6

我需要将参数作为 Xml 传递给我的存储过程。

我在中间层有一个 WCF 服务，它调用我的数据层，然后将请求转发到适当的存储过程。

设计是 WCF 服务负责构建 XML 以传递给存储库。

我只是想知道是要控制中间层的 Xml 中包含哪些参数，还是使用客户端构建的字典，然后将其转换为中间层的 Xml？

目前我选择了后者——例如：

 public TestQueryResponseMessage TestQuery(TestQueryRequestMessage message)
 {
        var result = Repository.ExecuteQuery("TestQuery", ParamsToXml(message.Body.Params));

        return new TestQueryResponseMessage
        {
            Body = new TestQueryResponse
            {
                TopicItems = result;
            }
        }
    }


private string ParamsToXml(Dictionary<string, string> nvc)
{
        //TODO: Refactor
        StringBuilder sb = new StringBuilder();

        sb.Append("<params>");
        foreach (KeyValuePair<string, string> param in nvc)
        {
            sb.Append("<param>");
            sb.Append("<" + param.Key + ">");
            sb.Append(param.Value);
            sb.Append("</" + param.Key + ">");
            sb.Append("</param>");
        }
        sb.Append("</params>");

        return sb.ToString();
}

但是我可能需要第一种方式。例如

public TestQueryResponseMessage TestQuery(TestQueryRequestMessage message)
{
       string xml = string.Format("<params><TestParameter>{0}</TestParameter></params>",message.Body.TestParameter)

       var result = Repository.ExecuteQuery("TestQuery", xml);

      return new TestQueryResponseMessage
      {
          Body = new TestQueryResponse
          {
                    TopicItems = result;
          }
      }
}

蜂巢有什么推荐的？

4

4 回答 4

6

如果必须使用xml；然后我不会传递字典，而是使用一个代表该数据的类，并使用XmlSerializer它作为 xml 获取它：

[Serializable, XmlRoot("args")]
public class SomeArgs {
    [XmlElement("foo")] public string Foo { get; set; } 
    [XmlAttribute("bar")] public int Bar { get; set; }
}
...
SomeArgs args = new SomeArgs { Foo = "abc", Bar = 123 };
XmlSerializer ser = new XmlSerializer(typeof(SomeArgs));
StringWriter sw = new StringWriter();
ser.Serialize(sw, args);
string xml = sw.ToString();

这使得以面向对象的方式管理哪些参数适用于哪些查询变得更加容易。这也意味着您不必进行自己的 xml 转义...

于 2009-04-15T12:57:13.717 回答
3

一旦您使用 Bob The Janitor 的解决方案，您就拥有了 XML。

使用 XML 参数创建存储过程。然后，根据您拥有多少 XML 以及您使用它做什么，您可以使用 Xquery 或 OpenXML 来分​​解 XML 文档。提取数据并执行正确的操作。这个例子是基本的伪代码，但你应该明白。

CREATE PROCEDURE [usp_Customer_INS_By_XML]
@Customer_XML XML
AS
BEGIN
EXEC sp_xml_preparedocument @xmldoc OUTPUT, @Customer_XML

--OPEN XML example of inserting multiple customers into a Table.
INSERT INTO CUSTOMER
(
First_Name
Middle_Name
Last_Name
)
SELECT
First_Name
,Middle_Name
,Last_Name
FROM OPENXML (@xmldoc, '/ArrayOfCustomers[1]/Customer',2)
WITH(
 First_Name VARCHAR(50)
,Middle_Name VARCHR(50)
,Last_Name VARCHAR(50)
)

EXEC sp_xml_removedocument @xmldoc
END
于 2009-04-15T15:57:42.970 回答
1

您可以像这样使用对象序列化类

 public class Serialization
    {
        /// <summary>
        /// Serializes the object.
        /// </summary>
        /// <param name="myObject">My object.</param>
        /// <returns></returns>
        public static XmlDocument SerializeObject(Object myObject)
        {
            XmlDocument XmlObject = new XmlDocument();
            String XmlizedString = string.Empty;

            try
            {                
                MemoryStream memoryStream = new MemoryStream();
                XmlSerializer xs = new XmlSerializer(myObject.GetType());
                XmlTextWriter xmlTextWriter = new XmlTextWriter(memoryStream, Encoding.UTF8);
                xs.Serialize(xmlTextWriter, myObject);
                memoryStream = (MemoryStream)xmlTextWriter.BaseStream;
                XmlizedString = UTF8ByteArrayToString(memoryStream.ToArray());                
            }
            catch (Exception e)
            {
                System.Console.WriteLine(e);
            }
            XmlObject.LoadXml(XmlizedString);
            return XmlObject;            
        }

        /// <summary>
        /// Deserializes the object.
        /// </summary>
        /// <typeparam name="T"></typeparam>
        /// <param name="XmlizedString">The p xmlized string.</param>
        /// <returns></returns>
        public static T DeserializeObject<T>(String XmlizedString)
        {
            XmlSerializer xs = new XmlSerializer(typeof(T));
            MemoryStream memoryStream = new MemoryStream(StringToUTF8ByteArray(XmlizedString));
            //XmlTextWriter xmlTextWriter = new XmlTextWriter(memoryStream, Encoding.UTF8);
            Object myObject = xs.Deserialize(memoryStream);
            return (T)myObject;
        } 

        /// <summary>
        /// To convert a Byte Array of Unicode values (UTF-8 encoded) to a complete String.
        /// </summary>
        /// <param name="characters">Unicode Byte Array to be converted to String</param>
        /// <returns>String converted from Unicode Byte Array</returns>
        private static String UTF8ByteArrayToString(Byte[] characters)
        {
            UTF8Encoding encoding = new UTF8Encoding();
            String constructedString = encoding.GetString(characters);
            return (constructedString);
        }



        /// <summary>
        /// Converts the String to UTF8 Byte array and is used in De serialization
        /// </summary>
        /// <param name="pXmlString"></param>
        /// <returns></returns>
        private static Byte[] StringToUTF8ByteArray(String pXmlString)
        {
            UTF8Encoding encoding = new UTF8Encoding();
            Byte[] byteArray = encoding.GetBytes(pXmlString);
            return byteArray;
        } 
    }

那么您不必手动构建 XML，而且您可以将它与任何项目一起使用以使用 XSLT 对其进行转换

于 2009-04-15T04:14:56.920 回答
0

我会将 xml 构造代码放在域对象中。这样，您只需从 Web 服务或数据层调用 obj.GetXML()。

于 2009-04-14T15:24:17.660 回答