使用Node和graphql将上传文件流式传输到Azure blob存储

Question

我需要使用nodejs和apollo server express将映像存储到Azure blob存储中。我使用createBlockBlobFromStream进行存储，但它需要流长度。我怎样才能找到流的长度？有没有更好的方法可以使用graphql将图像存储到azureblob存储中

uploadImageFunc由解析程序通过传递参数来调用

const fs = require('fs')
const azure = require('azure-storage')
const blobService = azure.createBlobService("connection string")

const uploadImageFunc = async function(args) {
return args.file.then(file => {
const {createReadStream, filename, mimetype} = file
const fileStream = createReadStream()

blobService.createBlockBlobFromStream('container-name',filename,fileStream,"fileStream.length",(error,response) => {
if(!error){
console.log(response)
}
})
fileStream.pipe(fs.createWriteStream(__dirname+`/uploadedFiles/${filename}`))
return file;

});
}
module.exports = uploadImageFunc;

感谢的回复

Answer 1

实际上，流大小存在于请求的内容长度中。req.headers['content-length']具有作为字符串的内容长度。使用parseInt将其转换为int，并将其传递给createBlockBlobFromStream函数。

const fs = require('fs')
const { v4: uuidv4 } = require('uuid');
const azure = require('azure-storage')
const blobService = azure.createBlobService(process.env.AZURE_STORAGE_CONNECTION_STRING)
const contestEntryModel = require('../models/contestEntryModel')

const uploadImageFunc = async function(args,{req, res}) {
return args.file.then ( file => {
const {createReadStream, filename, mimetype} = file
let streamSize = parseInt(req.headers['content-length'])
const fileStream = createReadStream()
const newFilename = uuidv4()

blobService.createBlockBlobFromStream('<container name>',`${args.contestId}/${newFilename}`,fileStream,streamSize,(error,response) => {
if(!error){
let entry = {
url:`<blobstorage url>/${args.contestId}/${newFilename}`,
participant: req.userId
}
contestEntryModel.findOneAndUpdate({contestId: args.contestId},{$push:{entries:entry}},(err, doc) => {
console.log("updated")
})
}
})

return true;

});

}
module.exports = uploadImageFunc;