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我使用 JsonConverter VBA 模块将 Excel 中的数据导出为标准 JSON 格式。JSON 输出类似于以下内容：-

[{
    "Manufacturer": "Ford",
    "Code": 5551234,
    "Model": "Escort"
    "Status": "Available"
  },
  {
    "Manufacturer": "Ford",
    "Code": 5551335,
    "Model": "Mondeo"
    "Status": "Out of stock"
  },
  {
    "Manufacturer": "Ford",
    "Code": 5551240,
    "Model": "Fiesta"
    "Status": "Available"
  },
  {
    "Manufacturer": "BMW",
    "Code": 5552567,
    "Model": "M1"
    "Status": "Available"
  },
  {
    "Manufacturer": "BMW",
    "Code": 5552328,
    "Model": "M2"
    "Status": "Available"
  },
  {
    "Manufacturer": "BMW",
    "Code": 5552573,
    "Model": "M3"
    "Status": "Out of stock"
  }
]

我使用的 VBA 代码如下：-

Sub Excel2JSON()

Dim excelRange As Range
Dim jsonItems As New Collection
Dim jsonDictionary As New Dictionary
Dim jsonFileObject As New FileSystemObject
Dim jsonFileExport As TextStream
Dim i As Long
Dim cell As Variant

Set excelRange = Cells(1, 1).CurrentRegion

For i = 2 To excelRange.Rows.Count
    jsonDictionary("Manufacturer") = Cells(i, 1)
    jsonDictionary("Code") = Cells(i, 2)
    jsonDictionary("Model") = Cells(i, 3)
    jsonDictionary("Status") = Cells(i, 4)

    jsonItems.Add jsonDictionary
    Set jsonDictionary = Nothing
Next i

Set jsonFileExport = jsonFileObject.CreateTextFile(".../cardata.json", True)
jsonFileExport.WriteLine (JsonConverter.ConvertToJson(jsonItems, Whitespace:=3))

End Sub

然后，我使用以下脚本将该 JSON 导入 HTML 表：-

$(document).ready(function() {
  $.ajax({
    url: "cardata.json",
    method: "GET",
    dataType: "json",
    success: function(data) {
      var $tbody = $("table#data tbody");
      $.each(data, function(i, data) {
        var $tr = $("<tr></tr>");
        $tr.appendTo($tbody);
        var $td = $("<td></td>");
        $td.html(data.Manufacturer)
          .appendTo($tr);
        $td = $("<td></td>");
        $td.html(data.Code)
          .appendTo($tr);
        $td = $("<td></td>");
        $td.html(data.Model)
          .appendTo($tr);
        $td = $("<td></td>");
        $td.html(data.Status)
          .appendTo($tr);
      });
    },
  });
});

问题/问题是双重的。我想将 JSON 导入 2 个不同的 HTML 表而不是一个（一个表用于福特，另一个用于宝马等）。我知道下面的 JSON 将与 2 个不同的表一起使用，但是我无法修改 VBA 代码来创建如下所示的 JSON：-

{
  "Ford": [{
      "Code": 5551234,
      "Model": "Escort"
      "Status": "Available"
    },
    {
      "Code": 5551335,
      "Model": "Mondeo",
      "Status": "Out of stock"
    },
    {
      "Code": 5551240,
      "Model": "Fiesta",
      "Status": "Available"
    }
  ],
  "BMW": [{
      "Code": 5552567,
      "Model": "M1",
      "Status": "Available"
    },
    {
      "Code": 5552328,
      "Model": "M2",
      "Status": "Available"
    },
    {
      "Code": 5552573,
      "Model": "M3",
      "Status": "Out of stock"
    }
  ]
}

另一种方法是保持 JSON 相同，但更改脚本将其导入 2 个不同 HTML 表的方式。

4

2 回答 2

1

下面我使用 jQuery 以您想要的格式更改了您的 JSON。

 $(document).ready(function(){

 $.ajax({
          url: "cardata.json",
          method: "GET",
          dataType: "json",
          success: function(data){
            var manufacturers = {};

            data.forEach(function(element) {

               if(!manufacturers[element.Manufacturer])
                   manufacturers[element.Manufacturer] = new Array();

                   var obj = {};
                   obj.Code = element.Code;
                   obj.Model = element.Model;
                   obj.Status = element.Status;
                   manufacturers[element.Manufacturer].push(obj);
            });


           console.log(manufacturers); //As object
           console.log(JSON.stringify(manufacturers)); // In JSON

          }
      });
    });
于 2018-12-08T09:25:59.867 回答
0

使用 Ash 提供的代码，表格现在可以正确填充。有可能我已经附加了很长一段时间的数据，但它正在工作。

$(document).ready(function() {
    $.ajax({
        url: "cardata.txt",
        method: "GET",
        dataType: "json",
        success: function(data) {
            var vendors = {};

            data.forEach(function(element) {

                if (!manufacturers[element.Manufacturer])
                    manufacturers[element.Manufacturer] = new Array();

                var obj = {};
                obj.Code = element.Code;
                obj.Model = element.Model;
                obj.Status = element.Status;
                manufacturers[element.Manufacturer].push(obj);
            });

            var $tbody = $("table#BMW tbody");
            $.each(manufacturers.BMW, function(i, data) {
                var $tr = $("<tr></tr>");
                $tr.appendTo($tbody);
                var $td = $("<td></td>");
                $td.html(data.Code)
                    .appendTo($tr);
                $td = $("<td></td>");
                $td.html(data.Model)
                    .appendTo($tr);
                $td = $("<td></td>");
                $td.html(data.Status)
                    .appendTo($tr);
            });

            var $tbody = $("table#Ford tbody");
            $.each(manufacturers.Ford, function(i, data) {
                var $tr = $("<tr></tr>");
                $tr.appendTo($tbody);
                var $td = $("<td></td>");
                $td.html(data.Code)
                    .appendTo($tr);
                $td = $("<td></td>");
                $td.html(data.Model)
                    .appendTo($tr);
                $td = $("<td></td>");
                $td.html(data.Status)
                    .appendTo($tr);
            });

            console.log(manufacturers); //As object
            console.log(JSON.stringify(manufacturers)); // In JSON

        }
    });
});

于 2018-12-10T13:06:21.563 回答