Question

我有一个类似以下的JSON响应

{
    "msg": "1",
    "code": "2",
    "data": [
        {
            "a": "3",
            "b": "4"
        }
    ],
    "ts": "5"
}

我想创建一个通用类

public class DTWSResponse<T>
{
    public string msg { get; set; }
    public string code { get; set; }
    public T data { get; set; }
    public long ts { get; set; }
}

因此此类将映射每个变量。但是data部分可以是通用的，即它可能具有不同的格式，而不是2个变量a和b。

所以我创建了另一个类

public class DTProf
{
    public string a { get; set; }
    public string b { get; set; }
}

在我的代码中，我叫

DTWSResponse<DTProf> prof = JsonConvert.DeserializeObject<DTWSResponse<DTProf>>(json);

但是我遇到了以下错误

An exception of type 'Newtonsoft.Json.JsonSerializationException' occurred in Newtonsoft.Json.dll but was not handled in user code

Additional information: Cannot deserialize the current JSON array (e.g. [1,2,3]) into type 'DataTransfer.DTProfile' because the type requires a JSON object (e.g. {"name":"value"}) to deserialize correctly.

To fix this error either change the JSON to a JSON object (e.g. {"name":"value"}) or change the deserialized type to an array or a type that implements a collection interface (e.g. ICollection, IList) like List<T> that can be deserialized from a JSON array. JsonArrayAttribute can also be added to the type to force it to deserialize from a JSON array.

Path 'data', line 1, position 40.

有什么想法吗？

Answer 1

为泛型类型参数使用正确的类型

显示的JSON具有data属性的集合。因此，请使用集合作为类型参数。无需更改通用类。

var prof = JsonConvert.DeserializeObject<DTWSResponse<IList<DTProf>>>(json);
var a = prof.data[0].a;

Answer 2

将data设为通用列表，就可以了...

public class DTWSResponse<T>
{
    public string msg { get; set; }
    public string code { get; set; }
    public IList<T> data { get; set; }
    public long ts { get; set; }
}

Newtonsoft.DeserializeObject泛型类

2 个答案: