在 c# 中解析 Json 数据以转换为 xml

本文关键字：转换 xml 数据 Json | 更新日期: 2023-09-27 18:37:04

在我的 asp.net 应用程序中，我必须将xml格式的数据传递给存储过程。

我拥有的数据采用 Json 格式，如下所示：

{
  "total": 10,
  "page": 1,
  "records": 100,
  "rows": [
    {
      "id": 0,
      "cell": [
        "217121",
        "1001",
        "CAMRY",
        "2532",
        "2532",
        "0",
        "36",
        "8",
        "13",
        "0.03",
        "0.15",
        "0.15",
        "0.10",
        "0.14"
      ]
    },
    {
      "id": 1,
      "cell": [
        "217121",
        "1001",
        "CAMRY",
        "2540",
        "2540",
        "0",
        "6",
        "18",
        "13",
        "0.29",
        "0.14",
        "0.14",
        "0.21",
        "0.27"
      ]
    },
    {
      "id": 2,
      "cell": [
        "217121",
        "1001",
        "CAMRY",
        "2546",
        "2546",
        "0",
        "9",
        "3",
        "5",
        "0.13",
        "0.35",
        "0.35",
        "0.24",
        "-0.32"
      ]
    },
    {
      "id": 3,
      "cell": [
        "217121",
        "1001",
        "CAMRY",
        "2548",
        "2548",
        "0",
        "29",
        "8",
        "14",
        "0.30",
        "0.16",
        "0.16",
        "0.23",
        "-0.41"
      ]
    },
    {
      "id": 4,
      "cell": [
        "217121",
        "1001",
        "CAMRY",
        "2550",
        "2550",
        "0",
        "31",
        "17",
        "7",
        "0.12",
        "0.10",
        "0.10",
        "0.11",
        "-0.14"
      ]
    },
    {
      "id": 5,
      "cell": [
        "217121",
        "1001",
        "CAMRY",
        "2554",
        "2554",
        "0",
        "20",
        "37",
        "3",
        "0.13",
        "0.10",
        "0.10",
        "0.11",
        "-0.62"
      ]
    }
  ]
}

在我的按钮单击中，我想解析它并将其转换为 xml。我正在使用命名空间 

using Newtonsoft.Json.Linq;

并使用以下代码，

protected void TEST_Click(object sender, EventArgs e)
{
    var modelJson = hdnModelObject.Value;
    var obj=JObject.Parse(modelJson);
}

但不知道如何循环它。 卡在这一行。如果有任何疑问，那就太好了。问候

在 c# 中解析 Json 数据以转换为 xml

如何从文档中阅读有关 Json.Net

反序列化XmlNode

JsonConvert 上的第二个帮助程序方法是 DeserializeXmlNode（）。这 方法采用 JSON 文本并将其反序列化为 XmlNode。

因为有效的 XML 必须有一个 JSON 传递给的根元素 反序列化XmlNode在根JSON对象中应该有一个属性。 如果根 JSON 对象具有多个属性，则重载 还取一个元素名称应该使用。具有该根元素的根元素 名称将入到反序列化的 XmlNode 中。

你的json有问题，你需要添加一个根元素，所以你可以做这样的事情。NETFiddle：

这是您当前的字符串格式的 json：

var json =
            @"{""total"":10,""page"":1,""records"":100,""rows"":[{""id"":0,""cell"":[""217121"",""1001"",""CAMRY"",""2532"",""2532"",""0"",""36"",""8"",""13"",""0.03"",""0.15"",""0.15"",""0.10"",""0.14""]},{""id"":1,""cell"":[""217121"",""1001"",""CAMRY"",""2540"",""2540"",""0"",""6"",""18"",""13"",""0.29"",""0.14"",""0.14"",""0.21"",""0.27""]},{""id"":2,""cell"":[""217121"",""1001"",""CAMRY"",""2546"",""2546"",""0"",""9"",""3"",""5"",""0.13"",""0.35"",""0.35"",""0.24"",""-0.32""]},{""id"":3,""cell"":[""217121"",""1001"",""CAMRY"",""2548"",""2548"",""0"",""29"",""8"",""14"",""0.30"",""0.16"",""0.16"",""0.23"",""-0.41""]},{""id"":4,""cell"":[""217121"",""1001"",""CAMRY"",""2550"",""2550"",""0"",""31"",""17"",""7"",""0.12"",""0.10"",""0.10"",""0.11"",""-0.14""]},{""id"":5,""cell"":[""217121"",""1001"",""CAMRY"",""2554"",""2554"",""0"",""20"",""37"",""3"",""0.13"",""0.10"",""0.10"",""0.11"",""-0.62""]}]}";

这是包装在根元素中的 json 字符串：

var jsonWithRoot = string.Format("{{'root': {0}}}",json);

这是您的 XMLDocument：

XmlDocument doc = (XmlDocument)JsonConvert.DeserializeXmlNode(jsonWithRoot);

控制台应用程序示例：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Xml;
using Newtonsoft.Json;
using System.IO;

public class Program
{
    public static void Main()
    {
        var json =
            @"{""total"":10,""page"":1,""records"":100,""rows"":[{""id"":0,""cell"":[""217121"",""1001"",""CAMRY"",""2532"",""2532"",""0"",""36"",""8"",""13"",""0.03"",""0.15"",""0.15"",""0.10"",""0.14""]},{""id"":1,""cell"":[""217121"",""1001"",""CAMRY"",""2540"",""2540"",""0"",""6"",""18"",""13"",""0.29"",""0.14"",""0.14"",""0.21"",""0.27""]},{""id"":2,""cell"":[""217121"",""1001"",""CAMRY"",""2546"",""2546"",""0"",""9"",""3"",""5"",""0.13"",""0.35"",""0.35"",""0.24"",""-0.32""]},{""id"":3,""cell"":[""217121"",""1001"",""CAMRY"",""2548"",""2548"",""0"",""29"",""8"",""14"",""0.30"",""0.16"",""0.16"",""0.23"",""-0.41""]},{""id"":4,""cell"":[""217121"",""1001"",""CAMRY"",""2550"",""2550"",""0"",""31"",""17"",""7"",""0.12"",""0.10"",""0.10"",""0.11"",""-0.14""]},{""id"":5,""cell"":[""217121"",""1001"",""CAMRY"",""2554"",""2554"",""0"",""20"",""37"",""3"",""0.13"",""0.10"",""0.10"",""0.11"",""-0.62""]}]}";

          var jsonWithRoot = string.Format("{{'root': {0}}}",json);
          XmlDocument doc = (XmlDocument)JsonConvert.DeserializeXmlNode(jsonWithRoot);
          using (var stringWriter = new StringWriter())
           using (var xmlTextWriter = XmlWriter.Create(stringWriter))
           {
             doc.WriteTo(xmlTextWriter);
             xmlTextWriter.Flush();
             Console.Write(stringWriter.GetStringBuilder().ToString());
           }
    }
}

您可以使用 JArray 来解析您的 json 字符串，并迭代数组。

JArray a = JArray.Parse(json);
foreach(var obj in a) {
    //iterator the array
}
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