具有自定义命名空间的 XmlSerializer答案

【问题标题】：XmlSerializer with custom namespace具有自定义命名空间的 XmlSerializer
【发布时间】：2015-08-09 04:26:08
【问题描述】：

我想使用 xml 来使用一个安静的 Web 服务。我必须创建这个模板才能创建一个有效的请求：

<WS_IN_GetAccountCredit xmlns="http://schemas.datacontract.org/2004/07/WcfWebService">
  <GetAccountCreditParams>
    <Password>String content</Password>
    <UserName>String content</UserName>
  </GetAccountCreditParams>
  <WSIdentity>
    <WS_PassWord>String content</WS_PassWord>
    <WS_UserName>String content</WS_UserName>
  </WSIdentity>
</WS_IN_GetAccountCredit>

我的序列化方法是这样的：

XmlSerializerNamespaces xmlNameSpace = new XmlSerializerNamespaces();
xmlNameSpace.Add("", "http://schemas.datacontract.org/2004/07/WcfWebService");

XmlSerializer xmlSerializer = new XmlSerializer(instance.GetType());

using (StringWriter textWriter = new StringWriter())
{
    xmlSerializer.Serialize(textWriter, instance, xmlNameSpace); ; return textWriter.ToString();
}

我的输出如下所示：

<?xml version="1.0" encoding="utf-16"?>
<WS_IN_GetAccountCredit xmlns:xmlns="http://schemas.datacontract.org/2004/07/WcfWebService">
  <WSIdentity>
    <WS_UserName>String content</WS_UserName>
    <WS_PassWord>String content</WS_PassWord>
  </WSIdentity>
  <GetAccountCreditParams>
    <UserName>String content</UserName>
    <Password>String content</Password>
  </GetAccountCreditParams>
</WS_IN_GetAccountCredit>

如您所见，命名空间和 xml 版本的格式错误。我还找到了this、this 和this 文章，但它们都不能解决我的问题。

如何创建一个有效的请求？

【问题讨论】：

你能展示一下WS_IN_GetAccountCredit 的样子吗？
是的诅咒：你在这里： public class WS_IN_GetAccountCredit { private WS_IN_WebServiceIdentity wsIdentity;私人 WS_IN_GetAccountCreditParams getAccountCreditParams；公共 WS_IN_WebServiceIdentity WSIdentity { set { this.wsIdentity = value; } 得到 { 返回 this.wsIdentity; } } 公共 WS_IN_GetAccountCreditParams GetAccountCreditParams { 设置 { this.getAccountCreditParams = value; } 得到 { 返回 this.getAccountCreditParams; } }
也可以使用XmlElements Order属性来设置顺序。
但是由于这是显然一个 WCF Web 服务 - 你不能只为它创建一个 客户端 然后正常调用该方法吗WCF 客户端类上的 C# 方法？为什么需要“手工组装”XML？？？
我当然可以。但由于某些原因，我不得不将它作为一种宁静的服务来使用。

标签： c# xmlserializer

【解决方案1】：

我们开始吧：

    [XmlRoot(ElementName = "WS_IN_GetAccountCredit", Namespace = "http://schemas.datacontract.org/2004/07/WcfWebService")]
    public class WS_IN_GetAccountCredit
    {
        private WS_IN_WebServiceIdentity wsIdentity;
        private WS_IN_GetAccountCreditParams getAccountCreditParams;
        public WS_IN_WebServiceIdentity WSIdentity { set { this.wsIdentity = value; } get { return this.wsIdentity; } }
        public WS_IN_GetAccountCreditParams GetAccountCreditParams
        {
            set { this.getAccountCreditParams = value; }
            get { return this.getAccountCreditParams; }
        }


    }
    public class WS_IN_WebServiceIdentity
    {
        public string UserName { get; set; }
        public string Password { get; set; }
    }

    public class WS_IN_GetAccountCreditParams
    {
        public string UserName { get; set; }
        public string Password { get; set; }
    }



            var namespaces = new XmlSerializerNamespaces();
            namespaces.Add("", "http://schemas.datacontract.org/2004/07/WcfWebService");
            var ser = new XmlSerializer(typeof(WS_IN_GetAccountCredit));
            using (var writer = new StringWriter())
            {
                ser.Serialize(writer, new WS_IN_GetAccountCredit
                {
                    GetAccountCreditParams = new WS_IN_GetAccountCreditParams { Password = "pass", UserName = "use" },
                    WSIdentity = new WS_IN_WebServiceIdentity { Password = "pass", UserName = "use" }
                },
                    namespaces);
                var xml = writer.ToString();
            }

结果是：

<?xml version="1.0" encoding="utf-16"?>
<WS_IN_GetAccountCredit xmlns="http://schemas.datacontract.org/2004/07/WcfWebService">
  <WSIdentity>
    <UserName>use</UserName>
    <Password>pass</Password>
  </WSIdentity>
  <GetAccountCreditParams>
    <UserName>use</UserName>
    <Password>pass</Password>
  </GetAccountCreditParams>
</WS_IN_GetAccountCredit>

【讨论】：