还有一个使用Typescript的解决方案:

let updatedArray = [];
for (let el of this.oldArray) {
    if (el !== elementToRemove) {
        updated.push(el);
    }
}
this.oldArray = updated;

_.pull(array,'a'); 

使用lib lodash https://lodash.com/docs/4.17.15#pull complelte代码:

import _ from 'lodash';
const allTagList = ['a','b','b']
_.pull(allTagList, b);
console.log(allTagList) // result: ['a']

PS: Lodash提供了大量的操作符，建议使用它来简化您的代码。https://lodash.com

我们可以使用筛选器和包含来实现逻辑

const checkAlpha2Code = ['BD', 'NZ', 'IN'] let countryAlpha2Code = ['US', 'CA', 'BD', 'NZ', 'AF' , 'AR' , 'BR'] /** * Returns the modified array countryAlpha2Code * after removing elements which matches with the checkAlpha2Code */ countryAlpha2Code = countryAlpha2Code.filter(alpha2code => { return !checkAlpha2Code.includes(alpha2code); }); console.log(countryAlpha2Code) // Output: [ 'US', 'CA', 'AF', 'AR', 'BR' ] // Resetting the values again countryAlpha2Code = ['US', 'CA', 'BD', 'NZ', 'AF' , 'AR' , 'BR'] /** * Returns the modified array countryAlpha2Code * which only matches elements with the checkAlpha2Code */ countryAlpha2Code = countryAlpha2Code.filter(alpha2code => { return checkAlpha2Code.includes(alpha2code); }); console.log(countryAlpha2Code) // Output: [ 'BD', 'NZ' ]

如果你需要从数组中删除一个给定的对象，并且你想确定以下情况，请使用此方法:

列表没有重新初始化 正确更新数组长度

    const objWithIdToRemove;
    const objIndex = this.objectsArray.findIndex(obj => obj.id === objWithIdToRemove);
    if (objIndex > -1) {
      this.objectsArray.splice(objIndex, 1);
    }

使用TypeScript扩展运算符(…)

// Your key
const key = 'two';

// Your array
const arr = [
    'one',
    'two',
    'three'
];

// Get either the index or -1
const index = arr.indexOf(key); // returns 0


// Despite a real index, or -1, use spread operator and Array.prototype.slice()    
const newArray = (index > -1) ? [
    ...arr.slice(0, index),
    ...arr.slice(index + 1)
] : arr;

还有一个使用Typescript的解决方案:

let updatedArray = [];
for (let el of this.oldArray) {
    if (el !== elementToRemove) {
        updated.push(el);
    }
}
this.oldArray = updated;