如何基于输入值数组从对象数组返回匹配结果

Question

我有一个像下面这样的对象数组

var item = [
    { "name": "John", "age": 30, "city": "New York1" },
    { "name": "John1", "age": 31, "city": "New York2" },
    { "name": "John2", "age": 32, "city": "New York3" },
    { "name": "John3", "age": 31, "city": "New York3" }
]

我想要的是从具有年龄属性值的对象数组中获得一些年龄 在[30,31]

中

因此，基本上，输入将是一个整数数组，例如var ageArray=[30,31];

示例：

输入：[30,31]

输出：以下对象age的总和

 { "name":"John", "age":30, "city":"New York1"},
 { "name":"John1", "age":31, "city":"New York2"},
 { "name":"John3", "age":31, "city":"New York3"}

所以这里是

92

我尝试为此使用过滤器，但不确定如何将filter与包含

一起使用

我尝试过的是

var result = item .filter(obj => {
                    return obj.age..includes(ages);
                })

有人可以帮我解决这个问题吗？

Answer 1

使用Array#reduce

const items = [
{ "name":"John", "age":30, "city":"New York1"},
{ "name":"John1", "age":31, "city":"New York2"},
{ "name":"John2", "age":32, "city":"New York3"},
{ "name":"John3", "age":31, "city":"New York3"}]


const ages = [30, 31];
const res = items.reduce((sum, {age})=>{
  return sum + (ages.includes(age) ? age : 0);
}, 0);

console.log(res);

为名称修改：

const items = [
{ "name":"John", "age":30, "city":"New York1"},
{ "name":"John1", "age":31, "city":"New York2"},
{ "name":"John2", "age":32, "city":"New York3"},
{ "name":"John3", "age":31, "city":"New York3"}]


const names = ["John", "john1"].map(n=>n.toLowerCase());
const res = items.reduce((sum, {name, age})=>{
  return sum + (names.includes(name.toLowerCase()) ? age : 0);
}, 0);

console.log(res);

Answer 2

您可以使用reduce

var item= [{ "name":"John", "age":30, "city":"New York1"},{ "name":"John1", "age":31, "city":"New York2"},{ "name":"John2", "age":32, "city":"New York3"},{ "name":"John3", "age":31, "city":"New York3"}]

var ageArray=[30,31];


let op = item.reduce((o,c)=>{
  if( ageArray.includes(c.age) ) 
   { o+=c.age }
  return o;
},0)

console.log(op)

Answer 3

您可以使用三个步骤

1. 只获取年龄
2. 使用Set
过滤年龄
3. 添加剩余值

var items = [{ name: "John", age: 30, city:" New York1" }, { name: "John1", age: 31, city: "New York2" }, { name: "John2", age: 32, city: "New York3" }, { name: "John3", age: 31, city: "New York3" }],
    ageArray = [30, 31],
    result = items
        .map(({ age }) => age)
        .filter(Set.prototype.has, new Set(ageArray))
        .reduce((a, b) => a + b, 0);

console.log(result);

对于不同的键进行过滤和添加的方法略有不同。

var items = [{ name: "John", age: 30, city:" New York1" }, { name: "John1", age: 31, city: "New York2" }, { name: "John2", age: 32, city: "New York3" }, { name: "John3", age: 31, city: "New York3" }],
    names = ['John', 'John2'],
    result = items
        .filter((s => ({ name }) => s.has(name))(new Set(names)))
        .map(({ age }) => age)
        .reduce((a, b) => a + b, 0);

console.log(result);

Answer 4

您可以使用reduce一行完成此操作：

const item = [{name:"John",age:30,city:"New York1"},{name:"John1",age:31,city:"New York2"},{name:"John2",age:32,city:"New York3"},{name:"John3",age:31,city:"New York3"}]
 ,ageArray = [30,31]
 ,total = item.reduce((sum, {age})=> sum += ageArray.includes(age) ? age : 0, 0);

console.log(total)

Answer 5

const items = [
   { "name":"John", "age":30, "city":"New York1"},
   { "name":"John1", "age":31, "city":"New York2"},
   { "name":"John2", "age":32, "city":"New York3"},
   { "name":"John3", "age":31, "city":"New York3"}
];
const ages = [30,31];
const result = items.filter(o => ages.find(o2 => o.age === o2)).map(o3 => o3.age);
console.log(result)
const reducer = (accumulator, currentValue) => accumulator + currentValue;
console.log(result.reduce(reducer))

Answer 6

        function getSum(total, { age }) {
            if(age===30 ||age===31) return total + age
            return total
        }
        var item = [
            { "name": "John", "age": 30, "city": "New York1" },
            { "name": "John1", "age": 31, "city": "New York2" },
            { "name": "John2", "age": 32, "city": "New York3" },
            { "name": "John3", "age": 31, "city": "New York3" }
        ];
        let sumItem = item.reduce(getSum,0)
        console.log(sumItem);

您可以使用过滤器并减少##

• 首先，您需要过滤数组
• 然后缩小数组以获得30,31,31之和