Question

我是Json的新手，我想从我的JObject中过滤掉不需要的信息。我有一个JObject如下：

 {
    "A": "sr",
    "B": {
      "B1": "some data",
      "B2": "some data,
      "Values": [
        {
          "C1": "some info",
          "C2": someinfo",
          "C3": {
            "D1": "some info",
            "D2": [
              "cat",
              "dog",
              "fish"
            ],
            "D3": "some info"
          },
          "C4": "some info",
          "C5": "some info"
        },
        {
          "C1": "some info",
          "C2": someinfo",
          "C3": {
            "D1": "some info",
            "D2": [
              "cat",
              "dog"
            ],
            "D3": "some info"
          },
          "C4": "some info",
          "C5": "some info"
        },
           {
          "C1": "some info",
          "C2": someinfo",
          "C3": {
            "D1": "some info",
            "D2": [
              "cat",
              "dog"
            ],
            "D3": "some info"
          },
          "C4": "some info",
          "C5": "some info"
        }
      ]
    }
  }

我希望扫描它，如果值[i] .C3.D2不包含＆＃34; fish＆＃34;，我想删除值[i]，所以我想要我的新（编辑后）JObject看起来像这样：

       {
    "A": "sr",
    "B": {
      "B1": "some data",
      "B2": "some data,
      "Values": [
        {
          "C1": "some info",
          "C2": someinfo",
          "C3": {
            "D1": "some info",
            "D2": [
              "cat",
              "dog",
              "fish"
            ],
            "D3": "some info"
          },
          "C4": "some info",
          "C5": "some info"
        },
           {
          "C1": "some info",
          "C2": someinfo",
          "C3": {
            "D1": "some info",
            "D2": [
              "cat",
              "fish"
            ],
            "D3": "some info"
          },
          "C4": "some info",
          "C5": "some info"
        }
      ]
    }
  }

最好和最干净的方法是什么？

Answer 1

示例过滤器奇数：

public class MyClass{
    public string A{get; set;}
    public List<int> B{get; set;}
 }

string js= "{"A" : "1","B" : [1,2,3,4,5,6]}";

MyClass obj  = new System.Web.Script.Serialization.JavaScriptSerializer().Deserialize<MyClass>(js);

obj.B= obj.B.Where(t => t % 2 == 0).ToList();

Answer 2

一些凌乱的Linq会做到这一点。选择Values到D2包含＆＃34; fish＆＃34;然后替换原始对象上的Values。

string jsondata = @"{ ""A"":""sr"", ""B"": { ""B1"":""some data"", ""B2"":""some data"", ""Values"": [{ ""C1"":""some info"", ""C2"":""someinfo"", ""C3"": { ""D1"":""some info"", ""D2"": [ ""cat"", ""dog"", ""fish"" ], ""D3"":""some info"" }, ""C4"":""some info"", ""C5"":""some info"" }, { ""C1"":""some info"", ""C2"":""someinfo"", ""C3"": { ""D1"":""some info"", ""D2"": [ ""cat"", ""dog"" ], ""D3"":""some info"" }, ""C4"":""some info"", ""C5"":""some info"" }, { ""C1"":""some info"", ""C2"": ""someinfo"", ""C3"": { ""D1"":""some info"", ""D2"": [ ""cat"", ""dog"" ], ""D3"":""some info"" }, ""C4"":""some info"", ""C5"":""some info"" } ] } } ";
JObject jobj = JObject.Parse(jsondata);
var filteredValues = jobj["B"]["Values"].Select(j => j).Where(j => j["C3"]["D2"].ToArray().Contains("fish")); 
jobj["B"]["Values"] = JToken.FromObject(filteredValues);

Console.WriteLine(JsonConvert.SerializeObject(jobj));

从Jobject c＃中删除属性

2 个答案: