you are doing because you are asking for trouble.

gnarf 2010-06-18 12:02:28

Clarification: If you are using readfile() to load a http:// url, then the resulting page will already be parsed with php (that is - it's a string containing html code)

Joel L 2010-06-18 12:04:07

I have tried using the include instead of read file, but the hosting services hates it and doesn't load the required file.Strangely the readfile() does load up the PHP code and executes it properly, unless the file is in the same location as the calling file. It is very confusing behaviour!

Alpha1 2010-06-18 12:13:26

@Alpha1 hosting services doesn't hate anything. It isn't human. It's a program. You have just experienced an error of some sort, and you have to repair it, not doing another way, backasswards one.

Col. Shrapnel 2010-06-18 12:28:34

Answer 2

A:

You need to pass parameters explicitly.

<?php
$title = 'Some Title';
$source_file = 'content.php';
readfile('http:...../format.php?title='.urlencode($title).'&source_file='.urlencode($source_file));
?>

.

<html> ...
<title>
<?php
echo $_GET['title'];
?>
</title>
...
<?php
include($_GET['source_file']); // Don't do that! 
                               // Never use include with argument that came from outside
?>
...

Please DON'T use include() with parameter that came from outside (as shown above). This is a gaping security hole.

Kamil Szot 2010-06-18 12:04:29

`s/potential/a gaping/`

gnarf 2010-06-18 12:36:02

Cool .. a web app that p0wns itself!

Tim Post 2010-06-18 13:17:12

You all are right. I stated danger of such solution too mildly. Now corrected.

Kamil Szot 2010-06-18 20:03:26

For future reference - this is a horrible "solution"

Joel L 2010-06-25 22:48:37

To downvoters: If you know how can I stress more dangers of this solution to future vieweres please let me know. I won't delete this answer because it adressess the question asked. However misguided the question itself is.

Kamil Szot 2010-06-26 00:40:21

I would still remove it. The question author is - for whatever reason - unable to use include() properly, and using a HTTP request is definitely *not* a solution.

Joel L 2010-06-26 13:52:38

I don't agree. I can imagine perfectly valid scenario of using readfile() to access some other page on different location. The trouble is just this include() with argument that came from outside.

Kamil Szot 2010-06-29 06:58:14

Quite strange answer, contains only bad code and full of warnings but not a sign of right solution. I don't think answers like `Shoot yourself in a leg. Don't do that! That's dangerous!!!` are ever sensible.

Col. Shrapnel 2010-06-29 07:04:31

Author of the question asks how he can make a hole in a coin that he has under his foot using a gun. I explain how he can do it in this arrangement, warn him that he will also get a hole in his foot during this process and advise him not to do it. If you want to give him better answer please explain to him step by step how to move his leg. I'll be the first to upvote it.

Kamil Szot 2010-06-29 12:23:56

I did not write this answer with evil intent. I just wanted to answer question within boundaries stated. I was little careless and I gave dangerous idea probably to someone who has no capacity yet to fully understand how dangerous it is and I'm sorry for that, but I see no point in removing the answer now. Let it serve as a warning to future viewers.

Kamil Szot 2010-06-29 12:24:57

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Pass a variable from the source file to an included file in PHP

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