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) => {   return a.zhangfu - b.zhangfu})