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我實際上有一個小問題:我想創建一個函數,它在數組中的一個對象中循環。功能返回錯誤信息不起作用
var contacts = [
{
"firstName": "Akira",
"lastName": "Laine",
"number": "0543236543",
"likes": ["Pizza", "Coding", "Brownie Points"]
},
{
"firstName": "Harry",
"lastName": "Potter",
"number": "0994372684",
"likes": ["Hogwarts", "Magic", "Hagrid"]
},
{
"firstName": "Sherlock",
"lastName": "Holmes",
"number": "0487345643",
"likes": ["Intriguing Cases", "Violin"]
},
{
"firstName": "Kristian",
"lastName": "Vos",
"number": "unknown",
"likes": ["Javascript", "Gaming", "Foxes"]
}
];
function lookUpProfile(firstName, prop){
for (var i = 0; i < contacts.length; i++) {
for (var prop1 in contacts[i]) {
if (contacts[i][prop1] == firstName) {
if (contacts[i].hasOwnProperty(prop)) {
return contacts[i][prop];
}
}
}
}
}
lookUpProfile("Sherlock", "likes");
而且我想返回兩個錯誤:「沒有這樣的接觸」和「沒有這樣的分類」(我commentated這裏我把「回報‘沒有這樣的聯繫’」的一部分,但它是這麼想的工作.. )。
嗨!所以爲了確保,你想要lookUpProfile返回匹配的第一個聯繫人?或者,在這個特定的例子中,返回「沒有這種聯繫」? – MacPrawn
*「不起作用」*不是有用的問題描述。你的代碼目前的問題是什麼? –