联系人信息列表不返回“错误”，即使它返回信息

contact info list not returning "error" even though it is returning the info

本文关键字：返回 信息 列表 联系人 错误      更新时间：2023-09-26

我正在做一个 codecademy.com 的JavaScript课程，我的联系信息完成时遇到问题。 它正在返回，但我收到一个错误，通知我它不会为史蒂夫返回。我一直在绞尽脑汁地研究这个问题，以至于我无法直视。

链接如下http://www.codecademy.com/courses/javascript-beginner-en-3bmfN/0/7

var friends = new Object();
friends.bill = new Object();
friends.steve = new Object();
friends.steve.firstName = "steve";
friends.steve.lastName = "jobs";
friends.steve.number = "317-222-3344";
friends.steve.address = ["one apple way","california","us"];
friends.bill.firstName = "bill";
friends.bill.lastName = "gates";
friends.bill.number = "327-332-3322";
friends.bill.address = ["one microsoft way","washington","us"];
var list = function (name){
    for(var key in name){
    console.log(friends[key]);
   }
};
var search = function(name){
    for(var key in friends){
    if(friends[key].firstName === name){
        console.log(key);
        return search;
     }
   }
};
search("bill");
search("steve");
list(friends);

测试正在查找区分大小写的名称，并且首先查找"Steve"。将每个名字更改为标题大小写，您将通过。

你返回的搜索是你的函数。只需将其更改为您想要的即可。

return friends[key];
var friends = new Object();
friends.bill = new Object();
friends.steve = new Object();
friends.steve.firstName = "Steve";
friends.steve.lastName = "jobs";
friends.steve.number = "317-222-3344";
friends.steve.address = ["one apple way","california","us"];
friends.bill.firstName = "Bill";
friends.bill.lastName = "gates";
friends.bill.number = "327-332-3322";
friends.bill.address = ["one microsoft way","washington","us"];
var list = function (name){
    for(var key in name){
    console.log(friends[key]);
   }
};
var search = function(name){
    for(var key in friends){
    if(friends[key].firstName === name){
        var f = friends[key];
        console.log(f.firstName,f.lastName,f.number,f.address);
        return f;
     }
   }
};
search("bill");
search("Steve");
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