PHP无法将变量传递给包含的文件

Question

I tried to find a solution for this one:

file1.php

$name = "Jacob";
include ("file2.php");

file2.php

Hi there! <?php echo $name ?>

Output

Hi There! Notice: Undefined variable: name in /volume1/web/test/file2.php on line 9 

Need help please ://

Answer 1

Your code works fine.

If you get that error your file1.php certainly does not define $name in the global scope.

Fyi, quoted from the docs:

When a file is included, the code it contains inherits the variable scope of the line on which the include occurs. Any variables available at that line in the calling file will be available within the called file, from that point forward. However, all functions and classes defined in the included file have the global scope.

My testcase with php-5.3.3-pl1-gentoo(cli) in case it's a PHP bug in whatever version you are using - but I doubt that:

[thiefmaster@hades:~]> cat inc1.php
<?php
$name = 'bleh';
include('inc2.php');
?>
[thiefmaster@hades:~]> cat inc2.php
Hi there! <?php echo $name; ?>
[thiefmaster@hades:~]> php inc1.php
Hi there! bleh

Answer 2

Make sure your URL ends in file1.php and not file2.php or some other file that includes file2.php . Also, try reducing both files to exactly what you posted here (plus a <?php at the beginning of file1.php ) to rule out surrounding code.

Answer 3

I had the same problem and fixed it like this:

$name = "Jacob";
include ("file2.php");

file2.php

$name = $name; //add this line
Hi there! <?php echo $name ?>

I can't explain why it should work like this. Just it worked.

Answer 4

I sometimes get cuzzled (short for confuzzled) and forget to use the base file for checking a project online.

For instance, if using a base file called index.php with an include inc.php, make sure to launch index.php and not the include file.

The inc.php file will show errors in a browser because it's not linked to the index.php variables, whereas the index.php file contains both the variables and includes.

Answer 5

I use an environment variable instead. It gets around this problem.

Ie

$_ENV["name"] = "Jacob";

and inside file2.php

echo $_ENV["name"];

Answer 6

如果你想打印/使用$name变量，那么你必须首先在file2声明该变量，就像$name="" ，然后你可以轻松地使用它，而不需要声明，你将得到Undefined Variable错误。

Answer 7

Try to use this code

global $name;    
$name = "Jacob";    
include ("file2.php");

** and inside file2.php

global $name;
echo "Hi there! ".$name;

I think that will work