Ziparchive extracto函数将第二个参数视为文件夹

Question

I tried to extract a zip file using ziparchive, here's my code:

$path = "/test/admin";
$zip = new ZipArchive;
if ($zip->open("test.zip") === TRUE) {
    $zip->extractTo($path."/", array("index.php"));
    $zip->close();
}

but the output becomes /test/admin/index.php/. The file becomes a folder.

Can someone help me with this.