合并两个数组，包括每个数组的一些字段

Question

我想使用 map 函数合并两个数组，想法是创建一个新数组，其中一些字段取自第一个数组，一些字段取自第二个数组。 条件是字段name。

这里是列表 A：

[
    {"name": "tom", "id": "1", "date": "1654"},
    {"name": "jack", "id": "2", "date": "6544"},
    {"name": "sarah", "id": "3", "date": "987"},
    {"name": "william", "id": "4", "date": "654"},
    {"name": "ronaldo", "id": "5", "date": "12345"}
]

这里是列表 B：

[
    {"name": "tom", "age": "20", "school": "A", "password": "abcd"},
    {"name": "jack", "age": "25", "school": "B", "password": "1234"}
]

因此，它应该返回一个合并的版本，但只包含一些选定的字段：

[
    {"name": "tom", "age": "20", "school": "A", "exists": true, , "date": "1654"},
    {"name": "jack", "age": "25", "school": "B", "exists": true, "date": "6544"},
    {"name": "sarah", "age": "", "school": "", "exists": false, "date": "987"},
    {"name": "william", "age": "", "school": "", "exists": false, "date": "654"},
    {"name": "ronaldo", "age": "", "school": "", "exists": false, "date": "12345"}
]

这是我尝试使用 map 合并这两个数组，但不是很成功。有人可以帮我实现这一目标吗？

const alldata = listA.map(u => listB.filter(oo => u.name === oo.name));

Answer 1

我使用name 作为键来查找a 中的条目是否存在于b 中，并使用两个列表中的适当值创建了所需的列表。

a = [
    {"name": "tom", "id": "1", "date": "1654"},
    {"name": "jack", "id": "2", "date": "6544"},
    {"name": "sarah", "id": "3", "date": "987"},
    {"name": "william", "id": "4", "date": "654"},
    {"name": "ronaldo", "id": "5", "date": "12345"}
];
b=[
    {"name": "tom", "age": "20", "school": "A", "password": "abcd"},
    {"name": "jack", "age": "25", "school": "B", "password": "1234"}
];

c = a.map(a1 => {
      let b3 = b.find(b1=>b1.name === a1.name) || {};
      return {name: a1.name, age: b3.age || "", school: b3.school || "", exists: b3.name != undefined, date: a1.date}
   })
console.log(c)

Answer 2

有了这个答案，您不会创建新数组。

let a = [{"name": "tom","id": "1","date": "1654"},{"name": "jack","id": "2","date": "6544"},{"name": "sarah","id": "3","date": "987"},{"name": "william","id": "4","date": "654"},{"name": "ronaldo","id": "5","date": "12345"}],
    b = [{"name": "tom","age": "20","school": "A","password": "abcd"},{"name": "jack","age": "25","school": "B","password": "1234"}];
a.forEach((key, index)=>{
  let exists;
  b.forEach((bkey, bindex)=>{
    if(key.name==bkey.name){
      a[index] = {...a[index], ...b[bindex]};
      exists = bkey.age;
    }
  });
  a[index].exists = !!exists;
});
console.log(a);

Answer 3

你很亲密。试试这个：

let arrA = [
    {"name": "tom", "id": "1", "date": "1654"},
    {"name": "jack", "id": "2", "date": "6544"},
    {"name": "sarah", "id": "3", "date": "987"},
    {"name": "william", "id": "4", "date": "654"},
    {"name": "ronaldo", "id": "5", "date": "12345"}
];
	
let arrB = [
    {"name": "tom", "age": "20", "school": "A", "password": "abcd"},
    {"name": "jack", "age": "25", "school": "B", "password": "1234"}
];
	
let result = arrA.map(v => {
    let fr = arrB.filter(f => v.name == f.name ) ;
    let e = fr.length ? true : false;
    fr = fr.length ? fr[0] : {age: "", school: "", password: "" };
	
    //Construct the format you want on the final array
    return  {
        name: v.name, 
        age: fr.age, 
        school: fr.school, 
        exists: e, 
        date: v.date
    }
});
	
console.log( result );

Answer 4

使用map创建新数组并使用filter从第二个元素中查找匹配的元素

var a = [{
    "name": "tom",
    "id": "1",
    "date": "1654"
  },
  {
    "name": "jack",
    "id": "2",
    "date": "6544"
  },
  {
    "name": "sarah",
    "id": "3",
    "date": "987"
  },
  {
    "name": "william",
    "id": "4",
    "date": "654"
  },
  {
    "name": "ronaldo",
    "id": "5",
    "date": "12345"
  }
]
var b = [{
    "name": "tom",
    "age": "20",
    "school": "A",
    "password": "abcd"
  },
  {
    "name": "jack",
    "age": "25",
    "school": "B",
    "password": "1234"
  }
]

var c = a.map(function(item) {
  let d = b.filter(function(items) {
    return item.name === items.name;
  })
  if (d.length > 0) {
    item.age = d[0].age;
    item.school = d[0].school;
    item.password = d[0].password
  }
  return item;
})
console.log(c)

Answer 5

const list1 = [
    {"name": "tom", "id": "1", "date": "1654"},
    {"name": "jack", "id": "2", "date": "6544"},
    {"name": "sarah", "id": "3", "date": "987"},
    {"name": "william", "id": "4", "date": "654"},
    {"name": "ronaldo", "id": "5", "date": "12345"}
];

const list2 = [
    {"name": "tom", "age": "20", "school": "A", "password": "abcd"},
    {"name": "jack", "age": "25", "school": "B", "password": "1234"}
];

const merged = list1.map(item => {
    const list2Match = list2.filter(i => i.name === item.name)[0];
    const obj = list2Match ? {school: list2Match.school, age: list2Match.age, exists: true} : { school: '', age: '', exists: false};

    return {school:obj.school, age: obj.age, exists: obj.exists, name: item.name, date: item.date };
});