PHP - 上传到服务器后获取文件的文件路径

美好的一天。我正在将图像文件上传到我的 php 服务器。我已成功将文件上传到文件夹。我现在要做的是在组装 JSON 响应时，我想获取文件的新文件路径，以便我知道它以编程方式在哪里。

这是我到目前为止的 php 代码：

<?php
header('Content-Type: text/html; charset=utf-8');
    if (
        !isset($_FILES['fileToUpload']['error']) ||
        is_array($_FILES['fileToUpload']['error'])
    ) {
        die(json_encode(array(
            'success' => false,
            'status' => "Invalid Parameters. - 1",
            'files' => $_FILES 
        )));
    }
    // Check $_FILES['fileToUpload']['error'] value.
    switch ($_FILES['fileToUpload']['error']) {
        case UPLOAD_ERR_OK:
            break;
        case UPLOAD_ERR_NO_FILE:
            die(json_encode(array(
                'success' => false,
                'status' => "No file sent. - 2",
                'files' => $_FILES
            )));
        case UPLOAD_ERR_INI_SIZE:
        case UPLOAD_ERR_FORM_SIZE:
            die(json_encode(array(
                'success' => false,
                'status' => "Exceeded filesize limit. - 3",
                'files' => $_FILES
            )));
        default:
            die(json_encode(array(
                'success' => false,
                'status' => "Unknown errors. - 4",
                'files' => $_FILES
            )));
    }
    // You should also check filesize here. 
    if ($_FILES['fileToUpload']['size'] > 1000000) {
            die(json_encode(array(
                'success' => false,
                'status' => "Exceeded File Size Limit. - 5",
                'files' => $_FILES
            )));
    }
    // DO NOT TRUST $_FILES['fileToUpload']['mime'] VALUE !!
    // Check MIME Type by yourself.
    $finfo = new finfo(FILEINFO_MIME_TYPE); 
    if (false === $ext = array_search(
        $finfo->file($_FILES['fileToUpload']['tmp_name']),
        array(
            'jpg' => 'image/jpeg',
            'png' => 'image/png',
            'gif' => 'image/gif',
        ),
        true
    )) {
        die(json_encode(array(
            'success' => false,
            'status' => "Invalid file format. - 6",
            'files' => $_FILES
        )));
    }
    // You should name it uniquely.
    // DO NOT USE $_FILES['fileToUpload']['name'] WITHOUT ANY VALIDATION !!
    // On this example, obtain safe unique name from its binary data.
    if (!move_uploaded_file(
        $_FILES['fileToUpload']['tmp_name'],
        sprintf('./gallery/%s.%s',
            sha1_file($_FILES['fileToUpload']['tmp_name']),
            $ext
        )
    )) {
        die(json_encode(array(
            'success' => false,
            'status' => "Failed to move uploaded file. - 7",
            'files' => $_FILES
        )));
    }
        $filePathFull = sprintf('./gallery/%s.%s',
            sha1_file($_FILES['fileToUpload']['tmp_name']);
        $uploaddir = '/gallery/';
        $uploadfile = $uploaddir . basename($_FILES['fileToUpload']['tmp_name']);

        die(json_encode(array(
            'success' => true,
            'status' => "File is uploaded successfully. - 8",
            'files' => $_FILES,
            'filePath' => $uploadfile
        )));
?>

我正在尝试获取/gallery/ +文件名的新文件路径，但是，当我在 ExtJS 部分中处理它时，我得到了一个无效的 JSON 对象。

谁能帮我解决问题？我也尝试了sprintf('./gallery/%s.%s', sha1_file($_FILES['fileToUpload']['tmp_name']);但这似乎不起作用。

任何帮助都非常感谢。

编辑：

到目前为止，这是有效的：

$uploaddir = '/gallery/';
$uploadFile = $uploaddir . basename($_FILES['fileToUpload']['tmp_name']);
die(json_encode(array(
    'success' => true,
    'status' => "File is uploaded successfully. - 8",
    'filePath' => $uploadFile
)));

但是，我没有文件扩展名，我得到类似的东西：/gallery/phptnLZJm但是在检查时，实际上上传了图像，但文件名已关闭。