Question

我正在研究C＃项目，我想出了一个我要回来的场景 JSON对象，数据集为2个数据表（表和表1）。

如何迭代数据，以便可以将数据复制到2个不同的变量中。

Ajax致电： -

 $(function () {
        $.ajax({
            type: 'POST',
            dataType: 'json',
            contentType: 'application/json',
            url: '@Url.Action("DrawChart")',
            data: '{}',
            success: function (chartsdataSet) {             
                var chartsdata = chartsdataSet.Table[0];
     }
    });

JSON对象： -

 {
      "Table": [
        {
          "PlanName": "Half Yearly",
          "Paymentamount": 17925,
          "SettledAmount": 13300,
          "PackageAmount": 12479
        },
        {
          "PlanName": "Quaretly",
          "Paymentamount": 7900,
          "SettledAmount": 6000,
          "PackageAmount": 8000
        },
        {
          "PlanName": "Yearly",
          "Paymentamount": 8200,
          "SettledAmount": 9000,
          "PackageAmount": 5198
        }
      ],
      "Table1": [
        {
          "year": 2017,
          "monthid": 1,
          "MonthName": "Jan",
          "Paymentamount": 7900,
          "SettledAmount": 6000,
          "PackageAmount": 8000
        },
        {
          "year": 2017,
          "monthid": 2,
          "MonthName": "Feb",
          "Paymentamount": 1500,
          "SettledAmount": 1300,
          "PackageAmount": 1900
        }
      ]
    }

Answer 1

您收到JSON数据。如果要将其用作Javascript对象，则应使用JSON.parse(chartsdataSet)

 $(function() {
     $.ajax({
         type: 'POST',
         dataType: 'json',
         contentType: 'application/json',
         url: '@Url.Action("DrawChart")',
         data: '{}',
         success: function(chartsdataSet) {
             var object = JSON.parse(chartsdataSet);
             var chartsdata = object.Table[0];
         }
     });
 })

迭代多维json

1 个答案: