我看到许多抱怨，删除方法不是内置的。考虑使用Set而不是array——它有内置的添加和删除方法。

Let departments是一个数组。您要从此数组中删除一项。

departments: string[] = [];

 removeDepartment(name: string): void {
    this.departments = this.departments.filter(item => item != name);
  }

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;

下面是一个简单的一行代码，用于按属性从对象数组中删除对象。

delete this.items[this.items.findIndex(item => item.item_id == item_id)];

or

this.items = this.items.filter(item => item.item_id !== item.item_id);

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' ]

let foo_object; // Itemitem(object here) to remove
this.foo_objects = this.foo_objects.filter(obj => return obj !== foo_object);