Json反序列化器或正则表达式或Json解析以在c#中转换Json字符串

Question

我有以下 Json 文件，我想将其转换为不同的格式。新格式如下所示。 我们可以为此使用 Json Parsing 吗？或者我们可以使用正则表达式来更改修改json字符串。

我认为我们还可以使用自定义 json 反序列化器。有人可以帮我用一些简单的方法来做到这一点吗？

当前格式

 {
"PRETEST":
{"response":
{"a":{"source":"VeD","name":"a","value":null},
"b":{"source":"VeD","name":"b","value":"2XX"},
"c":{"source":"VeD","name":"c","value":"4933011630372431565180"},
"d":{"source":"VeD","name":"d","value":null},
"e":{"source":"VeD","name":"e","value":"EHD453REN00000004"},
"f":{"source":"VeD","name":"f","value":"HU55"},
"g":{"source":"VeD","name":"g","value":"453"},
"h":{"source":"VeD","name":"h","value":null},
"i":{"source":"VeD","name":"i","value":null}},
"httpcode":200},
"TEST":
{"response":
{"a":{"source":"VeD","name":"a","value":null},
"b":{"source":"VeD","name":"b","value":"3XX"},
"c":{"source":"VeD","name":"c","value":"5933011630372431565180"},
"d":{"source":"VeD","name":"d","value":null},
"e":{"source":"VeD","name":"e","value":"FHD453REN00000004"},
"f":{"source":"VeD","name":"f","value":"HU55"},
"g":{"source":"VeD","name":"g","value":"433"},
"h":{"source":"VeD","name":"h","value":null},
"i":{"source":"VeD","name":"i","value":null}},
"httpcode":200},
"INT":
{"response":
{"a":{"source":"VeD","name":"a","value":null},
"b":{"source":"VeD","name":"b","value":"4XX"},
"c":{"source":"VeD","name":"c","value":"1933011630372431565180"},
"d":{"source":"VeD","name":"d","value":null},
"e":{"source":"VeD","name":"e","value":"KHD453REN00000004"},
"f":{"source":"VeD","name":"f","value":"KU55"},
"g":{"source":"VeD","name":"g","value":"253"},
"h":{"source":"VeD","name":"h","value":null},
"i":{"source":"VeD","name":"i","value":null}},
"httpcode":200}}

新格式

{
"PRETEST":
{"response":
{"a"null,
"b":"2XX",
"c":"4933011630372431565180",
"d"::null,
"e":"EHD453REN00000004",
"f":"HU55",
"g":"453",
"h"::null,
"i"::null},
"httpcode":200},
"TEST":
{"response":
{"a":null,
"b""3XX"
"c":"5933011630372431565180",
"d":null,
"e""FHD453REN00000004",
"f""HU55",
"g""433",
"h":null,
"i":null},
"httpcode":200},
"INT":
{"response":
{"a":null,
"b""4XX",
"c":"1933011630372431565180",
"d":null,
"e":"KHD453REN00000004",
"f":"KU55",
"g":"253",
"h":null,
"i":null},
"httpcode":200},
"PREPROD":
{"response":
{"a":null,
"b":"5XX",
"c":"8933011630372431565180",
"d":null,
"e":"EHD453REN00000004",
"f":"HU55",
"g":"453",
"h":null,
"i":null},
"httpcode":200}}

Answer 1

这是一个正则表达式的解决方案。

using System;
using System.Text.RegularExpressions;

public class Example
{
   public static void Main()
   {
      string pattern = @"\{[^{}]+value\":([^{}]+)\}";
      string replacement = "$1";
      Regex rgx = new Regex(pattern);
      string input = YOUR_JSON_STRING;
      string result = rgx.Replace(input, replacement);
      Console.WriteLine(result);
   }
}

Answer 2

是的。你可以。访问这个网站（http://jsonutils.com/）并使用相应的数据注释制作相应的类，最后使用NewtonSoft（http://www.newtonsoft.com/json）渲染您的json。