如何从对象数组中找到多个属性值？

Question

我有一个数组的对象，

customer1 = [
    {"key": "name",
     "value": "Peter"},
    {"key": "age",
     "value": 23},
    {"key": "address",
     "value": "xyz St, abcd"},
    {"key": "points",
     "value": 234}
    ]

我想从这个对象中找出年龄和地址，推荐的和最佳的方法是什么？对于真正的应用程序，我可能在这个数组中有20-40个键值对象，其中我可能想要访问5-10个值。

我现在所做的就是循环这个对象，并使用条件来查找和赋值给我的变量。但是在这种方法中，我必须编写多个if表达式(5-10)。

例如,

let name: string;
let points: number;

for (var item of customer1) {
    if (item.key === "name") {
        name = item.value;
    } else if (item.key === "points") {
        points = item.value;
    }};

Answer 1

您可以在Array.filter()和Array.map()方法的帮助下实现这一需求。

现场演示

​

const customer1 = [
  { "key": "name", "value": "Peter" },
  { "key": "age", "value": 23 },
  { "key": "address", "value": "xyz St, abcd" },
  { "key": "points", "value": 234 }
];

const searchKeys = ['age', 'address'];

const res = customer1.filter(obj => searchKeys.includes(obj.key)).map(({ value}) => value);

const [age, address] = res;

console.log(age);
console.log(address);

​

Answer 2

你总是要显式地写"name"，"points"，.因为您不能解析字符串值并使用它来标识同名变量。但是，为了避免使用多个其他语句，如果，也许开关case语句更合适和更易读？

interface CustomerProperty {
    key: string,
    value: string | number
}

let customer1: CustomerProperty[] = [
    {
        "key": "name",
        "value": "Peter"
    },
    {
        "key": "age",
        "value": 23},
    {
        "key": "address",
        "value": "xyz St, abcd"},
    {
        "key": "points",
        "value": 234
    }
];

let nameValue: string;
let pointsValue: number;
for (var item of customer1) {
    switch (item.key) {
        case "name":
            nameValue = item.value as string;
            break;
        case "points":
            pointsValue = item.value as number;
            break;
    }
};