JSON反序列化到一个对象 [英] Deserializing JSON into an object

问题描述

我有一些JSON：

{
    "foo" : [ 
        { "bar" : "baz" },
        { "bar" : "qux" }
    ]
}

和我想反序列化到一个集合这一点。我已经定义了这个类：

And I want to deserialize this into a collection. I have defined this class:

public class Foo
{
    public string bar { get; set; }
}

但是，下面的code不工作：

However, the following code does not work:

 JsonConvert.DeserializeObject<List<Foo>>(jsonString);

我如何反序列化JSON我？

How can I deserialize my JSON?

推荐答案

这JSON是不是富 JSON阵列。在code JsonConvert.DeserializeObject＆LT; T＆GT;（jsonString）将从根解析最多 JSON字符串，然后你键入 T 必须在JSON结构完全相同。解析器是不会去猜测的这的JSON成员应该重新present在列表＆LT;富方式＆gt; 你正在寻找

That JSON is not a Foo JSON array. The code JsonConvert.DeserializeObject<T>(jsonString) will parse the JSON string from the root on up, and your type T must match that JSON structure exactly. The parser is not going to guess which JSON member is supposed to represent the List<Foo> you're looking for.

您需要一个根对象时，重新$ P $从根元素psents的JSON。

You need a root object, that represents the JSON from the root element.

您可以轻松地让这从一个样本JSON生成的类的事。要做到这一点，复制您的JSON，然后点击编辑 - ＆GT;选择性粘贴 - ＆GT;粘贴JSON作为在Visual Studio中的类。

You can easily let the classes to do that be generated from a sample JSON. To do this, copy your JSON and click Edit -> Paste Special -> Paste JSON As Classes in Visual Studio.

另外，你可以做同样的 http://json2csharp.com ，产生或多或少相同的类。

Alternatively, you could do the same on http://json2csharp.com, which generates more or less the same classes.

您将看到该集合实际上是一个元素深度超过预期：

You'll see that the collection actually is one element deeper than expected:

public class Foo
{
    public string bar { get; set; }
}

public class RootObject
{
    public List<Foo> foo { get; set; }
}

现在，你可以反序列化从根本上JSON（并确保重命名 RootObject 有用的东西）：

Now you can deserialize the JSON from the root (and be sure to rename RootObject to something useful):

var rootObject = JsonConvert.DeserializeObject<RootObject>(jsonString);

和访问集合：

foreach (var foo in rootObject.foo)
{
    // foo is a `Foo`

}

您可以随时重命名的属性按照你的外壳会议及应用 JsonProperty 属性对他们说：

You can always rename properties to follow your casing convention and apply a JsonProperty attribute to them:

public class Foo
{
    [JsonProperty("bar")
    public string Bar { get; set; }
}

另外，还要确保JSON包含足够的样本数据。类解析器将不得不猜测基于JSON中发现的内容相应的C＃类型。

Also make sure that the JSON contains enough sample data. The class parser will have to guess the appropriate C# type based on the contents found in the JSON.

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