如何将PHP包含定义为字符串？

Question

我试过了:

$test = include 'test.php';

但那只是正常包含文件

Answer 1

您将需要查看输出缓冲功能.

//get anything that's in the output buffer, and empty the buffer
$oldContent = ob_get_clean();

//start buffering again
ob_start();

//include file, capturing output into the output buffer
include "test.php";

//get current output buffer (output from test.php)
$myContent = ob_get_clean();

//start output buffering again.
ob_start();

//put the old contents of the output buffer back
echo $oldContent;

编辑:

正如Jeremy指出的那样,输出缓冲区堆栈.所以理论上你可以做以下事情:

<?PHP
function return_output($file){
    ob_start();
    include $file;
    return ob_get_clean();
}
$content = return_output('some/file.php');

这应该等同于我更详细的原始解决方案.

但我没有费心去测试这个.

Answer 2

尝试类似的东西:

ob_start();
include('test.php');
$content = ob_get_clean();

Answer 3

试试file_get_contents().

此函数类似于file(),除了file_get_contents()以字符串形式返回文件.

Answer 4

解决方案#1:使用include(像函数一样工作):[我的最佳解决方案]

文件index.php:

<?php
$bar = 'BAR';
$php_file = include 'included.php';
print $php_file;
?>

文件included.php:

<?php
$foo = 'FOO';
return $foo.' '.$bar;
?>
<p>test HTML</p>

这将输出FOO BAR,但 注意:像函数一样工作,所以RETURN将内容传递回变量(<p>test HTML</p>将在上面丢失)

解决方案#2: op_buffer():

文件index.php:

<?php
$bar = 'BAR';
ob_start();
include 'included.php';

$test_file = ob_get_clean(); //note on ob_get_contents below
print $test_file;
?>

文件included.php:

<?php
$foo = 'FOO';
print $foo.' '.$bar;
?>
<p>test HTML</p>

如果你使用ob_get_contents()它会输出FOO BAR<p>test HTML</p> TWICE,请确保你使用ob_get_clean()

解决方案#3: file_get_contents():

文件index.php:

<?php
$bar = 'BAR';
$test_file = eval(file_get_contents('included.php'));

print $test_file;
?>

文件included.php:

$foo = 'FOO';
print $foo.' '.$bar;

这将输出FOO BAR,但注意:<?php当你通过eval()运行时,Include.php不应该有开始和结束标记