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我正在尝试遍历一个目录并显示里面的所有图片。我让它工作，直到我将 Jquery/ajax 添加到等式中。不知何故，路径被复制了，所以我得到了这个错误（你可以看到它在哪里被复制）：

Fatal error: Uncaught exception 'UnexpectedValueException' with message 'DirectoryIterator::__construct(images/gallery/album1,images/gallery/album1) [<a href='directoryiterator.--construct'>directoryiterator.--construct</a>]: The system cannot find the path specified

我一生都无法弄清楚为什么会这样……contruct(images/gallery/album1,images/gallery/album1) 应该是 construct(images/gallery/album1)

PHP：

$album = $_POST['album'];
$dir = new DirectoryIterator("images/gallery/$album");
foreach ($dir as $fileInfo)
{
    if($fileInfo->isDot()) continue;
    $pic = $fileInfo->getFilename();
    print "<div>
            <img src='images/gallery/$album/$pic'>
            </div>";
}

jQuery/ajax：

function albumChosen(id)
{
var id = id;
var album = $('a[id="'+id+'"]').attr("rel");
$.ajax({
    url: "PHPscripts/getAlbums.php",
    type: "POST",
    data: {'album' : album},
    success: function(data){
        $('#galleryList').html(data);
    }
});
return false;

}

如果我alert(album);正确显示'album1'。所以它与它如何从$_POST

编辑：我添加了建议的代码..

$album = $_POST['album'];
print_r($_POST);
var_dump($album);
$dir = new DirectoryIterator("images/gallery/$album");
//rest of code...

print_r($_POST)打印出来 Array ( [album] => album1 )
var_dump($album)打印出来string 'album1' (length=6)

4

1 回答 1

1

结果我只需要添加../到路径的前面。小路。所以这是完整的代码，我再次添加的只是../路径前面的代码。

$album = $_POST['album'];
$dir = new DirectoryIterator("../images/gallery/$album");
foreach ($dir as $fileInfo)
{
    if($fileInfo->isDot()) continue;
    $pic = $fileInfo->getFilename();

    print "<div>
                <img src='images/gallery/$album/$pic'>
            </div>";
}
于 2013-02-10T16:33:17.000 回答