我有一个对象数组,如:
types: [
{
id: 1,
name: "Hello"
},
{
id: 2,
name: "World"
},
{
id: 3,
name: "Jon Doe"
}
]
我也有一个像这样的简单数组:
selected_types = [1, 2]
所需结果应过滤“类型” 数组,并排除“ selected_types” 数组中存在的所有ID,如下所示:
final_types: [
{
id: 3,
name: "Jon Doe"
}
]
我完全不知道如何实现这一目标,但是下面是我的尝试:
this.types.filter(obj => {
for (let i = 0; i < this.selected_types.length; i++) {
if (obj.id !== selected_types[i]) {
final_types.push(attribute);
}
}
});
答案 0 :(得分:3)
只需使用this.types.filter(({id}) => !this.selected_types.includes(id))
:
let types = [{
id: 1,
name: "Hello"
},
{
id: 2,
name: "World"
},
{
id: 3,
name: "Jon Doe"
}
]
let selected_types = [1, 2];
let resArr = types.filter(({id}) => !selected_types.includes(id));
console.log(resArr);
答案 1 :(得分:0)
您可以实现Javascript的本机方法过滤器,该过滤器最终会返回一个新对象
let types = [{
id: 1,
name: "Hello"
},
{
id: 2,
name: "World"
},
{
id: 3,
name: "Jon Doe"
}
]
let selected_types = [1, 2];
types = types.filter(obj => {
if (selected_types.indexOf(obj.id) === -1) {
return obj
}
});