根據鍵從對象數組中刪除一個對象

Question

我有一個對象數組

arrayOfObj = [{All: true}, {Sports: true}, {Entertainment: true}]

我需要根據鍵刪除一個對象。 例如，從鍵為All的數組中刪除一個對象。

答案應該是：

arrayOfObj = [{Sports: true}, {Entertainment: true}]

我只需要使用es6語法（例如filter或find）來執行此操作

我試過的是：

const o = Object.keys(obj);
      this.categoriesSelected = this.categoriesSelected.filter(r => r.o !== o);

Answer 1

使用過濾器，查看Object.keys是否包含不需要的密鑰

 let arrayOfObj = [{All: true}, {Sports: true}, {Entertainment: true}] let unwantedKey = 'All'; let res = arrayOfObj.filter(e => ! Object.keys(e).includes(unwantedKey)); console.log(res); 

Answer 2

如果您的數組中沒有重復的對象，則可以使用destructuring assignment以一種優雅的方式進行操作

 const array = [{All: true}, {Sports: true}, {Entertainment: true}]; const [All, ...rest] = array; console.log(rest) // [{Sports: true}, {Entertainment: true}] 

更通用的方法是使用Array.prototype.filter方法：

 const array = [{All: true}, {All: true}, {Sports: true}, {Entertainment: true}]; const filterBy = (array, key) => array.filter(el => !el[key]) const result = filterBy(array, 'All'); console.log(result) // [{Sports: true}, {Entertainment: true}] 

Answer 3

您可以使用Array.prototype.filter和' in '運算符：

let arrayOfObj = [{All: true}, {Sports: true}, {Entertainment: true}]
let key= 'All';
let res = arrayOfObj.filter(o => !(key in o));

Answer 4

嘗試這個：

let arrayOfObj = [{ All: true }, { Sports: true }, { Entertainment: true }];

function removeProperty(arr, key) {
    return arr.filter((obj) => Object.getOwnPropertyNames(obj)[0] != key);        
}

console.log(removeProperty(arrayOfObj, 'All'));