javascript - javascript 的范围问题(node.js + mongodb)

Question

我的 ConvertKey 函数遇到一些问题，我怀疑这是由于范围问题造成的。基本上，我尝试从 mongo 数据库中检索记录并将其存储在计数变量中，但是当我尝试返回它时，我得到“未定义”。令人惊讶的是，console.log(nameSearch + count) 有效，而 return nameSearch + count 则无效。如果有人能帮助我，我将非常感激!

var dbUrl = "kidstartnow",
  collections = ["students", "studentsList"];
var db = require("mongojs").connect(dbUrl, collections);

function Student(name, src) {
  this.name = name.toLowerCase();

  //this function does not work
  this.key = convertKey(this.name);

  this.src = src;
  this.pointsTotal = 0;

  //inserts student into database
  var student = {name: this.name, key: this.key, pointsTotal: this.pointsTotal,
    src: this.src
  };
  db.students.insert(student);

  //converts name to a key by stripping white space and adding a number behind and ensures keys are unique
  //concatenates names together to form namesearch, and checks if entry exists in studentsList
  function convertKey(name) {

    var nameSearch = name.replace(/\s/g, ''),
      count = 1;



    db.studentsList.find({name: nameSearch}, function(err, student) {      
      //if nameSearch does not exist in studentsList, create entry and sets count to 1
      if(err || !student.length) {
        db.studentsList.insert({name: nameSearch, count: 1});
        count = 1;
        return nameSearch + count;
      }

      //if entry does exist, increments count by 1
      else {
        db.studentsList.update({name: nameSearch}, {$inc: {count: 1}}, function(err) {
          if(err) {
            console.log("Error incrementing records");
          }

          db.studentsList.find({name: nameSearch}, function(err, student) {
            count = student[0].count;
            //this works
            console.log(nameSearch + count)
            //but this doesn't
            return nameSearch + count;
          });
        });
      }
    });
  };
}

Answer 1

您将从回调返回到 db.studentsList.find，而不是从 convertKey 函数返回。

如果您希望从 db.studentsList.find 中返回值，则需要向 convertKey 提供回调，或者可能使用 Promise 库使 convertKey 成为延迟/ future 。否则，您的函数将在等待嵌套异步函数完成时立即返回。

回调允许您传递您正在查找的结果(例如callback(nameSearch + count))

编辑

每当我对函数的返回值有疑问时，我都会将大括号与注释匹配:

function convertKey(name) {

    var nameSearch = name.replace(/\s/g, ''),
      count = 1;

    db.studentsList.find({name: nameSearch}, function(err, student) {      
      //if nameSearch does not exist in studentsList, create entry and sets count to 1
      if(err || !student.length) {
        db.studentsList.insert({name: nameSearch, count: 1});
        count = 1;
        return nameSearch + count;
      } else {
        db.studentsList.update({name: nameSearch}, {$inc: {count: 1}}, function(err) {
          // ...
          db.studentsList.find({name: nameSearch}, function(err, student) {
             // ...
             return nameSearch + count;
          }); // end db.studentsList.find
        }); // end db.studentsList.update
      } // end else
    }); // end db.studentsList.find

    /**
     * Notice, no return value here...
     */
  }; // end convertKey