如何通过 javascript 中的键“内部连接”两个 arrays？

Question

我有两个 arrays 包含这样的对象：

const array1 = 
[
{
    "id": 1,
    "mass": 149,
    "height": 180,
    "dob": "2003-09-04"
},
{
    "id": 2,
    "mass": 140,
    "height": 175,
    "dob": "2000-02-12",
},
{
    "id": 3,
    "mass": 143,
    "height": 170,
    "dob": "2001-11-04" 
}
]

const array2 = 
[
{
    "id": 1,
    "name": "James",
    "sport": "Football"
},
{
    "id": 2,
    "name": "Adam",
    "sport": "Tennis"
}
]

我想合并它们，使组合数组仅包含id出现在 array1 和 array2 中的数据。

到目前为止，我使用的是 lodash _.merge ，它返回：

const merged = _.merge(array1, array2)

merged = [
{
"id": 1,
"mass": 149,
"height": 180,
"dob": "2003-09-04",
"name": "James",
"sport": "Football"
},
{
"id": 2,
"mass": 140,
"height": 175,
"dob": "2000-02-12",
"name": "Adam",
"sport": "Tennis"
},
{
"id": 3,
"mass": 143,
"height": 170,
"dob": "2001-11-04" ,
}
]

我只想merged以包含id出现在两者中的数据，即：

[{
"id": 1,
"mass": 149,
"height": 180,
"dob": "2003-09-04",
"name": "James",
"sport": "Football"
},
{
"id": 2,
"mass": 140,
"height": 175,
"dob": "2000-02-12",
"name": "Adam",
"sport": "Tennis"
}]

我相信 SQL 之类的等效 function 将是“内部连接”

任何帮助将不胜感激！

Answer 1

您可以使用 Array.filter 过滤您的第一个数组，以便在执行相同的合并之前仅包含第二个数组中存在的对象。

let merged = array1.filter(e => array2.some(f => f.id == e.id));

 const array1 = [ { "id": 1, "mass": 149, "height": 180, "dob": 2003-09-04 }, { "id": 2, "mass": 140, "height": 175, "dob": 2000-02-12, }, { "id": 3, "mass": 143, "height": 170, "dob": 2001-11-04 } ] const array2 = [ { "id": 1, "name": "James", "sport": "Football" }, { "id": 2, "name": "Adam", "sport": "Tennis" } ] let merged = array1.filter(e => array2.some(f => f.id == e.id)); merged = _.merge(merged, array2); console.log(merged);

 <script src="https://cdn.jsdelivr.net/npm/lodash@4.17.21/lodash.min.js"></script>

Answer 2

您需要将它们从一个到另一个 map。 使用 reduce 的基本思想，我们通过 id 跟踪项目。

 const array1 = [{ "id": 1, "mass": 149, "height": 180, "dob": '2003-09-04' }, { "id": 2, "mass": 140, "height": 175, "dob": '2000-02-12', }, { "id": 3, "mass": 143, "height": 170, "dob": '2001-11-04' } ] const array2 = [{ "id": 1, "name": "James", "sport": "Football" }, { "id": 2, "name": "Adam", "sport": "Tennis" } ] const result = Object.values([...array1, ...array2].reduce((acc, userData) => { // have you seen the id before? If yes, use it, if no use and empty object const data = acc[userData.id] || {}; //merge the objects together and store the updated data by the id acc[userData.id] = {...data, ...userData }; return acc }, {})); console.log(result);

现在这将包括所有内容。 如果您只想要两者中的项目，则可以在两个循环中进行。

 const array1 = [{ "id": 1, "mass": 149, "height": 180, "dob": '2003-09-04' }, { "id": 2, "mass": 140, "height": 175, "dob": '2000-02-12', }, { "id": 3, "mass": 143, "height": 170, "dob": '2001-11-04' } ] const array2 = [{ "id": 1, "name": "James", "sport": "Football" }, { "id": 2, "name": "Adam", "sport": "Tennis" } ] // create a look up by the id const array1MappedById = array1.reduce((acc, userData) => { acc[userData.id] = userData; return acc }, {}); // Loop over the second array and make a new array with the merged data const result = array2.reduce((arr, userData) => { const array1Data = array1MappedById[userData.id]; if (array1Data) { arr.push({...array1Data, ...userData}); } return arr; }, []); console.log(result);

Answer 3

这将合并 id 相同的数据

const merged = array2.reduce((arr, current) => {
    let array1Data = {...array1}[current.id];
    if (array1Data) {
      arr.push({...array1Data, ...current});
    }
    return arr;
}, []);

console.log(merged);