使用RestSharp构建XML请求

Gle*_*mas 6 c# xml rest restsharp

我正在尝试使用RestSharp和C#使用REST API.我正在使用的API文档提供了一个示例XML请求:

<?xml version='1.0' encoding='UTF-8'?>  
<messages>  
 <accountreference>EX0000000</accountreference> 
 <from>07700900654</from>
 <message>  
  <to>07700900123</to>  
  <type>SMS</type>  
  <body>Hello Mr Sands.</body>
 </message>  
 <message>  
  <to>07700900124</to>  
  <type>SMS</type>  
  <body>Hello Mr Mayo.</body>
 </message>  
</messages>
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我正在努力了解如何以他们想要的格式构建请求(多个元素称为"消息")

我已经为RestSharp创建了这些类来序列化:

public class messages
{
    public string accountreference { get; set; }

    public string from { get; set; }

    public message message { get; set; }
}

public class message
{
    public string to { get; set; }

    public string body { get; set; }
}
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这是我的RestSharp代码:

var client = new RestClient("http://api.url.com/v1.0")
                         {
                             Authenticator =
                                 new HttpBasicAuthenticator(
                                 UserName,
                                 Password)
                         };

var request = new RestRequest("theresource", Method.POST) { RequestFormat = DataFormat.Xml };

request.AddBody(
    new messages
        {
            accountreference = Configuration.AccountReference,
            from = Configuration.From,
            message =
                new message { to = Configuration.Message.To, body = Configuration.Message.Body }
        });

var response = client.Execute(request);
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当我只有一个消息元素时,这很有效,但是我不知道如何创建多个消息元素而不将它们嵌套在数组中,这不适用于API.

Vol*_*soy 11

默认情况下,RestSharp使用自己的序列化程序,但它也包含DotNetSerializer,因此您可以通过更改序列化程序来实现目标:

var request = new RestRequest("theresource", Method.POST) 
{ 
    RequestFormat = DataFormat.Xml, 
    XmlSerializer = new RestSharp.Serializers.DotNetXmlSerializer() 
};
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然后,您可以使用消息对象列表并使用XmlElement属性进行装饰:

public class messages
{
    public string accountreference { get; set; }

    public string from { get; set; }

    [XmlElement("message")]
    public List<message> messageList { get; set; }
}


public class message
{
    public string to { get; set; }

    public string body { get; set; }
}
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然后,您可以更改最后一位以添加多条消息:

request.AddBody(
    new messages
    {
        accountreference = "ref",
        from = "from",
        messageList = new List<message>() {
                new message { to = "to1", body = "body1" },
                new message { to = "to2", body = "body2" }
                }
    }); 
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会产生(我通过检查request.Parameters [0] .Value得到了XML):

<?xml version="1.0" encoding="utf-8"?>
<messages>
  <accountreference>ref</accountreference>
  <from>from</from>
  <message>
    <to>to1</to>
    <body>body1</body>
  </message>
  <message>
    <to>to2</to>
    <body>body2</body>
  </message>
</messages>
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我想这是你一直在寻找的XML格式.