使用 javascript 下载 Google Drive 文件

Question

当我单击 Google Drive 选择器但无法下载文件时，我想使用 javascript 将 Google Drive 文件下载到我的服务器。 我一直在寻找 4 天，但问题是相同的，下面是我正在使用的代码。

 function pickerCallback(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(); 
      }
    }


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 = $.param({ url: url, title: title, type: type, filesize: filesize, fileext: fileext, id: id});
        request.open("POST", "ajax-db-files/copy_drive_file.php?" + datatable, true);
        request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
        request.send("datatable=" + datatable);

    }

和 php 文件是：

$upload_path='sss';
     if (isset($_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 = $upload_path.'/'.$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: " . $type . "");
    header("Content-type: application/octet-stream");
    header("Content-Disposition: attachment; filename=\"".$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.'/';
    $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();

Answer 1

TLDR；

您需要序列化datatable为URL编码字符串第一。

---

解释

var datatable = [url, title, type, filesize, fileext,id];

在这里，您将数据表创建为数组

request.open("POST", "ajax-db-files/copy_drive_file.php?" + datatable, true);

request.send("datatable=" + datatable);

然后，在这两行中，您尝试将数组直接连接到字符串，通过在数组上调用.join(',')将数组转换为字符串。

您需要的是将数组序列化为 URL 编码的表示法。 如果你使用 jQuery，试试var datatable = $.param({ url: url, title: title, type: type, filesize: filesize, fileext: fileext, id: id]); 反而。