首頁猿問如果兩個鍵重復，如何過濾掉對象

如果兩個鍵重復，如何過濾掉對象

JavaScript

慕標琳琳 2022-05-22 18:02:39

const obj = [ { "id":"1", "name":"a", "email":"[email protected]", "expiryType":"premium" }, { "id":"2", "name":"b", "email":"[email protected]", "expiryType":"gold" }, { "id":"3", "name":"b", "email":"[email protected]", "expiryType":"premium" }, ]有人可以幫我如何過濾掉電子郵件相同但我想保留過期類型為高級的對象嗎？如何使用 Javascript 實現這一點預期輸出為const obj = [ { "id":"1", "name":"a", "email":"[email protected]", "expiryType":"premium" }, { "id":"3", "name":"b", "email":"[email protected]", "expiryType":"premium" }, ]

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3 回答

小唯快跑啊

TA貢獻1863條經驗獲得超2個贊

假設您想保留最近一年的條目，您可以保留電子郵件地址和您所見年份的地圖?？丛u論：

// The new list

const filtered = [];

// Known emails

const known = new Map();

// Loop through...

for (const entry of obj) {

// Get this email and expiry

const {email, expiryYear} = entry;

// Get the previous info if any

const previous = known.get(email);

if (previous) {

// If the previous one is older than this one,

// replace it with this one

if (previous.expiryYear < expiryYear) {

filtered[previous.index] = entry;

}

} else {

// Add this to the known list and the filtered array

known.set(email, {

index: filtered.length,

expiryYear

});

filtered.push(entry);

}

const obj = [

{

"id":"1",

"name":"a",

"email":"[email protected]",

"expiryYear":"2020"

{

"id":"2",

"name":"a",

"email":"[email protected]",

"expiryYear":"2019"

{

"id":"3",

"name":"b",

"email":"[email protected]",

"expiryYear":"2020"

];

// The new list

const filtered = [];

// Known emails

const known = new Map();

// Loop through...

for (const entry of obj) {

// Get this email and expiry

const {email, expiryYear} = entry;

// Get the previous info if any

const previous = known.get(email);

if (previous) {

// If the previous one is older than this one,

// replace it with this one

if (previous.expiryYear < expiryYear) {

filtered[previous.index] = entry;

}

} else {

// Add this to the known list and the filtered array

known.set(email, {

index: filtered.length,

expiryYear

});

filtered.push(entry);

}

console.log(filtered);

這樣做的好處是不會不斷地重新掃描新列表以查找已知條目。

反對回復 2022-05-22

汪汪一只貓

TA貢獻1898條經驗獲得超8個贊

您可以根據您想要的唯一鍵過濾掉整個對象，如下所示。

const obj =

[

{

"id": "1",

"name": "a",

"email": "[email protected]",

"expiryType": "premium"

{

"id": "2",

"name": "b",

"email": "[email protected]",

"expiryType": "gold"

{

"id": "3",

"name": "b",

"email": "[email protected]",

"expiryType": "premium"

}

]

function arrayUnique(arr, uniqueKey) {

const flagList = []

return arr.filter(function(item) {

if (flagList.findIndex(flagItem => flagItem[uniqueKey] === item[uniqueKey]) === -1) {

flagList.push(item)

return true

}

})

}

方法調用....

let newObj = arrayUnique(obj,'email')

輸出：

newObj = [

{

"id": "1",

"name": "a",

"email": "[email protected]",

"expiryType": "premium"

{

"id": "3",

"name": "b",

"email": "[email protected]",

"expiryType": "premium"

}

]

反對回復 2022-05-22

MYYA

TA貢獻1868條經驗獲得超4個贊

你可以用 2 個循環簡單地做到這一點。也許不是禁食，而是最簡單的：

function deleteDouble(array, objectKey) {

for (var i = 0; i < array.length; i++) {

for (var j = 0; j < array.length; j++) {

if (i == j) {

continue;

}

if (array[i][objectKey] == array[j][objectKey]) {

array.splice(i, 1);

i = 0;

j = 0;

break;

}

return array;

}

deleteDouble(obj, "email");

反對回復 2022-05-22

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