如何在C#中获取JSON中的值
本文关键字:JSON 获取 | 更新日期: 2023-09-27 18:22:47
我有json字符串作为:
{
"data": [
{
"id": "100000045402409_310121622373595",
"from": {
"name": "Ritesh Ranjan",
"id": "100000045402409"
},
"message": "greatttttttttttttt ab jaooooooooo",
"picture": "http://external.ak.fbcdn.net/safe_image.php?d=AQAGY5rsr5AeM5PI&w=90&h=90&url=http'u00253A'u00252F'u00252Fwww.ndtv.com'u00252Fnews'u00252Fimages'u00252Ftopstory_thumbnail'u00252FChidambaram_2G_120.jpg",
"link": "http://www.ndtv.com/article/india/2g-scam-chidambaram-verdict-expected-shortly-huge-implications-for-govt-173168",
"name": "2G scam: Chidambaram verdict expected shortly, huge implications for govt",
"caption": "www.ndtv.com",
"description": "A Delhi court handling the 2G spectrum allocation scam trial is likely to decide today whether Union Home Minister P Chidambaram should be made a co-accused in the case for allegedly allowing former Telecom Minister A Raja to gift mobile network licenses and scarce second-generation or 2G spectrum a...",
"icon": "http://static.ak.fbcdn.net/rsrc.php/v1/yD/r/aS8ecmYRys0.gif",
"type": "link",
"application": {
"name": "Links",
"id": "2309869772"
},
"created_time": "2012-02-04T11:02:22+0000",
"updated_time": "2012-02-04T11:02:22+0000"
},
{
"id": "100003303253347_132959650157476",
"from": {
"name": "Suman Dey",
"id": "100003303253347"
},
"message": "Check out this article I was reading on biNu. 2G verdict: Chidambaram off the hook, government exhales",
"type": "status",
"application": {
"name": "biNu",
"canvas_name": "binuapp",
"namespace": "binuapp",
"id": "378628085054"
},
"created_time": "2012-02-04T10:54:19+0000",
"updated_time": "2012-02-04T10:54:19+0000"
},
.....
//Continued...
现在我想用c#解析它
我用过:
WebClient client = new WebClient();
string Json = client.DownloadString("https://graph.facebook.com/search?q=2g+verdict+Chidambaram&type=post");
System.IO.StreamWriter SW = new System.IO.StreamWriter(JsonDestFile);
SW.WriteLine(Json);
System.IO.StreamWriter SW1 = new System.IO.StreamWriter(ValuesDestFile);
JObject o = JObject.Parse(Json);
var postTitles = from p in o["data"].Children()["from"]
select p["name"].Values<string>();
foreach (var item in postTitles)
{
SW1.WriteLine(item);
}
SW1.WriteLine(name);
但我根本无法获得任何name
值。
它给我错误:Cannot access child value on Newtonsoft.Json.Linq.JValue.
请建议我如何解析上面的json以获取id, name, id (from one) , message
的值
我开始工作了。。。
var postTitles = from p in JO["data"].Children()
select new
{
Names = (string)p["from"]["name"],
Msg = (string)p["message"],
};
使用这个LINQ,我可以访问所需的数据。
我还没有使用LINQ to JSON API,但如果你不坚持使用它,你可以简单地创建一个类,为JSON负载中的数据建模,然后使用以下方法:
Newtonsoft.Json.JsonConvert.DeserializeObject<YourDataModelClass>()
您需要取消json字符串的分隔,如下所示
public static T Deserialise<T>(string json)
{
using (var ms = new MemoryStream(Encoding.Unicode.GetBytes(json)))
{
var serialiser = new DataContractJsonSerializer(typeof(T));
return (T)serialiser.ReadObject(ms);
}
}
退货类型是您的
public class MyData
{
public string id { get; set;}
public string name{ get; set;}
public string message{ get; set;}
}
您可以查看完整的详细信息:在C#中解析JSON
这也适用于
JObject hh = JObject.Parse(jsonstring);
string name = (string)hh["Name"];