使用RestSharp反序列化XML响应

cro*_*les 5 c# xml api restsharp

我已经阅读了关于这个主题已经提出的各种问题,我仍然没有更接近解决我的问题.

我试图反序列化这个xml响应:

 <?xml version="1.0" encoding="ISO-8859-1" ?>
   <SubmissionResult>
     <Result>ACCEPTED</Result>
       <SubmissionID>
         <RefNo>77587-1425386500</RefNo>
         <Submitted>9</Submitted>
         <ValidNo>7</ValidNo>
         <InvalidNo>2</InvalidNo>
         <InvalidNo_1>08452782055</InvalidNo_1>
         <InvalidNo_2>01234567890</InvalidNo_1>
         <TextvID>77587-1425386500</TextvID>
       </SubmissionID>
     <Credits>999999</Credits>
   </SubmissionResult> 
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使用这些类:

[XmlRoot ("SubmissionResult")]
public class SubmissionResult
{
    [XmlElement ("Result")]
    public string Result { get; set; }
    public SubmissionID SubmissionID { get; set; }
    [XmlElement("Credits")]
    public int Credits { get; set; }
}

public class SubmissionID
{
    [XmlElement("RefNo")]    
    public int RefNo { get; set; }
    [XmlElement("Submitted")]    
    public int Submitted { get; set; }
    [XmlElement("ValidNo")]    
    public int ValidNo { get; set; }
    [XmlElement("TextVID")]    
    public string TextVID { get; set; }
}
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但是我只返回null和0值,我认为这是默认值.

以下是获取结果并希望反序列化的代码:

try
{
    var request = new RestRequest();
    request.RequestFormat = DataFormat.Xml;
    request.Resource = APIURL;
    request.RootElement = "SubmissionResult";

    SubmissionResult r = Execute<SubmissionResult>(request);
}

public static T Execute<T>(RestRequest request) where T : new()
{
     var client = new RestClient();
         client.BaseUrl = new  Uri("https://www.textvertising.co.uk/_admin/api", UriKind.Absolute);

     var response = client.Execute<T>(request);

     return response.Data;
}
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非常感谢您提供的任何帮助.

CS

Vol*_*soy 5

我做了一些改变,可以让它工作:

可能是问题中的错字,但我必须更改的是已发布的 XML 无效:

<InvalidNo_2>01234567890</InvalidNo_1>
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我知道您想使用 .NET XMLSerializer,因为您使用 XmlRoot 和 XmlElement 注释,因此您必须像这样覆盖 RestSharp 附带的默认注释:

request.XmlSerializer = new RestSharp.Serializers.DotNetXmlSerializer();
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此外,我不得不删除与 .NET 序列化程序不能很好地配合使用的 RootElement 设置,因此删除了以下行:

request.RootElement = "SubmissionResult";   
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所以我的版本变成了:

var request = new RestRequest();
request.XmlSerializer = new RestSharp.Serializers.DotNetXmlSerializer();
request.RequestFormat = DataFormat.Xml;
// request.Resource = APIURL;

SubmissionResult r = Execute<SubmissionResult>(request);
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最后我注意到 RefNo 是类中的整数,但返回的值不是 (77587-1425386500) 所以我把它变成了一个字符串:

[XmlElement("RefNo")]
public string RefNo { get; set; }
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我在 mocky.io 上创建了一个模拟响应,并对此进行了测试,它似乎工作正常:

public static T Execute<T>(RestRequest request) where T : new()
{
    var client = new RestClient();
    client.BaseUrl = new Uri("http://www.mocky.io/v2/56cd88660f00009800d61ff8", UriKind.Absolute);

    var response = client.Execute<T>(request);

    return response.Data;
}
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