警告:fopen()[function.fopen]:文件名不能为空

Question

我使用本教程http://papermashup.com/caching-dynamic-php-pages-easily/来缓存页面

<?php {
$cachefile = $_SERVER['DOCUMENT_ROOT'].'cache.html';
$cachetime = 4 * 60;
// Serve from the cache if it is younger than $cachetime
if (file_exists($cachefile) && time() - $cachetime < filemtime($cachefile)) {
    include($cachefile);
} else {
ob_start(); // Start the output buffer
 ?>

/* Heres where you put your page content */


<?php 
// Cache the contents to a file
$cached = fopen($cacheFile, 'w');
fwrite($cached, ob_get_contents());
fclose($cached);
ob_end_flush(); // Send the output to the browser
}
?>

但我得到以下错误

Warning: fopen() [function.fopen]: Filename cannot be empty in

Warning: fwrite(): supplied argument is not a valid stream resource in

Warning: fclose(): supplied argument is not a valid stream resource in

文件的路径是正确的.如果我编辑文件我的自我包括但我再次得到错误

Answer 1

您的变量名称的大小写有问题.PHP变量名称区分大小写.更改cacheFile到cachefile(与小˚F代替).

改变这个:

$cached = fopen($cacheFile, 'w');

对此:

$cached = fopen($cachefile, 'w');