Question

所以我有以下节点代码用于将视频上传到我的节点服务器：

    var fs = require('fs');

var videoExtensions = ['mp4','flv', 'mov'];
//Media object
function Media(file, targetDirectory) {
   this.file = file;
   this.targetDir = targetDirectory;
}

Media.prototype.isVideo = function () {
    return this.file.mimetype.indexOf('video') >= 0;
};
Media.prototype.getName = function () {
    return this.file.originalname.substr(0, this.file.originalname.indexOf('.'))
};

router.route('/moduleUpload')
    .post(function (request, response) {
        var media = new Media(request.files.file, '../user_resources/module/'+request.body.module_id+'/');
        if(!fs.existsSync('../user_resources/module/'+request.body.module_id+'/')){
            fs.mkdirSync('../user_resources/module/'+request.body.module_id+'/', 0766, function(err){
                if(err){
                    console.log(err);
                    response.send("ERROR! Can't make the directory! \n");    // echo the result back
                }
            });
        }
        convertVideos(media);
        response.status(200).json('user_resources/module/' + request.body.module_id + '/' + request.files.file.name);

    });


function convertVideos (media){
    var ffmpeg = require('fluent-ffmpeg');
    videoExtensions.forEach(function(extension){
        var proc = new ffmpeg({source: media.file.path, nolog: false})
            .withVideoCodec('libx264')
            .withVideoBitrate(800)
            .withAudioCodec('libvo_aacenc')
            .withAudioBitrate('128k')
            .withAudioChannels(2)
            .toFormat(extension)
            .saveToFile(media.targetDir+media.getName()+'.'+extension,
            function (retcode, error) {
                console.log('file has been converted succesfully');
            });
    });

}

现在不是使用直接路径加载视频，而是希望使用节点

加载它

但是我不太清楚如何做到这一点

使用直接路径我会做这样的事情：

$scope.videos.push(
{
    sources: [
        {src: $sce.trustAsResourceUrl($scope.component.video_mp4_path), type: "video/mp4"}
    ]
}

其中video_mp4_path变量将是视频的直接路径，即：myproject/resources/video.mp4

但是我需要以某种方式调用节点而不是即时路径。

正如我所说，我不太确定如何做到这一点可能有人指出我正确的方向

Answer 1

你可以使用express和Multer上传，列表和播放，我发现它相对简单，工作可靠。在我的例子中，我使用Angular和常规浏览器进行播放，但应该采用相同的原则。

以下代码摘录对我有用：

上传

// router.post('/web_video_upload', function(req, res) { //Log the request details Debug console.log(req.body); //Send a resposne res.send('Video Uploading'); console.dir(req.files); }); // POST: video upload route // multer approach var multer = require('multer'); app.use(multer({ //Set dstination directory dest: path.resolve(__dirname, 'public', 'uploaded_videos'), //Rename file rename: function (fieldname, filename) { //Add the current date and time in ISO format, removing the last 'Z' character this usually //includes var dateNow = new Date(); return filename + "_" + dateNow.toISOString().slice(0,-1) }, //Log start of file upload onFileUploadStart: function (file) { console.log(file.originalname + ' is starting ...') }, //Log end of file upload onFileUploadComplete: function (file) { console.log(file.originalname + ' uploaded to ' + file.path) done=true; } }));

列出视频

// GET: route to return list of upload videos router.get('/video_list', function(req, res) { //Log the request details console.log(req.body); // Get the path for the uploaded_video directory - in a real app the video list would likely be taken from // a database index table, but this is fine for us for now var _p; _p = path.resolve(__dirname, 'public', 'uploaded_videos'); //Find all the files in the diectory and add to a JSON list to return var resp = []; fs.readdir(_p, function(err, list) { //Check if the list is undefined or empty first and if so just return if ( typeof list == 'undefined' || !list ) { return; } for (var i = list.length - 1; i >= 0; i--) { // For each file in the directory add an id and filename to the response resp.push( {"index": i, "file_name": list[i]} ); } // Set the response to be sent res.json(resp); }); });

以上内容将返回服务器目录中所有视频的JSON列表。如果您的文件名是视频的路径，您可以在角度视图中使用它来创建视频表并使用HTML5视频标记播放它们（几乎所有现代浏览器都支持）。

节点js＆amp;使用节点的视频加载视频

1 个答案: