我连接到第三方Web服务,该服务返回一个复杂的
JSON对象,该对象仅包含我实际需要的一些信息.
基本上,我只需要“值”中的数组.从该数组中,我只需要“Id”,“Title”和“Status”属性.
我想将这些属性放入名为Project的c#类中.这是我的班级：
public class Project
{
public String Id { get; set; }
public String Title { get; set; }
public String Status { get; set; }
}
我正在尝试使用此代码来读取JSON并执行转换：
using (WebResponse response = request.GetResponse())
{
using (StreamReader reader = new StreamReader(response.GetResponseStream()))
{
var serializer = new JsonSerializer();
var jsonTextReader = new JsonTextReader(reader);
returnValue = serializer.Deserialize<Project>(jsonTextReader);
}
}
示例JSON：
{
"odata.Metadata":"http://school.edu/Api/1/$Metadata#Projects","odata.count":"3","value":[
{
"odata.id":"http://school.edu/Api/1/Projects('123')","RelatedProjects@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('123')/RelatedProjects","Tags@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('123')/Tags","TimedEvents@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('123')/Categories","ep@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('123')/ep","#CreateLike":{
"target":"http://school.edu/Api/1/Projects('123')/CreateLike"
},"#CreateShortcut":{
"target":"http://school.edu/Api/1/Projects('123')/CreateShortcut"
},"#Play":{
"target":"http://school.edu/Play/123"
},"#SendInvitation":{
"target":"http://school.edu/Api/1/Projects('123')/SendInvitation"
},"#CopyProject":{
"target":"http://school.edu/Api/1/Projects('123')/CopyProject"
},"#AddVideoPodcast":{
"target":"http://school.edu/Api/1/Projects('123')/AddVideoPodcast"
},"#AddEP":{
"target":"http://school.edu/Api/1/Projects('123')/AddEP"
},"Id":"123","Title":"Test Title 1","Status":"Viewable"
},{
"odata.id":"http://school.edu/Api/1/Projects('456')","RelatedProjects@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('456')/RelatedProjects","Tags@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('456')/Tags","TimedEvents@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('456')/Categories","ep@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('456')/ep","#CreateLike":{
"target":"http://school.edu/Api/1/Projects('456')/CreateLike"
},"#CreateShortcut":{
"target":"http://school.edu/Api/1/Projects('456')/CreateShortcut"
},"#Play":{
"target":"http://school.edu/Play/456"
},"#SendInvitation":{
"target":"http://school.edu/Api/1/Projects('456')/SendInvitation"
},"#CopyProject":{
"target":"http://school.edu/Api/1/Projects('456')/CopyProject"
},"#AddVideoPodcast":{
"target":"http://school.edu/Api/1/Projects('456')/AddVideoPodcast"
},"#AddEP":{
"target":"http://school.edu/Api/1/Projects('456')/AddEP"
},"Id":"456","Title":"Test Title 2",{
"odata.id":"http://school.edu/Api/1/Projects('789')","RelatedProjects@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('789')/RelatedProjects","Tags@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('789')/Tags","TimedEvents@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('789')/Categories","ep@odata.navigationLinkUrl":"http://school.edu/Api/1/Projects('789')/ep","#CreateLike":{
"target":"http://school.edu/Api/1/Projects('789')/CreateLike"
},"#CreateShortcut":{
"target":"http://school.edu/Api/1/Projects('789')/CreateShortcut"
},"#Play":{
"target":"http://school.edu/Play/789"
},"#SendInvitation":{
"target":"http://school.edu/Api/1/Projects('789')/SendInvitation"
},"#CopyProject":{
"target":"http://school.edu/Api/1/Projects('789')/CopyProject"
},"#AddVideoPodcast":{
"target":"http://school.edu/Api/1/Projects('789')/AddVideoPodcast"
},"#AddEP":{
"target":"http://school.edu/Api/1/Projects('789')/AddEP"
},"Id":"789","Title":"Test Title 3","Status":"Viewable"
}
],"odata.nextLink":"http://school.edu/Api/1/Folders('xyz')/Projects?$skip=10&$top=10"
}
我只是得到一个null对象.但是在调试器中,我可以看到它正在从Web服务中提取所有JSON数据.
如何从JSON获得我需要的东西,构建我的c#对象,并忽略所有其余的东西？
解决方法
如果你可以使用Json.NET(Newtonsoft json),你可以像这样利用Linq-to-Json [1]
//using Newtonsoft.Json.Linq;
var jsonString = File.ReadAllText(@"C:YourDirectory\file.json"); //source
var projects = new List<Project>(); //Your result
JObject data = JObject.Parse(jsonString);
foreach (var value in data["value"])
{
projects.Add(new Project
{
Id = value["Id"].ToString(),Status = value["Status"].ToString(),Title = value["Title"].ToString()
});
}
或者,您也可以像这样反序列化JObject [2]
var jsonReader = data["value"].CreateReader(); projects = new JsonSerializer().Deserialize<List<Project>>(jsonReader);
两者都很好,但哪一个更好？
第二种方法意味着更少的代码(特别是,如果在Project类中有许多属性,则必须编写许多代码行来映射代码[1]中的每个属性).
但第一种方法的表现要好很多倍！对于给定的json数据,代码[1]大约在1毫秒内运行,而代码[2]需要超过100毫秒！
更新
在James Newton-King(编写Json.NET :)的输入之后,还有另一种更优雅的方式来做到这一点[3]
projects = data["value"].ToObject<List<Project>>();