如何比较两个数组并创建具有不同键值对的新数组？

Question

我有两个数组，需要获取特定的键值对并分配给第三个数组，它还会处理重复条目。

第一个数组：

[ 
  {id: 2, name: "HSBC", status: "YES"},
  {id: 3, name: "Morgan Stanley", status: "Pending"}
]

第二个数组：

[
   {id: 1, name: "Deutsche Bank", category: "EQUITIES"},
   {id: 2, name: "HSBC", category: "EQUITIES"},
   {id: 3, name: "Morgan Stanley", category: "EQUITIES"},
   {id: 4, name: "Credit Suisse", category: "EQUITIES"},
]

从上面的两个数组中，我需要获取名称和状态字段并创建新数组。 如下所示：

[
  { brokerName: "HSBC", statis: "YES"},
  { brokerName: "Morgan Stanley", statis: "Pending"},
  { brokerName: "Deutsche Bank", statis: ""},
  { brokerName: "Credit Suisse", statis: ""},
]

TIA。 堆栈闪电战

Answer 1

let arr1 = [{id: 1, name: "name", status: "YES"}];
let arr2 = [{id: 1, name: "name", category: "EQUITIES"}];
let sumbit = [];

for (let i=0; i<arr1.length; i++) {
    let is_there = false;
    for (let j=0; j<arr2.length; j++) {
        if (arr1[i].name == arr[j].name) {
            submit.push({brokerName: arr1[i].name, statis: arr1[i].status});
            is_there = true;
        }
    }
    if (!is_there) submit.push({brokerName: arr1[i].name, statis: ""})
}

你可以只写代码：D

Answer 2

这是一个例子：

 const array1 = [ {id: 2, name: "HSBC", status: "YES"}, {id: 3, name: "Morgan Stanley", status: "Pending"} ]; const array2 = [ {id: 1, name: "Deutsche Bank", category: "EQUITIES"}, {id: 2, name: "HSBC", category: "EQUITIES"}, {id: 3, name: "Morgan Stanley", category: "EQUITIES"}, {id: 4, name: "Credit Suisse", category: "EQUITIES"}, ]; // returns { name, status } recoreds from any number of arrays function getBrokersDataFromArrays(...arrays) { const allBrokers = new Map(); // collect all the available data for all the brokers arrays.flat().forEach(broker => { if (allBrokers.has(broker.name)) { allBrokers.set(broker.name, { ...allBrokers.get(broker.name), ...broker }); } else { allBrokers.set(broker.name, broker); } }); // return only { name, status } fields from the collected brokers return Array.from(allBrokers.values()).map(broker => ( { name: broker.name, status: broker.status } )); } const brokersData = getBrokersDataFromArrays(array1, array2); console.log(brokersData);

Answer 3

let array1 = [ 
    {id: 2, name: "HSBC", status: "YES"},
    {id: 3, name: "Morgan Stanley", status: "Pending"}
  ]

let array2 = [
    {id: 1, name: "Deutsche Bank", category: "EQUITIES"},
    {id: 2, name: "HSBC", category: "EQUITIES"},
    {id: 3, name: "Morgan Stanley", category: "EQUITIES"},
    {id: 4, name: "Credit Suisse", category: "EQUITIES"},
 ]
   
 // filter data which are present into both array
    const result1 = array1.filter(o => array2.some(({ name }) => o.name === name));
    
    // filter data which are present into array1 but not into array2
    const result2 = array1.filter((o1) => {
        return !array2.some((o2) => {
            return o1.name === o2.name;
        });
    })
    
    // filter data which are present into array2 but not into array1
    const result3 = array2.filter((o1) => {
        return !array1.some((o2) => {
            return o1.name === o2.name;
        });
    })
    
    // concat all result
    let result = (result1.concat(result2)).concat(result3)
    
    let finalArr = []
    for (let obj of result) {
        finalArr.push({ brokerName: obj.name, status: obj.status ? obj.status : "" })
    }