JSON DeserializeObject 返回 null

Question

我有一個 JSON 響應，其格式如下：

{
    "list": {
        "historySteps": [
            {
                "actualUserDisplayName": "John Doe",
                "actualUserID": "SiteOwner",
                "invokerType": "USER",
                "step": 10,
                "timestampFormatted": "09/23/2019 12:37:04 PM",
                "transitionName": "Work",
                "userDisplayName": "John Doe",
                "userID": "SiteOwner"
            },
            {
                "actualUserDisplayName": "John Doe",
                "actualUserID": "SiteOwner",
                "comments": "Zebra",
                "invokerType": "USER",
                "step": 9,
                "timestampFormatted": "09/23/2019 11:12:30 AM",
                "transitionName": "Return",
                "userDisplayName": "John Doe",
                "userID": "SiteOwner"
            },
            {
                "actualUserDisplayName": "John Doe",
                "actualUserID": "SiteOwner",
                "invokerType": "USER",
                "step": 1,
                "timestampFormatted": "09/23/2019 10:11:48 AM",
                "transitionName": "Initiate",
                "userDisplayName": "John Doe",
                "userID": "SiteOwner"
            }
        ]
    }
}

我有 C# 代碼為：

public historyList GetWorkFlowHistory(string workspaceId, string dmsId)
{
    try {
    var client = new RestClient(TenantUrl + $"/api/rest/v1/workspaces/{workspaceId}/items/{dmsId}/workflows/history");
    client.Proxy = Proxy;
    var request = new RestRequest(Method.GET);
    request.AddCookie("customer", CustomerToken);
    request.AddCookie("JSESSIONID", SessionId);
    request.AddHeader("cache-control", "no-cache");


    IRestResponse response = client.Execute(request);


    if (response.ErrorException != null)
        throw response.ErrorException;
    else if (response.StatusCode != HttpStatusCode.OK)
        throw new Exception(response.Content);

        var results = SimpleJson.SimpleJson.DeserializeObject<historyList>(response.Content);

        return results;

}
    catch (Exception ex)
    {
        Log.WriteLog("Exception in GetWorkFlowHistory() - " + ex.Message);
    }

    return null;

}

[XmlRoot(ElementName = "historySteps")]
public class HistorySteps
{
    [XmlElement(ElementName = "actualUserDisplayName")]
    public string ActualUserDisplayName { get; set; }
    [XmlElement(ElementName = "actualUserID")]
    public string ActualUserID { get; set; }
    [XmlElement(ElementName = "comments")]
    public string Comments { get; set; }
    [XmlElement(ElementName = "invokerType")]
    public string InvokerType { get; set; }
    [XmlElement(ElementName = "step")]
    public string Step { get; set; }
    [XmlElement(ElementName = "timestampFormatted")]
    public string TimestampFormatted { get; set; }
    [XmlElement(ElementName = "transitionName")]
    public string TransitionName { get; set; }
    [XmlElement(ElementName = "userDisplayName")]
    public string UserDisplayName { get; set; }
    [XmlElement(ElementName = "userID")]
    public string UserID { get; set; }
}

[XmlRoot(ElementName = "list")]
public class historyList
{
    [XmlElement(ElementName = "historySteps")]
    public HistorySteps HistorySteps { get; set; }

}

代碼一直運行到 response.Content 檢索內容的位置。 當我嘗試反序列化對象時，我無法弄清楚我做錯了什么。 它返回 historySteps = null。 我錯過了什么？ 我是否采取了正確的方法？

Answer 1

屬性public HistorySteps HistorySteps { get; set; } public HistorySteps HistorySteps { get; set; } public HistorySteps HistorySteps { get; set; }不是對public List<HistorySteps> HistorySteps { get; set; }的列表更改public List<HistorySteps> HistorySteps { get; set; } public List<HistorySteps> HistorySteps { get; set; } 。

因為 SimpleJson 非常...簡單，您必須將您的屬性完全命名為 json，並且您將需要一個包裝器 class 用於"list" json。

public class HistorySteps
{
    public string actualUserDisplayName { get; set; }
    public string actualUserID { get; set; }
    public string comments { get; set; }
    public string invokerType { get; set; }
    public string step { get; set; }
    public string timestampFormatted { get; set; }
    public string transitionName { get; set; }
    public string userDisplayName { get; set; }
    public string userID { get; set; }
}

public class historyList
{
    public List<HistorySteps> historySteps { get; set; }
}

public class MyResponse
{
    public historyList list { get; set; }
}

Answer 2

首先，您處理的是 json，而不是 xml。 您應該使用 DataContract/DataMember 或 JsonProperty 注釋您的屬性。

其次，你快到了，你只需要一個額外的 class 來包裹整個東西。 “list”是根 object中的一個屬性