Question

我正在尝试使用map函数遍历以下格式的对象：

Name: Hong Kong
TopLevelDomain: .hk
Alpha2Code: HK
Alpha3Code: HKG
CallingCodes: 852
Capital: City of Victoria
AltSpellings: HK, 香港
Region: Asia
Subregion: Eastern Asia
Population: 7324300
Latlng: 22.25, 114.16666666

来自下面的JavaScript对象：

const object = {
    "name": "Hong Kong",
    "topLevelDomain": [
        ".hk"
    ],
    "alpha2Code": "HK",
    "alpha3Code": "HKG",
    "callingCodes": [
        "852"
    ],
    "capital": "City of Victoria",
    "altSpellings": [
        "HK",
        "香港"
    ],
    "region": "Asia",
    "subregion": "Eastern Asia",
    "population": 7324300,
    "latlng": [
        22.25,
        114.16666666
    ],
    "demonym": "Chinese",
    "area": 1104.0,
    "gini": 53.3,
    "timezones": [
        "UTC+08:00"
    ],
    "borders": [
        "CHN"
    ],
    "nativeName": "香港",
    "numericCode": "344",
    "currencies": [
        {
            "code": "HKD",
            "name": "Hong Kong dollar",
            "symbol": "$"
        }
    ]
};

我的尝试是：

const object = { "name": "Hong Kong", "topLevelDomain": [ ".hk" ], "alpha2Code": "HK", "alpha3Code": "HKG", "callingCodes": [ "852" ], "capital": "City of Victoria", "altSpellings": [ "HK", "香港" ], "region": "Asia", "subregion": "Eastern Asia", "population": 7324300, "latlng": [ 22.25, 114.16666666 ], "demonym": "Chinese", "area": 1104.0, "gini": 53.3, "timezones": [ "UTC+08:00" ], "borders": [ "CHN" ], "nativeName": "香港", "numericCode": "344", "currencies": [ { "code": "HKD", "name": "Hong Kong dollar", "symbol": "$" } ] };
let display = [];
let print = Object.keys(object).map(function(elem){
    if(object[elem] == 0 || object[elem] ==""){
        display = "N/A";
    } else if (typeof object[elem] =='object'){
        display = object[elem].join(", ") ;
    } else {
        display = object[elem];        
    }
    return `${elem.charAt(0).toUpperCase()}${elem.slice(1)}: ${display}`;
})
console.log(print.join('\n'));

但是，出现了一个错误，我不知道如何处理。据我所知，.join（）可用于连接数组的元素。有想法吗？

TypeError: object[elem].join is not a function

Answer 1

因为您有嵌套的对象。 object [elem] .join不起作用join是一个数组函数，它将仅与Array一起使用。像下面一样

"currencies": [
        {
            "code": "HKD",
            "name": "Hong Kong dollar",
            "symbol": "$"
        }
    ]

如果用于数组，您将需要再添加一个，并且在存在嵌套对象时将需要使用递归函数

function print(object) {
  let objMap= Object.keys(object).map(function(elem){
    if(object[elem] == 0 || object[elem] ==""){
        display = "N/A";
    } else if (object[elem].constructor === Array){
        display = object[elem].join(", ") ;
    } else if (typeof object[elem] =='object'){
        display = print(object[elem]);
    } else {
        display = object[elem];        
    }
    return `${elem.charAt(0).toUpperCase()}${elem.slice(1)}: ${display}`;
  });
  console.log(objMap.join('\n'));
}

Answer 2

您的代码工作正常。所有数组都是javascript中的对象，因此typeof可能无济于事。

.join是一个数组方法，为确保在有效数组上运行，我将使用Array.isArray（）而不是typeof。

如果传入的值是Array，则

Array.isArray（）返回true；否则，则返回false。

在处理嵌套对象/数组时，我同意上面的Shashi。

当您必须评估正在使用的数据类型时，一个简单的while循环将起作用。

.join（）函数在map函数中无法正常工作

2 个答案: