从谷歌驱动全光照选取器下载的文件已损坏(files downloaded from google d

我用我的web应用程序谷歌选择器，允许用户从谷歌自己的驱动器浏览并选择文件。 一旦他作出选择，选择器返回关于选定的文件，包括文件名和URL的各种数据。 我的目标是将选定的文件下载到服务器。 如果我将URL传递到我的后端脚本，它下载文件，但他们都在这里破坏是我的代码：

回调函数：

function onPickerAction(data) {
    if (data.action === google.picker.Action.PICKED) {
        var id = data.docs[0].id;
        var request = new XMLHttpRequest();
        request.open('GET', 'https://www.googleapis.com/drive/v2/files/' + id);
        request.setRequestHeader('Authorization', 'Bearer ' + gapi.auth.getToken().access_token);
        request.addEventListener('load', function() {
            var item = JSON.parse(request.responseText);
            console.log(item);
            downloadFile(item);
        });
        request.send();
    }
}

这是将数据发送到我的PHP文件中的函数：

function downloadFile(item) {
    var request = new XMLHttpRequest();
    var mimeType = item['mimeType'];
    if (typeof item.exportLinks != 'undefined') {
        if (mimeType == 'application/vnd.google-apps.spreadsheet') {
            mimeType = 'application/vnd.openxmlformats-officedocument.spreadsheetml.sheet';
        }
        url = item['exportLinks'][mimeType];
        link = url;
    } else {
        lien = item['downloadUrl'].split("&");
        link = lien[0] + '&' + lien[1];
        url = item['downloadUrl'];
    }
    title = item['title'];
    type = mimeType;
    filesize = item['fileSize'];
    fileext = item['fileExtension'];
    id = item['id'];
    var datatable = [url, title, type, filesize, fileext,id];
    document.getElementById("myddlink").href=link; 
    request.open("POST", "downloadfile.php?" + datatable, true);
    request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
    request.send("datatable=" + datatable);
}

这是我的PHP函数下载文件：

        if (isset($_POST['exportFormat'])) {
        $pieces = explode(",", $_POST['exportFormat']);
        $url = $_POST['datatable'] . '&exportFormat=xlsx';
        $title = $pieces[1];
        $type = $pieces[2];
        $fileext = $pieces[0];
        $fileId = $pieces[5];
    }else {
        $url = $_POST['datatable'] . '&e=download';
        $pieces = explode(",", $_POST['gd']);
        $onlytitle = explode(".", $pieces[1]);
        $title = $onlytitle[0];
        $type = $pieces[2];
        $filesize = $pieces[3];
        $fileext = $pieces[4];
        $fileId = $pieces[5];
    }
    $fullPath = $upload_path . '/Myfiles1/' . $title . '.' . $fileext;
    header("Pragma: public");
    header("Expires: 0");
    header("Cache-Control: must-revalidate, post-check=0, pre-check=0");
    header("Cache-Control: public");
    header("Content-Description: File Transfer");
    header("Content-type: " . $type . "");
    header("Content-Disposition: attachment; filename=\"" . $title . '.' . $fileext . "\"");
    header("Content-Transfer-Encoding: binary");
    header("Content-Length: " . $filesize);
    // folder to save downloaded files to. must end with slash
    $destination_folder = $upload_path . '/Myfiles1/';
    $newfname = $destination_folder . basename($title . '.' . $fileext);
    $file = fopen($url, "rb");
    if ($file) {
        $newf = fopen($newfname, "wb");
        if ($newf)
            while (!feof($file)) {
                fwrite($newf, fread($file, 1024 * 8), 1024 * 8);
            }
    }
    if ($file) {
        fclose($file);
    }
    if ($newf) {
        fclose($newf);
    }
    ob_end_flush();

任何想法，请我如何可以下载文件，而不会破坏他们？

文章来源: files downloaded from google drive usin picker are corrupted