我有这样一个数组对象，我需要做排序操作

Question

drawerList:[
        {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:+122.33,price:228.253,about:3.3248,state:1},
        {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-12.20,price:68.53,about:3.3248,state:0},
        {url:'../../static/images/icon/coin-gram.png',name:'GRA',zhangfu:+121.13,price:68.3223022,about:3.3248,state:1},
        {url:'../../static/images/icon/coin-eth.png',name:'ETH',zhangfu:+12.83,price:68.25232,about:3.3248,state:1},
        {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-122.03,price:12.25322232,about:3.3248,state:0},
        {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:-12.13,price:68.253232,about:3.3248,state:0},
        {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-9.03,price:68.253232,about:3.3248,state:0},
      ],

我想分别在click事件中，根据zhangfu字段做从大到小和从小到大对数组做排序

请问，这个js方法应该怎么写？ 我尝试了几种写法，好像都不太对

Answer 1

var drawerList = [
    {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:+122.33,price:228.253,about:3.3248,state:1},
    {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-12.20,price:68.53,about:3.3248,state:0},
    {url:'../../static/images/icon/coin-gram.png',name:'GRA',zhangfu:+121.13,price:68.3223022,about:3.3248,state:1},
    {url:'../../static/images/icon/coin-eth.png',name:'ETH',zhangfu:+12.83,price:68.25232,about:3.3248,state:1},
    {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-122.03,price:12.25322232,about:3.3248,state:0},
    {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:-12.13,price:68.253232,about:3.3248,state:0},
    {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-9.03,price:68.253232,about:3.3248,state:0},
]

function sortList(a, b) {
    return b.zhangfu - a.zhangfu
}
drawerList.sort(sortList)
console.log(drawerList)

Answer 2

let obj = {
    drawerList:[
        {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:+122.33,price:228.253,about:3.3248,state:1},
        {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-12.20,price:68.53,about:3.3248,state:0},
        {url:'../../static/images/icon/coin-gram.png',name:'GRA',zhangfu:+121.13,price:68.3223022,about:3.3248,state:1},
        {url:'../../static/images/icon/coin-eth.png',name:'ETH',zhangfu:+12.83,price:68.25232,about:3.3248,state:1},
        {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-122.03,price:12.25322232,about:3.3248,state:0},
        {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:-12.13,price:68.253232,about:3.3248,state:0},
        {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-9.03,price:68.253232,about:3.3248,state:0},
      ]
}

//记得不要污染源数据
let arr = obj.drawerList.concat();
//升序
arr.sort(function(i,j){
    return i.zhangfu - j.zhangfu
})

Answer 3

let data = [
    {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:+122.33,price:228.253,about:3.3248,state:1},
    {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-12.20,price:68.53,about:3.3248,state:0},
    {url:'../../static/images/icon/coin-gram.png',name:'GRA',zhangfu:+121.13,price:68.3223022,about:3.3248,state:1},
    {url:'../../static/images/icon/coin-eth.png',name:'ETH',zhangfu:+12.83,price:68.25232,about:3.3248,state:1},
    {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-122.03,price:12.25322232,about:3.3248,state:0},
    {url:'../../static/images/icon/coin-etc.png',name:'ETC',zhangfu:-12.13,price:68.253232,about:3.3248,state:0},
    {url:'../../static/images/icon/coin-bat.png',name:'BAT',zhangfu:-9.03,price:68.253232,about:3.3248,state:0}
]
// 使用slice拷贝一下数组然后再排序
let result1 = data.slice().sort((a, b) => a.zhangfu - b.zhangfu) // 从小到大
let result2 = data.slice().sort((a, b) => b.zhangfu - a.zhangfu) // 从大到小

Answer 4

drawerList.sort((a, b) => a.zhangfu - b.zhangfu)

Answer 5

drawerList.sort((a,b)=>{return a.zhangfu-b.zhangfu;})

Answer 6

// 从小到大
drawerList.sort((a, b) => {
    return a.zhangfu-b.zhangfu
});

// 从大到小
drawerList.sort((a, b) => {
    return b.zhangfu-a.zhangfu
});

Answer 7

万能的 sort 和给力啊，不要考虑那一堆排序方法了

drawerList = ary.sort((a, b) => a.zhangfu - b.zhangfu)