解析c#中的JSON字符串

Question

好的，有一些关于这方面的帖子。。。我还没有让他们工作。

我有一个简单的对象

{
    "Projects": "Projects",
    "Dashboard": "Dashboard",
    "Applications": "Applications",
    "Plans": "Plans",
    "Logout": "Logout"
}

我正在尝试反序列化它。我有Newtonsoft可用。

这是我所拥有的不起作用的东西。

var languageString = "{"Projects":"Projects","Dashboard":"Dashboard","Applications":"Applications","Plans":"Plans","Logout":"Logout"}";
var objOut = new { name = "", lastname = "" };
var result = JsonConvert.DeserializeAnonymousType(languageString, objOut);

我一直在匿名对象上收到一个错误。

如有任何帮助，我们将不胜感激。

感谢

Answer 1

要反序列化对象，需要确保用于反序列化对象的类或匿名类型与源JSON字符串匹配。

如果您有一个JSON { "name": "john", "lastname": "smith" }，那么您的代码应该可以工作：

// I expect no one to have concerns about the double quots (""): it's
// another way of escaping quots in C# (you should also use " but it's
// uglier, isn't it? ;)
string jsonText = @"{ ""name"": ""john"", ""lastname"": ""smith"" }";
var sampleType = new { name = "", lastname = "" };
var deserializedObject = JsonConvert.DeserializeAnonymousType(jsonText, sampleType );

在你的情况下

您需要一个匿名类型实例示例，如下所示：

var sampleType = new 
{
    Projects = "",
    Dashboard = "",
    Applications = "",
    Plans = "",
    Logout = ""
}

或者使用动态对象呢

您还可以使用dynamic将对象反序列化为ExpandoObject，并且不需要示例类型！

dynamic deserializedObject = 
       JsonConvert.DeserializeObject(languageString, new ExpandoObjectConverter());

请记住，这是鸭子类型：直到运行时，您才知道反序列化对象是否具有预期的属性。顺便说一句，文档丰富的代码可以做到这一点，如果您希望反序列化的对象具有类似lastname的属性，则可以使用deserializedObject.lastname直接访问它！

Answer 2

 var languageString = @"{""Projects"":""Projects"",""Dashboard"":""Dashboard"",""Applications"":""Applications"",""Plans"":""Plans"",""Logout"":""Logout""}";
        var objOut = new { Projects = "", Dashboard = "", Applications = "", Plans = "", Logout = "" };
        var result = JsonConvert.DeserializeAnonymousType(languageString, objOut);

这对你有用。

使用您的代码，我没有得到任何错误。请详细解释您的错误

Answer 3

我在JSON字符串中看不到name或lastname

逐字字符串：

var languageString = @"{""Projects"":""Projects"",""Dashboard"":""Dashboard"",""Applications"":""Applications"",""Plans"":""Plans"",""Logout"":""Logout""}";

反向冲击：

var languageString = "{"Projects":"Projects","Dashboard":"Dashboard","Applications":"Applications","Plans":"Plans","Logout":"Logout"}";

现在，当你正确处理了这个字符串后，如果你愿意反序列化JSON，你应该创建一个合适的类来匹配它

var objOut = new
{
    Projects = "",
    Dashboard = "",
    Applications = "",
    Plans = "",
    Logout = ""
};

解决所有这些问题后，您将在匿名类的对象实例中检索JSON值。

Answer 4

在我看来，反序列化为强类型类通常更容易。考虑到您提供的JSON，您的反序列化类将如下所示：

public class RootObject
{
    public string Projects { get; set; }
    public string Dashboard { get; set; }
    public string Applications { get; set; }
    public string Plans { get; set; }
    public string Logout { get; set; }
}

尽管这个JSON相当简单，但如果它更复杂，您可以使用http://json2csharp.com以从JSON自动生成C#类。

那么你所需要做的就是：

var languageString = "{"Projects":"Projects","Dashboard":"Dashboard","Applications":"Applications","Plans":"Plans","Logout":"Logout"}";
var result = JsonConvert.DeserializeObject<RootObject>(languageString);