PHP's文件路径组件函数返回错误的结果

您可以使用数组符号获取完整文件路径的一部分，如

echo $path_parts['dirname'], "'n";
echo $path_parts['basename'], "'n";
echo $path_parts['extension'], "'n";
echo $path_parts['filename'], "'n"; // since PHP 5.2.0

但是除了"basename"命名不正确(因为它返回整个文件名)，'basename'和'filename'版本只返回文件名的第一个字符。如果文件名是Archive.zip，我执行以下操作:

echo "Here's the zip filename: " . $zipFilename . "<br>";
echo "Here's the basename: " . $zipFilename['basename'] . "<br>";
echo "Here's the 'filename': " . $zipFilename['filename'];

:

Here's the zip filename: Archive.zip
Here's the basename: A
Here's the 'filename': A

如果我在完整路径上尝试，上载/Archive.zip，我得到:

Here's the zip filename: uploads/Archive.zip
Here's the basename: u
Here's the 'filename': u

知道为什么吗?谢谢。