如何等待节点的回调函数执行

Question

我正在尝试使用node上传文件/存储，并将文件保存在Gird FS中。下面是代码示例。上传功能工作得很好。由于节点的异步特性，响应将立即发送回浏览器。我希望在回调函数执行之前保持响应。注意:我也尝试在回调函数内部发送最终响应，但仍然不起作用。

请抛出一些想法。

const multer = require("multer");
const GridFsStorage = require("multer-gridfs-storage");
const mongoose = require("mongoose");
mongoose.connect(CONFIG.dbConfig, {useNewUrlParser: true, useUnifiedTopology: true} );

var storage = new GridFsStorage({
  url: CONFIG.dbConfig,
  options: { useNewUrlParser: true, useUnifiedTopology: true },
  file: (req, file) => {
    return {
      bucketName: "xyz",
      filename: `${file.originalname}`
    };
  }
});

const viewPage = (req, res) => {
  try {
    
    var uploadFile = multer({ storage: storage }).single("file");
     uploadFile(req, res, function(err) {
      if (req.file == undefined) {
        return res.send(`You must select a file.`);
      }

      /* code to save anthother Schema */
      fileDetails.save(function(err) {
        if (err) throw err;
      });
      
    });
    
    return res.send(`File has been uploaded.`);
  } catch (error) {
    console.log(error);
    return res.send(`Error when trying upload image: ${error}`);
  }
};

module.exports = {
  getPage:  viewPage
};

Answer 1

您将不得不使用async并等待

// convert multer function to async await
function uploadFile(req) {
    return new Promise(resolve => {
       var uploadFile = multer({ storage: storage }).single("file");
       if (req.file == undefined) {
          return resolve([`You must select a file.`, null]);
       }else {
          // do something
          return resolve([null, data])
       }
    });
}
const viewPage = async (req, res) => {
  try {
     let [err, data] = await uploadFile(req)
      // check error and send response
      // convert do async await similarly for below fileDetails.save  
 
      /* code to save anthother Schema (change this to async await) */
      fileDetails.save(function(err) {
        if (err) throw err;
      });
      
    });
    
    return res.send(`File has been uploaded.`);
  } catch (error) {
    console.log(error);
    return res.send(`Error when trying upload image: ${error}`);
  }
};