首頁猿問從一個對象中刪除與其他對象項匹配的項

從一個對象中刪除與其他對象項匹配的項

JavaScript

MMMHUHU 2022-10-08 17:53:21

我有一個對象結構如下"designProjects": [ { "projectNumber": "number1", "name": "test1" }, { "projectNumber": "number2", "name": "test2" }, { "projectNumber": "number3", "name": "test3" },]我有另一個具有如下結構的對象"allProjects": [ { "project": { "name": "test1", "number": "number1" }, "employee": { "displayName": "name1" }, "projectRoleName": "Editor" }, { "project": { "name": "test2", "number": "number2" }, "employee": { "displayName": "name2" }, "projectRoleName": "Editor" },]我有點像下面這樣的結果"designProjects": [ { "projectNumber": "number3", "name": "test3" },]這里的結果 designProjects只有一個，因為項目編號和名稱與對象project數組匹配allprojects。有沒有辦法我們可以在react js. 任何建議或想法都會非常感謝我，非常感謝提前

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2 回答

回首憶惘然

TA貢獻1847條經驗獲得超11個贊

您只需與結合.filter使用.some，例如：

let d = {

"designProjects": [

{

"projectNumber": "number1",

"name": "test1"

{

"projectNumber": "number2",

"name": "test2"

{

"projectNumber": "number3",

"name": "test3"

]

}

let a = {

"allProjects": [

{

"project": {

"name": "test1",

"number": "number1"

"employee": {

"displayName": "name1"

"projectRoleName": "Editor"

{

"project": {

"name": "test2",

"number": "number2"

"employee": {

"displayName": "name2"

"projectRoleName": "Editor"

]

};

console.log(

d.designProjects.filter((designProject) => {

return !a.allProjects.some((project) => designProject.projectNumber === project.project.number && designProject.name === project.project.name);

})

);

反對回復 2022-10-08

慕村225694

TA貢獻1880條經驗獲得超4個贊

您可以結合過濾器和一些

filter用于根據條件返回一個新的過濾數組

some將成為條件并在找到匹配項后立即返回

const designProjects = [{

"projectNumber": "number1",

"name": "test1"

{

"projectNumber": "number2",

"name": "test2"

{

"projectNumber": "number3",

"name": "test3"

];

const allProjects = [{

"project": {

"name": "test1",

"number": "number1"

"employee": {

"displayName": "name1"

"projectRoleName": "Editor"

{

"project": {

"name": "test2",

"number": "number2"

"employee": {

"displayName": "name2"

"projectRoleName": "Editor"

]

const cleaned = designProjects.filter((x) => {

return !allProjects.some(y => y.project.number === x.projectNumber);

});

console.info(cleaned);

反對回復 2022-10-08

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從一個對象中刪除與其他對象項匹配的項

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