Multer gridfs存储将上载的文件引用到新集合

Question

我使用multer gridfs存储上传了文件，我想将其引用到新创建的收藏集（电影）。 我引用的已上传文件的fileID与已上传文件不同，即使电影收藏没有保存到数据库中也是如此。 即使电影收藏未保存在数据库中，我引用的已上传文件的fileID与已上传文件也不相同

 //movie schema const mongoose = require('mongoose'); const Schema = mongoose.Schema; const MovieSchema = new Schema({ description: { type: String, }, category: { type: String, }, token: { type: String, }, fileID: { type: Schema.Types.ObjectId, ref: "contents.files" } }); module.exports = movies = mongoose.model('movies', MovieSchema); let gfs; conn.once('open', () => { gfs = Grid(conn.db, mongoose.mongo); gfs.collection('contents'); }); const storage = new GridFsStorage({ url: config.db, file: (req, file) => { return new Promise((resolve, reject) => { const filename = req.body.fileName + path.extname(file.originalname); const Description = req.body.Description const fileInfo = { filename: filename, bucketName: 'contents', metadata: req.body, } resolve(fileInfo, Description); }); } }); const upload = multer({ storage }); router.get('/', (req, res) => { res.render('index'); console.log(req.body) }); //** uploading file to the db */ router.post('/', upload.any(), (req, res) => { const movie = new movies({ description: "test", category: "test", fileID: gfs.files.id }) movie.save() }); 

fileID与上载的文件ID不同，并且集合未保存在DB上。fileID与上载的文件ID不同，并且集合未保存在DB上。fileID与上载的文件ID不同，并且集合也没有保存在数据库中

Answer 1

它不起作用，因为您使它变得比所需的复杂。 文件中的引用是不必要的。 您可以使用此代码完成相同的操作。

 //movie schema const mongoose = require('mongoose'); const Schema = mongoose.Schema; const MovieSchema = new Schema({ description: { type: String, }, category: { type: String, }, token: { type: String, }, fileID: { type: Schema.Types.ObjectId, // There is no need to create references here } }); module.exports = movies = mongoose.model('movies', MovieSchema); const storage = new GridFsStorage({ url: config.db, file: (req, file) => { return new Promise((resolve, reject) => { // if you are using a separate collection to store data // there is no need to save this information on the metadata // because you'll probably never use it const filename = req.body.fileName + path.extname(file.originalname); const fileInfo = { filename: filename, bucketName: 'contents' } resolve(fileInfo); }); } }); const upload = multer({ storage }); router.get('/', (req, res) => { res.render('index'); console.log(req.body) }); //** uploading file to the db */ router.post('/', upload.any(), (req, res) => { const movie = new movies({ description: req.body.Description, category: req.body.Category, // Grab the file id that was stored in the database by the storage engine as the reference to your file fileID: req.file._id }) movie.save() }); 

您不应该使用uploads.any() 。 这将接受任何传入的文件。 取而代之的是，您应该使用array ， single或field，具体取决于发送的文件数量以及用于发送文件的字段名称。

要稍后从数据库中读取该文件，您只需查询Movies集合，并且fileId字段具有标识符，您可以将其传递给GridfsBucket中的openDownloadStream并将该流传输到响应中，从而将与该文件一起存储的文件返回给用户id或在您的应用程序具有的任何业务逻辑中使用。