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Javascript 为什么变量显示为未定义?_Javascript_Node.js_Mongodb_Mongoose - Fatal编程技术网

Javascript 为什么变量显示为未定义?

Javascript 为什么变量显示为未定义?,javascript,node.js,mongodb,mongoose,Javascript,Node.js,Mongodb,Mongoose,我在一个朋友请求数组中查询一个用户,该数组可以工作,但是我使用userAdd进行的检查显示为未定义。有人知道为什么它显示为未定义的吗 代码如下: exports.searchPost = function(req, res, err) { User.find({$or:[ {firstName: req.body.firstName}, {lastName: req.body.lastName}, {emai

我在一个朋友请求数组中查询一个用户,该数组可以工作,但是我使用userAdd进行的检查显示为未定义。有人知道为什么它显示为未定义的吗

代码如下:

    exports.searchPost = function(req, res, err) {
    User.find({$or:[
            {firstName: req.body.firstName},
            {lastName: req.body.lastName},
            {email: req.body.email},
            {phone: req.body.phone}]
    }, function(err, users) {
        if(err) {

            return res.render('searchError', {title: 'Weblio'}); 
        } else {

            if(req.body.firstName=== '' && req.body.lastName==='' && req.body.email==='' && req.body.phone=== '') {
                //maybe  a diff page saying that is not a valid search page
                return res.render('searchError', {title: 'Weblio'});        
            } else {
                    var userAdd;
                    console.log('addman');
                    users.forEach(function(user)  {
                        User.findById(req.signedCookies.userid, 
                            {friendRequest: user._id}, function() {
                                if(user._id === true ) {
                                    console.log('addman1'); 
                                    return userAdd = true;
                                } else {
                                    console.log('addman2');
                                    return userAdd = false;
                                    console.log(user._id);  
                                }

                            })

                        }); 
                    console.log(userAdd);

                return res.render('searchResults', {title: 'Weblio',        
                    usersFound: users, 
                    userAdded: userAdd
                });

            } 
        }
    });
};

它是异步的,因此您需要使用回调方式来链接操作。您可以尝试一些节点模块,如或,以更好的语法实现它

顺便说一句,我认为您不需要遍历
users.forEach(function(user)

如果使用Step,则可以执行以下操作:

exports.searchPost = function(req, res, err) {
    User.find({$or:[
            {firstName: req.body.firstName},
            {lastName: req.body.lastName},
            {email: req.body.email},
            {phone: req.body.phone}]
    }, function(err, users) {
        if(err) {

            return res.render('searchError', {title: 'Weblio'}); 
        } else {

            if(req.body.firstName=== '' && req.body.lastName==='' && req.body.email==='' && req.body.phone=== '') {
                //maybe  a diff page saying that is not a valid search page
                return res.render('searchError', {title: 'Weblio'});        
            } else {
                var userAdd;
                console.log('addman');

                Step(
                    function(){
                        User.findById(req.signedCookies.userid, {friendRequest: user._id}, this);
                    },

                    function(err, user){
                        if(user !== null && user._id === true ) {
                            console.log('addman1'); 
                            userAdd = true;
                        } else {
                            console.log('addman2');
                            userAdd = false;
                            console.log(user._id);  
                        }

                        return res.render('searchResults', {title: 'Weblio',        
                            usersFound: users, 
                            userAdded: userAdd
                        });
                    }
                );
            } 
        }
    });
};

欢迎来到奇妙的异步世界!你需要在回调中。相关吗?还有,你有一个错误的
x
,你的选项卡很糟糕。是的,很抱歉,这实际上是一段更大的代码的片段,x在那里是偶然的,因为我从文件中删除了注释,对不起…我更新了代码以包含我不在回调中吗?我在users.foreach中的函数中。我如何实现异步?如果不是,那么我仍然无法使用上面所写的内容将userAdd设置为未定义。是的,我查看了异步,老实说,我在这里很难理解如何使用它。它是说user.\u id对于第二个函数是未定义的激动