javascript的作用域问题(node.js+;mongo.db)
我的convertKey函数遇到了一些问题,我怀疑这是由于范围问题造成的。基本上,我尝试从mongo数据库中检索一条记录并将其存储在count变量中,但当我尝试返回它时,会得到“未定义”。令人惊讶的是,javascript的作用域问题(node.js+;mongo.db),javascript,node.js,mongodb,Javascript,Node.js,Mongodb,我的convertKey函数遇到了一些问题,我怀疑这是由于范围问题造成的。基本上,我尝试从mongo数据库中检索一条记录并将其存储在count变量中,但当我尝试返回它时,会得到“未定义”。令人惊讶的是,console.log(nameSearch+count)起作用,而return nameSearch+count则不起作用。如果有人能在这方面帮助我,我将不胜感激 var dbUrl = "kidstartnow", collections = ["students", "studentsL
console.log(nameSearch+count)
起作用,而return nameSearch+count
则不起作用。如果有人能在这方面帮助我,我将不胜感激
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}, {$inc: {count: 1}}, function(err) {
if(err) {
console.log("Error incrementing records");
}
db.studentsList.find({name: nameSearch}, function(err, student) {
count = student[0].count;
//this works
console.log(nameSearch + count)
//but this doesn't
return nameSearch + count;
});
});
}
});
};
}
您正在从回调返回到
db.studentsList.find
,而不是从convertKey
函数返回
如果希望从db.studentsList.find
中返回值,则需要提供对convertKey
的回调,或者可能使用承诺库使convertKey
成为延迟/未来。否则,在等待嵌套异步函数完成时,函数将立即返回
回调允许您传递正在查找的结果(例如,回调(nameSearch+count)
)
编辑
每当我对函数的返回值有疑问时,我都会将大括号与注释相匹配:
function convertKey(name) {
var nameSearch = name.replace(/\s/g, ''),
count = 1;
db.studentsList.find({name: nameSearch}, function(err, student) {
//if nameSearch does not exist in studentsList, create entry and sets count to 1
if(err || !student.length) {
db.studentsList.insert({name: nameSearch, count: 1});
count = 1;
return nameSearch + count;
} else {
db.studentsList.update({name: nameSearch}, {$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
您正在从回调返回到
db.studentsList.find
,而不是从convertKey
函数返回
如果希望从db.studentsList.find
中返回值,则需要提供对convertKey
的回调,或者可能使用承诺库使convertKey
成为延迟/未来。否则,在等待嵌套异步函数完成时,函数将立即返回
回调允许您传递正在查找的结果(例如,回调(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相匹配:
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
谢谢你!只是一个简单的跟进:我尝试添加一个回调函数,并这样调用它,但它不起作用。你知道我哪里出错了吗this.key=convertKey(this.name,function(text){returntext;})@丹东,我想你需要好好读一读,这是一篇很好的文章。基本上,您需要在回调函数中包含依赖于
text
值的所有逻辑。这与您在传递给db.studentsList.find
的回调中所做的相同。谢谢!只是一个简单的跟进:我尝试添加一个回调函数,并这样调用它,但它不起作用。你知道我哪里出错了吗this.key=convertKey(this.name,function(text){returntext;})@丹东,我想你需要好好读一读,这是一篇很好的文章。基本上,您需要在回调函数中包含依赖于text
值的所有逻辑。这与您在传递给db.studentsList.find
的回调中所做的相同。