Multer gridfs存储引用已上载的文件到新集合

Question

使用multer gridfs存储上传的文件我想将其引用到新创建的集合(电影(中。引用到上传文件的fileID i与上传文件不相同，即使电影集没有保存到数据库中。引用到上传文件的文件ID i与上传文件不相同，即使电影集没有保存在数据库上

//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()
});

文件Id与上传的文件Id不同，并且集合未保存在DB上文件Id与上载的文件Id不相同，并且集合也未保存在数据库上文件Id不同于上传的文件Id，并且集合不保存在DB 上

Answer 1

它不起作用，因为你让它变得比需要的更复杂。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。相反，您应该使用array、single或字段，具体取决于您发送的文件数量和用于发送这些文件的字段名。

若要稍后从数据库中读取该文件，您只需查询Movies集合，并且fileId字段具有一个标识符，您可以将该标识符传递给openDownloadStream中的GridfsBucket，并将该流通过管道传输到响应，从而将使用该id存储的文件返回给用户，或在应用程序具有的任何业务逻辑中使用。