JavaScript的问题（node.js + mongo.db）

Question

我遇到了一些问题，我的convertKey函数，我怀疑这是由于范围问题。基本上，我尝试从我的mongo数据库中检索一条记录并将其存储在count变量中，但是当我尝试返回它时，我得到“undefined”。令人惊讶的是，console.log(nameSearch + count)工程，而return nameSearch + count没有。如果有人能帮助我，我真的很感激！

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

你从回调返回db.studentsList.find，而不是从你的convertKey功能。

如果你想从db.studentsList.find内返回的值，你要么需要提供一个回调convertKey或者可能使用无极库提convertKey一个deferred /未来。否则，您的函数将在等待嵌套异步函数完成时立即返回。

的回调允许你传递你正在寻找（如callback(nameSearch + count)）

编辑

结果每当我对函数的返回值的问题，我匹配大括号注释：

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