c#从json获取值
我有一个json文本,我想获得author name和description标记的值。不需要url和urltoimage等其他字段。 当我运行下面的代码时,它不提供任何字符串值。我想这里有些错误c#从json获取值,c#,json,json.net,C#,Json,Json.net,我有一个json文本,我想获得author name和description标记的值。不需要url和urltoimage等其他字段。 当我运行下面的代码时,它不提供任何字符串值。我想这里有些错误 { "status": "ok", "articles": [ { "source": { "id": "techcrunch", "name": "TechCrunch" }, "author": "Khaled \"Tito\" Hamze
{
"status": "ok",
"articles": [
{
"source": {
"id": "techcrunch",
"name": "TechCrunch"
},
"author": "Khaled \"Tito\" Hamze",
"title": "Crunch Report",
"description": "Your daily roundup of the biggest TechCrunch stories and startup news.",
"url": "https://techcrunch.com/video/crunchreport/",
"urlToImage": "https://tctechcrunch2011.files.wordpress.com/2015/03/tccrshowogo.jpg?w=500&h=200&crop=1",
"publishedAt": "2017-12-11T20:20:09Z"
},
{
"source": {
"id": "techcrunch",
"name": "TechCrunch"
},
"author": "Sarah Perez",
"title": "Facebook is trying to make the Poke happen again",
"description": "Facebook's \"Poke\" feature has never really gone away, but now the social network is giving it a more prominent placement - and is even considering expanding..",
"url": "https://techcrunch.com/2017/12/11/facebook-is-trying-to-make-the-poke-happen-again/",
"urlToImage": "https://tctechcrunch2011.files.wordpress.com/2017/12/facebook-poke-icon.jpg",
"publishedAt": "2017-12-11T20:02:30Z"
},
{
"source": {
"id": "techcrunch",
"name": "TechCrunch"
},
"author": "Sarah Perez",
"title": "Amazon Alexa can now wake you up to music",
"description": "This fall, Amazon made a play to become your new alarm clock with the introduction of a combination smart speaker and clock called the Echo Spot. Today, the..",
"url": "https://techcrunch.com/2017/12/11/amazon-alexa-can-now-wake-you-up-to-music/",
"urlToImage": "https://tctechcrunch2011.files.wordpress.com/2017/09/amazon-event-9270069.jpg",
"publishedAt": "2017-12-11T17:22:30Z"
},
{
"source": {
"id": "techcrunch",
"name": "TechCrunch"
},
"author": "Ingrid Lunden, Katie Roof",
"title": "Apple confirms Shazam acquisition; Snap and Spotify also expressed interest",
"description": "After we broke the story last week that Apple was acquiring London-based music and image recognition service Shazam, Apple confirmed the news today. It is..",
"url": "https://techcrunch.com/2017/12/11/apple-shazam-deal/",
"urlToImage": "https://tctechcrunch2011.files.wordpress.com/2017/12/shazam-app-icon-ios.jpg",
"publishedAt": "2017-12-11T15:59:31Z"
}
]}
如何得到这个?下面是我的代码,它根本不起作用
var data = (JObject)JsonConvert.DeserializeObject(myJSON);
string nameArticles= data["articles"].Value<string>();
MessageBox.Show(nameArticles);
public class Source
{
public string id { get; set; }
public string name { get; set; }
}
public class Article
{
public Source source { get; set; }
public string author { get; set; }
public string title { get; set; }
public string description { get; set; }
public string url { get; set; }
public string urlToImage { get; set; }
public DateTime publishedAt { get; set; }
}
Article art = new Article();
art = JsonConvert.DeserializeObject<Article>(myJSON);
MessageBox.Show(art.description.ToString());
var data=(JObject)JsonConvert.DeserializeObject(myJSON);
字符串名称articles=data[“articles”].Value();
MessageBox.Show(nameArticles);
公共类源
{
公共字符串id{get;set;}
公共字符串名称{get;set;}
}
公共类文章
{
公共源{get;set;}
公共字符串作者{get;set;}
公共字符串标题{get;set;}
公共字符串说明{get;set;}
公共字符串url{get;set;}
公共字符串urlToImage{get;set;}
public DateTime publishedAt{get;set;}
}
Article art=新文章();
art=JsonConvert.DeserializeObject(myJSON);
MessageBox.Show(艺术描述.ToString());
上面的代码返回对象未设置为实例错误 使用
示例:
String[] your_json_file = Directory.GetFiles("C:\your_folder", json_file_name.json", SearchOption.AllDirectories);
foreach (string s in your_json_file)
{
JObject JSON = JObject.Parse(File.ReadAllText(s));
string AutorName = (string)JSON["author"];
string Description = (string)JSON["description"];
}
请为JSON文件创建一个类,并为所有标记添加属性 然后编写如下代码:
public class exampleJson{
public string author {get;set;}
public string description {get;set;}
.....
}
var data = JsonConvert.DeserializeObject<exampleJson>(myJSON);
string authorName = data.author;
string descriptions = data.description ;
public类exampleJson{
公共字符串作者{get;set;}
公共字符串说明{get;set;}
.....
}
var data=JsonConvert.DeserializeObject(myJSON);
字符串authorName=data.author;
字符串描述=data.description;
假设您希望反序列化到具体的类(按照您问题中尝试的第二种方法),那么您需要一个包装类来保存整个对象,并反序列化到该类
目前,您正试图将整个对象序列化为一篇文章
,但只有该对象的文章
数组中的单个对象才能与文章
类中的结构匹配
您试图在对象的错误级别执行操作,并且忘记了articles
是一个列表(数组)
大概是这样的:
public class JSONResponse
{
public string status { get; set; }
public List<Article> articles { get; set; }
}
公共类JSONResponse
{
公共字符串状态{get;set;}
公共列表项目{get;set;}
}
及
JSONResponse-response=JsonConvert.DeserializeObject(myJSON);
然后,您可以使用一个普通循环来迭代
响应.articles
列表并提取作者姓名和描述。您的Json生成以下一组类
public class Source
{
public string id { get; set; }
public string name{get;set;}
}
public class Article
{
public Source source { get; set; }
public string author { get; set; }
public string title { get; set; }
public string description { get; set; }
public string url { get; set; }
public string urlToImage { get; set; }
public DateTime publishedAt { get; set; }
}
public class RootObject
{
public string status { get; set; }
public List<Article> articles { get; set; }
}
公共类源代码
{
公共字符串id{get;set;}
公共字符串名称{get;set;}
}
公共类文章
{
公共源{get;set;}
公共字符串作者{get;set;}
公共字符串标题{get;set;}
公共字符串说明{get;set;}
公共字符串url{get;set;}
公共字符串urlToImage{get;set;}
public DateTime publishedAt{get;set;}
}
公共类根对象
{
公共字符串状态{get;set;}
公共列表项目{get;set;}
}
因此,您可以通过以下方式对其进行反序列化
var data = JsonConvert.DeserializeObject<RootObject>(myJSON);
nameArticles=data.articles.FirstOrDefault().description;
MessageBox.Show(nameArticles);
var data=JsonConvert.DeserializeObject(myJSON);
nameArticles=data.articles.FirstOrDefault().description;
MessageBox.Show(nameArticles);
数据[“文章”]
可能是JArray
而不是字符串。您需要在前面提到的JArray
中迭代每个JObject
,提取author和description值
var data = (JObject)JsonConvert.DeserializeObject(myJSON);
var articles = data["articles"].Children();
foreach (var article in articles)
{
var author = article["author"].Value<string>();
var description = article["author"].Value<string>();
Console.WriteLine($"Author: " + author + ", Description: " + description);
}
var data=(JObject)JsonConvert.DeserializeObject(myJSON);
var articles=data[“articles”].Children();
foreach(条款中的var条款)
{
var author=article[“author”].Value();
var description=article[“author”].Value();
Console.WriteLine($“Author:+Author+”,Description:+Description);
}
这将有助于您开始做任何事情。如果您不想创建包装类,可以尝试下面的代码段,它使用to
示例json数据
string jsonString = "{\"displayName\":\"Alex Wu\",\"signInNames\":[{\"type\":\"emailAddress\",\"value\":\"AlexW@example.com\"},{\"type\":\"emailAddress\",\"value\":\"AlexW2@example.com\"}]}";
将json转换为jObject,并使用名为selectToken()的内置方法获取值。
谢谢
数据[“文章”]
可能是JArray
而不是字符串。您需要在前面提到的JArray
中迭代每个JObject
,提取author
和description
值。@phuzi您能展示一些示例代码吗?请看下面我的答案1。他们使用的是Json.NET2<代码>“author”和“description”
不是示例中所示的顶级属性。虽然反序列化到具体类可能有好处,但也完全有效。虽然反序列化到具体类可能有好处,也完全有效。虽然反序列化到具体类可能有好处,但也完全有效。@crashmstr这当然是一个同样有效的选择。然而,OP的代码似乎暗示他们想要在o类中反序列化,但却很难正确地做到这一点,所以我的答案是使用相同的方法构造的。你认为答案在任何方面都是错误的吗?他们有两块代码,一块在顶部使用LINQ到JSON,另一块试图反序列化到更远的类。所以他们真正想要的是什么是模棱两可的。我不认为你的答案是错误的,但我认为更好的措辞是,如果他们想反序列化到一个类,他们需要在整个结构的顶层进行反序列化。@Crashstr我添加了一点澄清。我认为这些更改使这是一个更好的答案。感谢这个解决方案,经过这么多的努力,找到了这个,它成功了。!
var json = "Your JSON string";
dynamic stuff = JsonConvert.DeserializeObject(json);
string name = stuff.status;
var arr = stuff.articles;
foreach (var a in arr)
{
var authorName = a.author;
}
string jsonString = "{\"displayName\":\"Alex Wu\",\"signInNames\":[{\"type\":\"emailAddress\",\"value\":\"AlexW@example.com\"},{\"type\":\"emailAddress\",\"value\":\"AlexW2@example.com\"}]}";
JObject jObject = JObject.Parse(jsonString);
string displayName = (string)jObject.SelectToken("displayName");
string type = (string)jObject.SelectToken("signInNames[0].type");
string value = (string)jObject.SelectToken("signInNames[0].value");
Console.WriteLine("{0}, {1}, {2}", displayName, type, value);
JArray signInNames = (JArray)jObject.SelectToken("signInNames");
foreach (JToken signInName in signInNames)
{
type = (string)signInName.SelectToken("type");
value = (string)signInName.SelectToken("value");
Console.WriteLine("{0}, {1}", type, value);
}