根据id从两个数组中查找同步用户，并返回数组A，其中数组b中有一列

Question

我试图找到两个数组之间同步用户的列表(objArray1 &objArray2)并返回来自objArray1的数据以及objArray2中的"aid"。我有下面的代码工作，但我想看看是否可以从objArray2返回"援助"，以及在下面的格式。

下面的代码示例

// SystemA
const objArray1 = [
{ id: "1", name: "John" },
{ id: "2", name: "Jack" },
{ id: "3", name: "Sam" },
{ id: "4", name: "Bill" },
];
// SystemB
const objArray2 = [
{ id: "1", name: "John", aid:"uuud2905555" },
{ id: "3", name: "Sam" }, aid:"uuud29029303"
{ id: "5", name: "Bob" }, aid:"uuud29435454"
];
const array1IDs = objArray1.map((item) => {
return item.id
})
// Result: array1IDs = ["1", "2", "3", "4"];
const array2IDs = objArray2.map((item) => {
return item.id
})
// Result: array2IDs = ["1", "3", "5"];
// FIND SYNCED USERS
// Compare the id value of each item in objArray1 with each item of objArray2
// Return the ones that match. 
const syncedUsers = objArray1.filter((item) => {
const found = objArray2.find((element) => element.id === item.id);
return found;
});

所需的JSON格式，请注意，所有匹配项应从objArray1返回，除了objArray2中的'aid'

{
"records": [
{
"id": {aid},                // from objArray2
"fields": {
"Name":{name},          // from objArray1  
"sid": {id}             // from objArray1
}
]
}

Answer 1

下面是实现预期目标的一种可能方法。

<<p>

代码片段/strong>

// method to create result as "JSON"
const createJSON = (arr1, arr2) => (
// use "stringify" to transform JavaScript object into JSON
JSON.stringify({
// set-up the "records" prop
records: arr2
.filter(          // filter to keep only those that have matching "id"
({ id }) => arr1.some(ob => ob.id === id)
)
.map(             // de-structure & rename props to match desired objective
({ id : sid, name : Name, aid: id }) => ({
id,
fields: {Name, sid}
})
)
})
);
// SystemA
const objArray1 = [
{ id: "1", name: "John" },
{ id: "2", name: "Jack" },
{ id: "3", name: "Sam" },
{ id: "4", name: "Bill" },
];
// SystemB
const objArray2 = [
{ id: "1", name: "John", aid:"uuud2905555" },
{ id: "3", name: "Sam", aid:"uuud29029303" },
{ id: "5", name: "Bob", aid:"uuud29435454" },
];
console.log('result as a JSON: ', createJSON(objArray1, objArray2));

.as-console-wrapper { max-height: 100% !important; top: 0 }

在上面的代码片段中添加内联注释。

---

编辑使用array-1中的name和id。UsedrestOfArr1Props

说明array-1对象的所有其他道具被包括在内。

const createJSON = (arr1, arr2) => (
JSON.stringify({
records: arr1
.filter(
({ id }) => arr2.some(ob => ob.id === id)
)
.map(
({ id : sid, name : Name, ...restOfArr1Props }) => ({
id: arr2.find(a2 => a2.id === sid)?.aid ?? 'missing aid',
fields: {
Name, sid, ...restOfArr1Props
},
})
)
})
);
// SystemA
const objArray1 = [
{ id: "1", name: "John", prop1: 'value11', prop2: 'value12' },
{ id: "2", name: "Jack", prop1: 'value21', prop2: 'value22' },
{ id: "3", name: "Sam", prop1: 'value31', prop2: 'value32' },
{ id: "4", name: "Bill", prop1: 'value41', prop2: 'value42' },
];
// SystemB
const objArray2 = [
{ id: "1", name: "John", aid:"uuud2905555" },
{ id: "3", name: "Sam", aid:"uuud29029303" },
{ id: "5", name: "Bob", aid:"uuud29435454" },
];
console.log('result as a JSON: ', createJSON(objArray1, objArray2));

.as-console-wrapper { max-height: 100% !important; top: 0 }

Answer 2

const objArray1 = [
{ id: '1', name: 'John', type: 'bully' },
{ id: '2', name: 'Jack', type: 'sporty' },
{ id: '3', name: 'Sam', type: 'kind' },
{ id: '4', name: 'Bill', type: 'poliet' },
];

const objArray2 = [
{ id: '1', name: 'John', aid: 'uuud2905555' },
{ id: '3', name: 'Sam', aid: 'uuud29029303' },
{ id: '5', name: 'Bob', aid: 'uuud29435454' },
];
const syncedUsers = { records: [] };
for (let user1 of objArray1) {
const foundUser = objArray2.find(user2 => user2.id === user1.id);
if (foundUser) {
syncedUsers.records.push({
id: foundUser.aid,
fields: { sid: user1.id, name: user1.name, ...user1 },
});
}
}
console.log(JSON.stringify(syncedUsers));