如何遍历C#中复杂的嵌套Json，并在同一Json中附加一些值集

Question

我需要遍历以下内容，并在同一JSON模型中添加一组附加值，形成最终输出。请帮助我如何递归地添加数据。

不应修改密钥的顺序。最终输出应该是相同的格式。我想将这个JSON反序列化为一些类，更新更多的数据，然后将其序列化回JSON。我们该怎么做？

读取Json文件：输入：

{
"id" : "abv.123",
"developName" : "ABV.123",
"pay25" : {
"0" : {
"data" : "group35",
"divison" : 0,
"pay25" : {
"0" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "name33"        
}
}
},
"1" : {
"data" : "group35",
"divison" : 0,
"pay25" : {
"0" : {
"data" : "group35",
"divison" : 0,
"pay25" : {
"0" : {
"data" : "label",
"divison" : 0,
"text" : "name55",
"color1" : "green"
}
}
},
"1" : {
"data" : "group35",
"divison" : 0,
"pay25" : {
"0" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "fab125"
},
"1" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "gbt333"
},
"2" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test53"
},
"3" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test55"
},
"4" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "teds422"
},
"5" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test66"
},
"6" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test982"
},
"7" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test322"
},
"8" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test633"
},
"9" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test622"
},
"10" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test733"
},
"11" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test71"
}
}
}
}
},
"2" : {
"data" : "group35",
"divison" : 0,
"pay25" : {
"0" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test9134234",
"deps1" : [ {
"datadiv123" : [ "1", "2" ],
"value" : true
} ]
},
"1" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test898932"
},
"2" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "tewst432"
}
}
},
"3" : {
"data" : "group35",
"divison" : 0,
"pay25" : {
"0" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test352322",
"deps1" : [ {
"datadiv123" : [ "test", "tesstt" ],
"value" : true
} ]
},
"1" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test523432"
},
"2" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test634fdg"
},
"3" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "tssg32432",
"validators" : [ {
"data" : "integer"
} ]
},
"4" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test1111"
},
"5" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test2222"
}
}
},
"4" : {
"data" : "group35",
"divison" : 0,
"pay25" : {
"0" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test55632"
}
}
},
"5" : {
"data" : "dynamicgrouptable",
"operation" : [ "add", "delete" ],
"dynamicValue" : {
"0" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test953312"
},
"1" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test953312622"
},
"2" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test9533126499"
},
"3" : {
"data" : "divisiondata",
"divison" : 0,
"divName" : "test953312644"
}
}
}
}
}

从JSON:中读取附加值集

{
"name33": {
"def1": "Value 1",
"tip3": "Tip 53",
"disp3": "develop 1",
"path" : "Displayname"
},
"fab125": {
"def1": "Value 5",
"tip3": "strr11",
"disp3": "develop 2",
"path" : "Displayname"
},
"test322": {
"def1": "Value 2",
"tip3": "dev 53",
"disp3": "develop 3",
"path" : "Displayname"
},
{....},
{.....},
"test953312": {
"def1": "Value 21",
"tip3": "dev 5311",
"disp3": "develop 334",
"path" : "Displayname"
},
"test953312622": {
"def1": "Value 212",
"tip3": "dev 53222",
"disp3": "develop 22113",
"path" : "Displayname"
},
{....}
}

输出应该与两个json合并，而不改变数据的顺序：

Answer 1

解决方案1：

您可以使用CCD_ 1或CCD_。由于您的数据不是正则的，因此反序列化会更加复杂。它可能不会输出额外的字段。

//deserialize
JObject jObj = JObject.Parse(jsonString);
JObject idObj = (JObject)jObj["id"];
JObject idObj0 = (JObject)idObj["0"]["id"]["0"]; // 0 can replace to N, loop can deal with it.
idObj0.add("value", "ewr");
idObj0.add("type", "bool");
//serialize
string resultJsonString = jObj.ToString();

参考：Json.NET修改Json

解决方案2：

如果您喜欢反序列化为类。您可以在.Net 5或.Net Core 3.1 中使用本机库System.Text.Json

public class Model
{
public string reports {get; set;};
public Dictionary<string, IdModel> id {get; set;};
}
public class IdModel
{
[JsonIgnore(Condition = JsonIgnoreCondition.WhenWritingNull)]
public string data {get; set;}
[JsonIgnore(Condition = JsonIgnoreCondition.WhenWritingNull)]
public string text {get; set;}
[JsonIgnore(Condition = JsonIgnoreCondition.WhenWritingNull)]
public string name {get; set;}
[JsonIgnore(Condition = JsonIgnoreCondition.WhenWritingNull)]
public string display {get; set;}
[JsonIgnore(Condition = JsonIgnoreCondition.WhenWritingNull)]
public string value {get; set;}
[JsonIgnore(Condition = JsonIgnoreCondition.WhenWritingNull)]
public string type {get; set;}
[JsonIgnore(Condition = JsonIgnoreCondition.WhenWritingNull)]
public Dictionary<string, IdModel> id {get; set;};
}
var data = JsonSerializer.Deserialize<Model>(jsonString);
data.id["0"].id["0"].value = "ewr";
data.id["0"].id["0"].type = "bool";
.......
string resultJsonString = JsonSerializer.Serialize<Model>(data);

参考：如何在.NET 中序列化和反序列化(封送和反封送(JSON

由于数据复杂，我将选择解决方案1。

这是你可以参考的代码

using Newtonsoft.Json.Linq;
var inputJObject = JObject.Parse(jsonString);
var appendDataJObject = JObject.Parse(appendDataString);
FindAndReplace(inputJObject, appendDataJObject);
string resultString = inputJObject.ToString();
public static void FindAndReplace(
JObject input, 
JObject appendData, 
string keyName = "pay25", 
string toReplaceFieldName = "divName")
{
foreach (var targetKey in ((JObject)input[keyName]))
{
var targetJObject = (JObject)targetKey.Value;
//check whether there is pay25 property in child, if yes, search child recursively
if (targetJObject.ContainsKey(keyName))
{
FindAndReplace(targetJObject, appendData);
}
else
{
//if no, check divName property existence
if (targetJObject.ContainsKey(toReplaceFieldName))
{
string divName = (string)targetJObject[toReplaceFieldName];
//check whether there is data has key = divName
if (appendData.ContainsKey(divName))
{
//append data
foreach (var data in (JObject)appendData[divName])
{
targetJObject.Add(data.Key, data.Value);
}
}
}
}
}
}

最后，我认为这是一个如何递归替换数据的问题。

Answer 2

你可以使用Newtonsoft。Json，它是C#中使用最多的库。因此，首先，您需要为您的数据创建一个模型。我想会是这样的：

public class MyModel{
public List<MyModel> id {get;set;}
.....
}
public class MyTopLevelModel : MyModel{
public string reports{get;set;}
}

所以你可以这样做：

MyTopLevelModel myTopLevelModel = JsonConvert.DeserializeObject<MyTopLevelModel>(json);
//then get the place where you want to place the extra data like 
myTopLevelModel.id[0].id[0].value = "whatever"

之后你可以使用

var result = JsonConvert.SerializeObject(myTopLevelModel);