Javascript 我可以把这个循环数组转换成Json对象吗？

我有这个数组

[“20152”、“15301”、“020053”、“2016-08-05”、“05:00”、“06:00”、“07150100190”]、[“20152”、“15301”、“020051”、“2016-08-05”、“03:00”、“04:00”、“07150100190”]、[“20152”、“15301”、“000040”、“2016-08-05”、“01:00”、“02:00”、“07150100190”]、[“20152”、“15301”、“020050”、“2016-08-05”、“07:00”、“08:00”、“07150100190”]

有可能把它改成

var timetable = [{
    "thakad"  : "20152", 
    "prodi"   : "15301",
    "mk"      : "020053",        
    "date"    : "2016-08-05", 
    "open"    : "05:00",
    "close"   : "06:00",
    "number"  : "07150100190"
},
{
    //and keep looping dynamically
}];

---

如果可以，如何在javascript中循环更改的数组？

您可以尝试类似的方法

var array = [["20152","15301","020053","2016-08-05","05:00","06:00","07150100190"],["20152","15301","020051","2016-08-05","03:00","04:00","07150100190"],["20152","15301","000040","2016-08-05","01:00","02:00","07150100190"],["20152","15301","020050","2016-08-05","07:00","08:00","07150100190"]]
var elems = ["thakad","prodi","mk","date","open","close","number"]


var array = array.map(function(arg){
  return (function(){
    var obj = {};
    for (var i = 0;i < elems.length; i++) {
      obj[elems[i]] = arg[i];
    }
    return obj;
  })()
});
array = JSON.stringify(array);
console.log(array);

<代码>var array=[[“20152”、“15301”、“02053”、“2016-08-05”、“05:00”、“06:00”、“07150100190”]，[“20152”、“15301”、“02051”、“2016-08-05”、“03:00”、“04:00”、“07150100190”]、[“20152”、“15301”、“000040”、“2016-08-05”、“01:00”、“02:00”、“07150100190”]，[“20152”、“15301”、“02050”、“2016-08-05”、“07:00”、“08:00”、“07150100190”]] 变量元素=[“thakad”、“prodi”、“mk”、“date”、“open”、“close”、“number”] var array=array.map（函数（arg）{ 返回（函数（）{ var obj={}； 对于（变量i=0；i您可以尝试类似的方法

var array = [["20152","15301","020053","2016-08-05","05:00","06:00","07150100190"],["20152","15301","020051","2016-08-05","03:00","04:00","07150100190"],["20152","15301","000040","2016-08-05","01:00","02:00","07150100190"],["20152","15301","020050","2016-08-05","07:00","08:00","07150100190"]]
var elems = ["thakad","prodi","mk","date","open","close","number"]


var array = array.map(function(arg){
  return (function(){
    var obj = {};
    for (var i = 0;i < elems.length; i++) {
      obj[elems[i]] = arg[i];
    }
    return obj;
  })()
});
array = JSON.stringify(array);
console.log(array);

var数组=[“20152”，“15301”，“020053”，“2016-08-05”，“05:00”，“06:00”，“07150100190”，“20152”，“15301”，“020051”，“2016-08-05”，“03:00”，“04:00”，“07150100190”，“20152”，“15301”，“000040”，“2016-08-05”，“01:00”，“02:00”，“07150100190”，“20152”，“15301”，“020050”，“2016-08-05”，“07:00”，“07150100190”]
变量元素=[“thakad”、“prodi”、“mk”、“date”、“open”、“close”、“number”]
var array=array.map（函数（arg）{
返回（函数（）{
var obj={}；
对于（变量i=0；i

您也可以像下面这样做：-

<script type="text/javascript">
    function test() {
        var json = [["20152", "15301", "020053", "2016-08-05", "05:00", "06:00", "07150100190"], ["20152", "15301", "020051", "2016-08-05", "03:00", "04:00", "07150100190"], ["20152", "15301", "000040", "2016-08-05", "01:00", "02:00", "07150100190"], ["20152", "15301", "020050", "2016-08-05", "07:00", "08:00", "07150100190"]];
        var arr = [];
        $(json).each(function (i) {
            arr.push({
                thakad: json[i][0],
                prodi: json[i][1],
                mk: json[i][2],
                date: json[i][3],
                open: json[i][4],
                close: json[i][5],
                number: json[i][6]
            });
        });         
        console.log(JSON.stringify(arr));
    }
</script>

功能测试（）{
var json=[“20152”，“15301”，“020053”，“2016-08-05”，“05:00”，“06:00”，“07150100190”，“20152”，“15301”，“020051”，“2016-08-05”，“03:00”，“04:00”，“07150100190”，“20152”，“15301”，“000040”，“2016-08-05”，“01:00”，“02:00”，“07150100190”，“20152”，“15301”，“020050”，“2016-08-05”，“07:00”，“08:00”，“07150100190”]；
var-arr=[]；
$（json）。每个（函数（i）{
arr.push({
thakad:json[i][0]，
prodi:json[i][1]，
mk:json[i][2]，
日期:json[i][3],，
open:json[i][4]，
关闭：json[i][5]，
编号:json[i][6]
});
});         
log（JSON.stringify（arr））；
}

您也可以像下面这样做：-

<script type="text/javascript">
    function test() {
        var json = [["20152", "15301", "020053", "2016-08-05", "05:00", "06:00", "07150100190"], ["20152", "15301", "020051", "2016-08-05", "03:00", "04:00", "07150100190"], ["20152", "15301", "000040", "2016-08-05", "01:00", "02:00", "07150100190"], ["20152", "15301", "020050", "2016-08-05", "07:00", "08:00", "07150100190"]];
        var arr = [];
        $(json).each(function (i) {
            arr.push({
                thakad: json[i][0],
                prodi: json[i][1],
                mk: json[i][2],
                date: json[i][3],
                open: json[i][4],
                close: json[i][5],
                number: json[i][6]
            });
        });         
        console.log(JSON.stringify(arr));
    }
</script>

功能测试（）{
var json=[“20152”，“15301”，“020053”，“2016-08-05”，“05:00”，“06:00”，“07150100190”，“20152”，“15301”，“020051”，“2016-08-05”，“03:00”，“04:00”，“07150100190”，“20152”，“15301”，“000040”，“2016-08-05”，“01:00”，“02:00”，“07150100190”，“20152”，“15301”，“020050”，“2016-08-05”，“07:00”，“08:00”，“07150100190”]；
var-arr=[]；
$（json）。每个（函数（i）{
arr.push({
thakad:json[i][0]，
prodi:json[i][1]，
mk:json[i][2]，
日期:json[i][3],，
open:json[i][4]，
关闭：json[i][5]，
编号:json[i][6]
});
});         
log（JSON.stringify（arr））；
}

ES2015版本：

const data = [["20152","15301","020053","2016-08-05","05:00","06:00","07150100190"],["20152","15301","020051","2016-08-05","03:00","04:00","07150100190"],["20152","15301","000040","2016-08-05","01:00","02:00","07150100190"],["20152","15301","020050","2016-08-05","07:00","08:00","07150100190"]]

const arr = data.map(([thakad, prodi, mk, date, open, close, number]) => 
    ({thakad, prodi, mk, date, open, close, number})
);

ES2015版本：

const data = [["20152","15301","020053","2016-08-05","05:00","06:00","07150100190"],["20152","15301","020051","2016-08-05","03:00","04:00","07150100190"],["20152","15301","000040","2016-08-05","01:00","02:00","07150100190"],["20152","15301","020050","2016-08-05","07:00","08:00","07150100190"]]

const arr = data.map(([thakad, prodi, mk, date, open, close, number]) => 
    ({thakad, prodi, mk, date, open, close, number})
);

试试这个

var t = [["20152","15301","020053","2016-08-05","05:00","06:00","07150100190"],["20153","15301","020051","2016-08-05","03:00","04:00","07150100190"],["20154","15301","000040","2016-08-05","01:00","02:00","07150100190"],["20155","15301","020050","2016-08-05","07:00","08:00","07150100190"]];
var res=[];
var keyArray =['thakad','prodi','mk','date','open','close','number'];
for(var i=0;i<t.length;i++){
  var temp={};
   for(var j=0;j<t[i].length;j++){
     temp[keyArray[j]] = t[i][j];
     res[i] = temp;
   }
}
console.log(JSON.stringify(res));

var t=[“20152”，“15301”，“020053”，“2016-08-05”，“05:00”，“06:00”，“07150100190”，“20153”，“15301”，“020051”，“2016-08-05”，“03:00”，“04:00”，“07150100190”，“20154”，“15301”，“000040”，“2016-08-05”，“01:00”，“02:00”，“07150100190”，“20155”，“15301”，“020050”，“2016-08-05”，“07:00”，“07150100190”]；
var-res=[]；
var-keyArray=['thakad'、'prodi'、'mk'、'date'、'open'、'close'、'number']；
对于（var i=0；i试试这个

var t = [["20152","15301","020053","2016-08-05","05:00","06:00","07150100190"],["20153","15301","020051","2016-08-05","03:00","04:00","07150100190"],["20154","15301","000040","2016-08-05","01:00","02:00","07150100190"],["20155","15301","020050","2016-08-05","07:00","08:00","07150100190"]];
var res=[];
var keyArray =['thakad','prodi','mk','date','open','close','number'];
for(var i=0;i<t.length;i++){
  var temp={};
   for(var j=0;j<t[i].length;j++){
     temp[keyArray[j]] = t[i][j];
     res[i] = temp;
   }
}
console.log(JSON.stringify(res));

var t=[“20152”，“15301”，“020053”，“2016-08-05”，“05:00”，“06:00”，“07150100190”，“20153”，“15301”，“020051”，“2016-08-05”，“03:00”，“04:00”，“07150100190”，“20154”，“15301”，“000040”，“2016-08-05”，“01:00”，“02:00”，“07150100190”，“20155”，“15301”，“020050”，“2016-08-05”，“07:00”，“07150100190”]；
var-res=[]；
var-keyArray=['thakad'、'prodi'、'mk'、'date'、'open'、'close'、'number']；
对于（var i=0；i您可以使用
array.map
和
array.reduce


var数组=[“20152”，“15301”，“020053”，“2016-08-05”，“05:00”，“06:00”，“07150100190”，“20152”，“15301”，“020051”，“2016-08-05”，“03:00”，“04:00”，“07150100190”，“20152”，“15301”，“000040”，“2016-08-05”，“01:00”，“02:00”，“07150100190”，“20152”，“15301”，“020050”，“2016-08-05”，“07:00”，“07150100190”]
变量键=[“thakad”、“prodi”、“mk”、“date”、“open”、“close”、“number”]
var r=array.map（函数（el）{
返回键。减少（功能（c、n、i）{
c[n]=el[i]
返回c；
},{});
});
console.log（r）
您可以结合使用
array.map
和
array.reduce


var数组=[“20152”，“15301”，“020053”，“2016-08-05”，“05:00”，“06:00”，“07150100190”，“20152”，“15301”，“020051”，“2016-08-05”，“03:00”，“04:00”，“07150100190”，“20152”，“15301”，“000040”，“2016-08-05”，“01:00”，“02:00”，“07150100190”，“20152”，“15301”，“020050”，“2016-08-05”，“07:00”，“07150100190”]
变量键=[“thakad”、“prodi”、“mk”、“date”、“open”、“close”、“number”]
var r=array.map（函数（el）{
返回键。减少（功能（c、n、i）{
c[n]=el[i]
返回c；
},{});
});
console.log（r）
好的。是的。你在它上面循环。你在每个循环中创建一个对象。那导致你问题的两个非常基本的JavaScript位呢？好的。是的。你在它上面循环。你在每个循环中创建一个对象。那导致你问题的两个非常基本的JavaScript位呢？