CODEWITHSUNDEEP
php
sof_user
sof_user

Reputation: 709

how to pass a variable through the require() or include() function of php?

when I use this:

require("diggstyle_code.php?page=$page_no");

the warning is :failed to open stream: No error in C:\xampp\htdocs\4ajax\gallery_core.php on line 198

and the error is:

Failed opening required 'diggstyle_code.php?page=1' (include_path='.;C:\xampp\php\PEAR') in C:\xampp\htdocs\4ajax\gallery_core.php on line 198

value of the variable $page_no is collected beforehand.

But if I omit the '?page=$page_no part' from the argument of the require function, then no error or warning is shown.

I need to pass the variable when I use the require() function.

Upvotes: 55

Views: 84199

Answers (9)

Pascal MARTIN
Pascal MARTIN

Reputation: 401022

require() and include() will open the file corresponding to the path/name they receive.

Which means that, with your code, you would have to have a file called diggstyle_code.php?page=1 on your disk. That's obviously not the case, so it fails.

Quoting the Variable scope page of the PHP Manual:

The scope of a variable is the context within which it is defined. For the most part all PHP variables only have a single scope. This single scope spans included and required files as well.

In your case, you don't need to pass the variable. If you have a variable in your current script, it will also exist in the script you include, outside of functions, which have their own scope.

In your main script, you should have:

$page_no = 10;
require 'diggstyle_code.php';

And in diggstyle_code.php:

echo $page_no;
// Or work with $page_no the way you have to

Remember that including/requiring a file is exactly the same as copy-pasting its content at the line it's required.

Upvotes: 83

Marty
Marty

Reputation: 39456

Though this question is old there's another option that I use which is missing from this thread. You can return a function from the required file which accepts the arguments you want to pass along:

return function(array $something) {
    print_r($something);
}

And call it with the arguments when you require it:

require('file.php')(['some', 'data']);

// or:

$context = require('file.php');
$context(['some', 'data']);

Upvotes: 6

linepogl
linepogl

Reputation: 9335

require() does not make an HTTP call. All it does is open the file from disk and include the code in the position of the call. So simple local variables are enough.

Upvotes: 1

Tormy Van Cool
Tormy Van Cool

Reputation: 801

if I've correctly understood, what you need is to call the file diggstyle_code.php passing an argument, so that no one can call that file and make it work, rather than your main file. Am I right?

Thus supposing that your "main.php" has the lines

require("diggstyle_code.php?page=$page_no");

it means that: if anyone calls "main.php" gets diggstyle_code.php running. But if anybody in any manner calls directly diggstyle_code.php he/she shoudl get nothing.

If I am right on my understanding, a way to achieve this, is to include into the main file a variable or a constant, that will be scoped by diggstyle_code.php

Thus for instance: 'main.php'

<?php
define("_VERIFICATION_", "y");
require("diggstyle_code.php");
?>

and now diggstyle_code.php

<?php
if ( _VERIFICATION_ == "y" ) {
//Here the code should be executed
} else {
// Something else
}
?>

Upvotes: 2

user3420355
user3420355

Reputation: 1

I had this problem and I noticed if you use http:// in your url then it doesn't work

Upvotes: -4

Blagovest Buyukliev
Blagovest Buyukliev

Reputation: 43508

require doesn't pull the file from the web server - it should refer to a file on the filesystem instead.

Calling include or require just tells PHP to paste the contents of the given file in your code at this place, nothing more than that.

Upvotes: 3

stephane k.
stephane k.

Reputation: 1788

If your variable is global, there's no need to "pass"it, it is there already: PHP variable scope.

The answer then is, don't do anything, if $page_no exists in the file in which you call require(), it will be available in the included file.

Upvotes: 1

krtek
krtek

Reputation: 26597

require, require_once, include and include_once try to include files from the filesystem in the current file.

Since there's no files named diggstyle_code.php?page=1, it's totally logical that PHP can't find it.

You can't pass values that way, however, any variable declared in the current file will be accessible in the included files.

Upvotes: -1

knittl
knittl

Reputation: 265241

this should work, but it's quite a dirty hack:

$_GET['page'] = $page_no;
require('diggstyle_code.php');

you probably want to refactor your code to use functions and/or objects and call them from your files instead of including them (spaghetti code alert)

Upvotes: 8

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