从Jobject删除属性c#
问题描述:
Im新的Json和我我想从我的JObject过滤不需要的信息。 我有一个JObject如下:从Jobject删除属性c#
{
"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"
}
]
}
}
,我想进行扫描,如果值[I] .C3.D2不包括 “鱼”,我想删除值[I],所以我想我的新(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"
}
]
}
}
什么是最好的和最干净的方式呢?
答
样品过滤奇数:
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();
答
有些凌乱的LINQ将做到这一点。选择Values
到一个新对象中,其中D2
包含“鱼”,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));
将您的jsons转换为对象列表,使用linq过滤并从对象再生成jsons。 –
怎么样?它已经是一个对象(JObject)。 – voldemort