Javascript 联系人信息列表未返回“错误”，即使它正在返回信息

我正在学习codecademy.com javascript课程，我的联系信息有问题。它正在返回，但我得到一个错误通知我，它不会为史蒂夫返回。这件事我绞尽脑汁，看不清楚

链接如下

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您返回的搜索是您的功能。 把它改成你想要的

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");

---

测试正在寻找区分大小写的名字，它首先寻找Steve。将每个名字更改为标题，您将通过。

请始终准确检查问题。你写信给史蒂夫和比尔。但是问题提到了比尔和史蒂夫。那么比尔应该到史蒂夫面前了解情况吗？不，重要的是第一个字母应该是大写

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");