js数组筛选过滤

Question

var arr = [
        {name:'xiaoming',id:"123"},
        {name:'xiaoming',id:"456"},
        {name:'xiaoming',id:"789"},
        {name:'xiaohua',id:"101112"},
        {name:'xiaowang',id:"131415"},
        {name:'xiaohong',id:"161718"}
    ];
    var obj = [];
    for(var i=0; i<arr.length; i++){
        obj[arr[i]['name']] = arr[i]['id'];
    }
 console.log(obj)

我想把'name'为'xiaoming'的id过滤出来

这是现在的结果,我本来想objarr[i].push() 但用不了这个方法，

不要用if或别的什么判断

，有没有直接暴力的枚举算法

[xiaoming: "789", xiaohua: "101112", xiaowang: "131415", xiaohong: "161718"]

我想实现的是[xiaoming: ['123','456','789']]
或者{name:'xiaoming',id:['123','456','789']}

Answer 1

引入jq文件

var arr = [
        {name:'xiaoming',id:"123"},
        {name:'xiaoming',id:"456"},
        {name:'xiaoming',id:"789"},
        {name:'xiaohua',id:"101112"},
        {name:'xiaowang',id:"131415"},
        {name:'xiaohong',id:"161718"}
    ];

var arrnew = [];
arrnew["xiaoming"] = $.map(arr,item => { if(item.name == "xiaoming") return item.id; });
console.log(arrnew);

或者

 var arrNew = { name: 'xiaoming', id: $.map(arr,item => { if(item.name == "xiaoming") return item.id; }) };
 console.log(arrNew);

Answer 2

这个用filter很好实现的
Nodejs中的一些小trick - QueenKing - SegmentFault https://segmentfault.com/a/11...

Answer 3

var arr = [
        {name:'xiaoming',id:"123"},
        {name:'xiaoming',id:"456"},
        {name:'xiaoming',id:"789"},
        {name:'xiaohua',id:"101112"},
        {name:'xiaowang',id:"131415"},
        {name:'xiaohong',id:"161718"}
    ];
var obj = {};
arr.forEach(person => {
    obj[person.name] = !obj[person.name] ?  [person.id] : [ ...obj[person.name], person.id];
})

没用if…用了三元表达式…

Answer 4

var arr = [
        {name:'xiaoming',id:"123"},
        {name:'xiaoming',id:"456"},
        {name:'xiaoming',id:"789"},
        {name:'xiaohua',id:"101112"},
        {name:'xiaowang',id:"131415"},
        {name:'xiaohong',id:"161718"}
    ];

var obj={};

obj.xiaoming=arr.filter(item=>item.name=='xiaoming').map(item=>item.id);

console.log(obj);

mycodepen

Answer 5

var arr = [
    {name:'xiaoming',id:"123"},
    {name:'xiaoming',id:"456"},
    {name:'xiaoming',id:"789"},
    {name:'xiaohua',id:"101112"},
    {name:'xiaowang',id:"131415"},
    {name:'xiaohong',id:"161718"}
];

res = arr.filter(e=>e.name==='xiaoming').reduce((all,e)=>{
    return {name: 'xiaoming',id:[...all.id , e.id]}
},{name:'xiaoming',id:[]})


console.log(res);

// output
// { name: 'xiaoming', id: [ '123', '456', '789' ] }

Answer 6

首先，你给的参考结果前半句 [xiaoming: ['123','456','789']] 语法上就有问题，我想你是要后半句 {xiaoming: ['123','456','789']} 。

实现过程如下（es6 语法）

const arr = [] // 内容省略
const obj = {}
arr.forEach(item => {
    const name = item.name
    const items = obj[name] || [] // 判断是否要初始化
    items.push(item.id)
    obj[name] = items
})

const xiaoming = obj['xiaoming'] // 过滤出你要的答案