问题
When I use file_get_contents() function on a local file, the result contains php code, though I need HTML only.
Content of the file being read:
<?php echo '<p>Hello</p>';?>
And the result of file_get_contents called from a different file located in the same folder:
<?php echo file_get_contents('test.php'); //returns the following: string(31) "Hello'; ?>"
If I read a file from an external server, it returns HTML - as I would expect. So the question is: how do I get HTML output from the local file? Thank you all.
回答1:
You can use the files url (not filepath), so it is processed by the server eg:
echo file_get_contents('http://website.com/test.php');
However include/require would be better, eg:
include 'test.php';
回答2:
file_get_contents() only reads the plain file contents, as it's name suggests.
If you want the PHP interpreted, wou will have to include() it.
The according documentation can be found here: http://uk3.php.net/manual/en/function.include.php
回答3:
brainbowler alredy answered is pretty well. Additionally is suggest you to read the following docs:
- http://de2.php.net/manual/de/function.include.php
- http://de2.php.net/manual/de/function.require.php.
The PHP functions include and require are much alike and will do what you are trying to achive (they handle the include paths differently tough). There are also include_once and require_once which will also include and execute your script (and output the generated/contained HTML). But only once. They are useful if you haver your include statement in a loop or whatever and dont want it to load 42 times but only once.
Looks like you need to spend some more time on the basics (we all did). ;)
来源：https://stackoverflow.com/questions/24908777/file-get-contents-returns-php-code