async函数then块无法获取正确返回值？Mongoose操作MongoDB问题

你使用mongoose向mongodb添加文档时，日志显示addQuestion中的q.save()执行成功并打印了新增文档，但调用它的addQuestionHandler的then块中获取到的返回值为undefined。相关代码及日志如下：

相关代码 #

addQuestion函数 #

async function addQuestion (description,hints,topic) {
  var q = null;
  dotenv.config();
  console.log(`adding question:`);
  console.log(`description: ${description}`);
  console.log(`hints: ${hints}`);
  console.log(`topic: ${topic}`);

  
  const Question = mongoose.model(topic, questionSchema,topic);
  
  q = Question({
      description: description,
      hints: hints,
      topic: topic
  });
  await q.save()
  .then((result) => {
    console.log(`added question ${result}`);
    return result;
  })
  .catch ( (err)=>{
    console.log(`error in adding question: ${err}`);
    return null;
  });
}

addQuestionHandler函数 #

addQuestionHandler: async function (req, res) {
  console.log(`adding question: request is ${req}`);
   

  await addQuestion(req.description, req.hints,req.topic)
  .then((question)=>{
    console.log(`added question ${question}`);
    console.log(`returning value ${question}`);
    return question;
  })
  .catch((error) => {
    console.log(`error in adding question: ${error}`);
  })
  }

成功日志 #

added question {
      description: 'description of question on area',
      hints: [ 'area hint 1', 'area hint 2' ],
      topic: 'areas',
      _id: new ObjectId("64f47a56e5b349ec88a8ca98"),
      __v: 0
    }

原因分析 #

addQuestion是async函数，但它没有显式返回值。虽然你在q.save().then()里return了result、catch里return了null，但这些return仅在对应的回调函数内生效，并没有作为整个addQuestion函数的返回值向外传递。async函数如果没有显式return，默认返回undefined，所以addQuestionHandler中await该函数后拿到的就是undefined。

修复方案 #

方案1：返回Promise链 #

直接return整个q.save()的Promise链，让async函数把这个Promise的最终结果作为自身返回值：

async function addQuestion (description,hints,topic) {
  dotenv.config();
  console.log(`adding question:`);
  console.log(`description: ${description}`);
  console.log(`hints: ${hints}`);
  console.log(`topic: ${topic}`);

  const Question = mongoose.model(topic, questionSchema,topic);
  
  const q = Question({
      description: description,
      hints: hints,
      topic: topic
  });
  // 返回整个Promise链
  return q.save()
  .then((result) => {
    console.log(`added question ${result}`);
    return result;
  })
  .catch ( (err)=>{
    console.log(`error in adding question: ${err}`);
    return null;
  });
}

方案2：改用纯async/await写法（更推荐） #

避免混用then/catch和await，代码逻辑更清晰：

async function addQuestion (description,hints,topic) {
  dotenv.config();
  console.log(`adding question:`);
  console.log(`description: ${description}`);
  console.log(`hints: ${hints}`);
  console.log(`topic: ${topic}`);

  const Question = mongoose.model(topic, questionSchema,topic);
  
  const q = Question({
      description: description,
      hints: hints,
      topic: topic
  });
  try {
    const result = await q.save();
    console.log(`added question ${result}`);
    return result;
  } catch (err) {
    console.log(`error in adding question: ${err}`);
    return null;
  }
}

内容的提问来源于stack exchange，提问作者Manu Chadha