三种方法

 

var mapColumn = {               "vdoing" : "访问深度",               "_visitorNumber": "访问量",               "_pageViews": "浏览量",               "_jumpOutRate": "跳出率",               "_avgAccessTime": "平均访问时长",               "_allTargetConvRate": "转化率",               "_orderTotalPrice": "总收益"                     };

 

一、原生

var target=[];  var targetl=[];for (var key in mapColumn) {       target.push(key);       targetl.push(mapColumn[key]);}  console.log(target); console.log(targetl);

 

二、Jquery

var m_list = [];var m_lists = [];var m_listall = [];$.each(mapColumn,function(key,value){    m_list.push(key);      //获取所有的key    m_lists.push(value);   //获取所有的value    m_listall.push({       //获取拼接        title:key,        name:value    })})console.log(m_list); console.log(m_lists);console.log(m_listall);

 

三、ES6字符串拼接

var data = [{              "vdoing" : "访问深度",               "_visitorNumber": "访问量",               "_pageViews": "浏览量",               "_jumpOutRate": "跳出率"                    },{              "vdoing" : "访问深度1",               "_visitorNumber": "访问量1",               "_pageViews": "浏览量1",               "_jumpOutRate": "跳出率1"                    }]; for (var i = 0; i < data.length; i++) {     /*es6模板字符串*/     $(".part1").append(`
     
         
      ${data[i].vdoing}         
      ${data[i]._visitorNumber}         
      ${data[i]._pageViews}         
      ${data[i]._jumpOutRate}     
     
`);}

 

---

 

从别的地方看到的不错的习题，实际工作中也常用到

一、从某数据库接口得到如下值：

{ rows: [  ["Lisa", 16, "Female", "2000-12-01"],  ["Bob", 22, "Male", "1996-01-21"] ], metaData: [  {name: "name", note: ''},  {name: "age", note: ''},  {name: "gender", note: ''},  {name: "birthday", note: ''} ]}

 

rows是数据，metaData是对数据的说明。现写一个函数，将上面的Object转化为期望的数组：

[ {name: "Lisa", age: 16, gender: "Female", birthday: "2000-12-01"}, {name: "Bob", age: 22, gender: "Male", birthday: "1996-01-21"},]

 

答案：两种，for循环和reduce

var temparry = [];var result = [];for (var k = 0; k < data.metaData.length; k++) {    var a = data.metaData[k].name;    temparry.push(a);}for (var i = 0; i < data.rows.length; i++) {    var ob = {};    for (var j = 0; j < temparry.length; j++) {        var name = temparry[j];        ob[name] = data.rows[i][j];    }    result.push(ob);}console.log(result);

var result = data.rows.reduce(function(prev1, cur1) {    console.log('prev1:' + prev1);    console.log('cur1:' + cur1);    prev1.push(data.metaData.reduce(function(prev, cur, index) {        prev[cur.name] = cur1[index];        return prev;    }, {}))    return prev1;}, []);//console.log(result);//console.log(result[0]);//console.log(result[1]);

 

二、数组

a = [{id: 10001, name: "Lisa", age: 16},{id: 10002, name: "Bob", age: 22},{id: 10003, name: "Alice", age: 20},];

 

数组

b = [{id: 10001, gender: "Female"},{id: 10002, name: "Bob King", birthday: "1996-01-22"},{id: 10005, name: "Tom", birthday: "2000-01-01"},];

 

写一个函数按id用b更新a,期望得到的结果为：

[{id: 10001, name: "Lisa", age: 16, gender: "Female"},{id: 10002, name: "Bob King", birthday: "1996-01-22", age: 22},{id: 10003, name: "Alice", age: 20},{id: 10005, name: "Tom", birthday: "2000-01-01"},]

 

这个自己写吧，不附答案了