无法根据给定的数组对数组对象进行排序 你能帮我吗!
const data = [ {
parent_email: 'testuser@mailinator.com',
childSurName: null,
childFirstName: null,
id: 22},
{
parent_email: 'p20@mailinator.com',
childSurName: null,
childFirstName: 'Sachidanand',
id: 31},
{
parent_email: 'aa@aa.aaa',
childSurName: null,
childFirstName: null,
id: 26}];
const order = [ 31,26 ];
const sorted = data.sort((a, b) => (
order.indexOf(a.id) - order.indexOf(b.id)
));
console.log(sorted)
我需要像第二个数组一样进行排序的结果,然后在其他数据之后:
[ {
parent_email: 'p20@mailinator.com',
childSurName: null,
childFirstName: 'Sachidanand',
id: 31},
{
parent_email: 'aa@aa.aaa',
childSurName: null,
childFirstName: null,
id: 26}
{
parent_email: 'testuser@mailinator.com',
childSurName: null,
childFirstName: null,
id: 22},
];
但是我得到的结果是没有得到正确的排序数据:
[{
parent_email:"testuser@mailinator.com",
childSurName:null,
childFirstName:null,
id:22
},
{
parent_email:"p20@mailinator.com",
childSurName:null,
childFirstName:"Sachidanand",
id:31
},
{
parent_email:"aa@aa.aaa",
childSurName:null,
childFirstName:null,
id:26
}]
答案 0 :(得分:2)
您可以在Map内使用Map
和place values along with priority
,然后进行排序
注意:-另外,不要忘记保持搜索值和键的类型相同
const data = [{parent_email: 'testuser@mailinator.com',childSurName: null,childFirstName: null,id: 22},{parent_email: 'p20@mailinator.com',childSurName: null,childFirstName: 'Sachidanand',id: 31},{parent_email: 'aa@aa.aaa',childSurName: null,childFirstName: null,id: 26}];
const order = ['31', '22'];
const mapper = new Map(order.map((v, i) => [parseInt(v), i + 1]))
const sorted = data.sort((a, b) => (
(mapper.get(a.id) || Infinity) - (mapper.get(b.id) || Infinity)
));
console.log(sorted)
答案 1 :(得分:0)
const sorted = data.sort((a, b) => order.indexOf(a.id.toString()) - order.indexOf(b.id.toString()) );
id
必须转换为toString
。
或者,更改
const order = [ '31','22' ];
到
const order = [ 31,22 ];
答案 2 :(得分:0)
问题是由于ID 22
引起的。它在order
数组中不存在,并且indexOf
函数为此ID返回-1
。这意味着top priority
,因此它出现在结果数组的第一位。
您应该处理order
数组中不存在的ID。
例如(用于排序),您可以执行以下操作:
const sorted = data.sort((a, b) => {
let aindex = order.indexOf(a.id);
let bindex = order.indexOf(b.id);
if(aindex === -1) return 1;
if(bindex === -1) return -1;
return aindex - bindex;
});