我如何从XML返回布尔值的值

本文关键字：布尔值 返回 XML | 更新日期: 2023-09-27 18:11:04

我想创建一个函数来输入值(funName)并检查XML文件属性(funName)然后输出XML文件属性(isEnable)布尔值true或false

如何修改这段代码?

My XML file

<itema>
   <itemb FunName="ABC" isEnable="true"></itemb>
   <itemb FunName="DEF" isEnable="false"></itemb>
</itema>

My Code 

public bool FunEnable(string funName , string isEnable)
{
    bool result = true;
    XmlDocument xDL = new XmlDocument();
    xDL.Load("C://XMLFile2.xml"); //Load XML file
    XmlNode xSingleNode = xDL.SelectSingleNode("//itemb");
    XmlAttributeCollection xAT = xSingleNode.Attributes; //read all Node attribute            
    for (int i = 0; i < xAT.Count; i++)
    {
        if (xAT.Item(i).Name == "isEnable")
        {
            Console.WriteLine(xAT.Item(i).Value); //read we want attribute content                    
        }
    }
    return result;
}

Thanks to lot

我如何从XML返回布尔值的值

你可以试试这个:

public static bool FunEnable(string funNam)
        {
            bool result = true;
            XmlDocument xDL = new XmlDocument();
            xDL.Load(@"C:/XMLFile2.xml"); //Load XML file
            XmlNodeList nodeList = xDL.SelectNodes("//itemb");
            foreach (XmlNode node in nodeList)
            {
                if (node.Attributes["FunName"].Value.Equals(funNam))
                {
                    result = Convert.ToBoolean(node.Attributes["isEnable"].Value);
                    break;
                }
            }
            Console.WriteLine("with funName = "+ funNam +" isEnable equal to : " + result);
            return result;
        }
<标题> 输出

with funName = ABC isEnable = True

使用LINQ to XML非常简单。您可以使用XDocument.Load加载文档，然后获取isEnable值，如下所示:

var result = doc.Descendants("itemb")
    .Where(e => (string)e.Attribute("FunName") == "ABC")
    .Select(e => (bool)e.Attribute("isEnable"))
    .Single();

您可以在这里看到一个工作演示:https://dotnetfiddle.net/MYTOl6

var xDoc = XDocument.Load(path);
bool result = (from itemb in xDoc.Descendants("itemb")
               where itemb.Attribute("FunName").Value == funcName
               select itemb.Attribute("isEnable").Value == "true")
               .FirstOrDefault();

与XML相比，我更喜欢Linq .

也许这个有用:

    public bool FunEnable(string funName, string isEnable)
    {
        bool result = true;
        XDocument xDL = XDocument.Load("C://XMLFile2.xml");
        var xSingleNode = from node in xDL.Descendants("itemb")
                          where node.Attribute("FunName").Value == funName
                          select node;
        if(xSingleNode.Count() > 0)
        {
            result = xSingleNode.ElementAt(0).Attribute("isEnable").Value == "true";
            //If there is at least one node with the given name, result is set to the first nodes "isEnable"-value
        }
        return result;
    }

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