Question

我有一个对象数组，

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

我需要根据键删除一个对象。例如，从密钥具有All的数组中删除对象。

答案应该是：

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

我只需要使用es6语法（如filter或find

）执行此操作我试过的是：

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

Answer 1

使用过滤器并查看Object.keys是否包含不需要的密钥

＆＃13;

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

＆＃13;

Answer 2

如果您的阵列没有重复的对象，那么您可以使用destructuring assignment

以优雅的方式完成

＆＃13;

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

＆＃13;

更通用的方法是使用Array.prototype.filter方法：

＆＃13;

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}]

＆＃13;

Answer 3

您可以使用Array.prototype.filter和“in”运算符：

onLangChange(langObj) {
        this.lanId = this.commonService.getCurrentLanguageCode()
        $( ".instartdate input, .inenddate input, .crstartdate input, .crenddate input" ).datepicker( "option", $.datepicker.regional[this.lanId]);

    })
}

Answer 4

试试这个：

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'));

根据键从对象数组中删除对象

4 个答案: