Question

我使用以下代码从OData XML及其工作中获取数据，但我不确定我是否完全理解它，所以也许有一种方法可以用简单的方式编写它？我需要的是获取第一个循环值= 0001和text = approve的属性值第二个值= 0002 text = reject

代码

XNamespace dns = "http://schemas.microsoft.com/ado/2007/08/dataservices";
      if (response.StatusCode == HttpStatusCode.OK)
      {
         string decisionOptions = ReadResponse(response);
         XDocument document = XDocument.Parse(decisionOptions);
         foreach (XElement element in document.Element(dns + "DecisionOptions").Elements(dns + "element"))
        {
            PropertyKeyRef decisionOption = new PropertyKeyRef();
            decisionOption.PropertyValue = element.Element(dns + "DecisionKey").Value;
            decisionOption.PropertyName = element.Element(dns + "DecisionText").Value;
             dat.Add(decisionOption);

        }
      }

XML

 <?xml version="1.0" encoding="utf-8" ?> 
- <d:DecisionOptions xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices" xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata">
- <d:element m:type="TAS.DecisionOption">
  <d:InstanceID>007</d:InstanceID> 
  <d:DecisionKey>0001</d:DecisionKey> 
  <d:DecisionText>Approve</d:DecisionText> 
  <d:CommentMandatory>false</d:CommentMandatory> 
  <d:Nature>POSITIVE</d:Nature> 
  </d:element>
- <d:element m:type="TAS.DecisionOption">
  <d:InstanceID>007</d:InstanceID> 
  <d:DecisionKey>0002</d:DecisionKey> 
  <d:DecisionText>Reject</d:DecisionText> 
  <d:CommentMandatory>true</d:CommentMandatory> 
  <d:Nature>NEGATIVE</d:Nature> 
  </d:element>
  </d:DecisionOptions>

Answer 1

您已经以最简单的方式实现了它。 LINQ的一点点可能会提高可读性（远离foreach循环），但它只是你所写内容的syntactic sugar。

XNamespace dns = "http://schemas.microsoft.com/ado/2007/08/dataservices";
XDocument document = XDocument.Load("database.xml");

PropertyKeyRef decisionOption = new PropertyKeyRef();
decisionOption.PropertyValue = document.Descendants(dns + "DecisionKey")
                                   .Select(node => node.Value).First();
decisionOption.PropertyName = document.Descendants(dns + "DecisionText")
                                   .Select(node => node.Value).First();

dat.Add(decisionOption);

Answer 2

这里如何使用LINQ

以简单的方式完成

namespace ConsoleApplication7
{
    class Program
    {
        static void Main(string[] args)
        {
            XDocument xdoc = XDocument.Load("test.xml");
            XNamespace dns = "http://schemas.microsoft.com/ado/2007/08/dataservices";
            //in xml every element should have it's namespace for this reason  I have to concatenate namespace with the name of element  
            var elementsRes = xdoc.Root.Elements(dns+"element").Select((elt) => new PropertyKeyRef { PropertyName = elt.Element(dns+"DecisionKey").Value.ToString(),PropertyValue = elt.Element(dns+"DecisionText").Value.ToString() }).ToList();
            foreach (var item in elementsRes)
            {
                //your code for the result  
            }


        }
    }
    public  class PropertyKeyRef
    {
        public string PropertyName
        { get; set;  }
        public string PropertyValue
        { get; set; }
    }
}

以简单的方式从XML获取数据

2 个答案: