XML選值的問題解决办法
日期：2014-05-18 浏览次数：20947 次
XML選值的問題
現要取得CaseID="Case001"且UserID="User002"的各元素的值該如何做？
------解决方案--------------------
XmlDocument doc = new XmlDocument();
doc.Load(@"D:\Project\C#Test\WindowsApplication1\WindowsApplication1\XMLFile1.xml");
XmlNodeList nodeList = doc.SelectNodes("//TestCase//Case[@CaseID='Case001']//User[@UserID='User002']//*");
List<string> values = new List<string>();
foreach (XmlElement element in nodeList)
{
if (!values.Contains(element.InnerText))
{
Debug.WriteLine(string.Format("ElementName:{0} Value:{1}",element.Name ,element.InnerText));
values.Add(element.InnerText);
}
}
------解决方案--------------------
- XML code
<?xml version="1.0" standalone="yes" ?>
<TestCase>
<Case CaseID="Case001">
<User UserID="User001">
<FunctionName>FName1</FunctionName>
<ActType>Create</ActType>
<RequestNumber>3</RequestNumber>
<FrequencyNumber>2</FrequencyNumber>
<DelayNumber>1</DelayNumber>
</User>
<User UserID="User002">
<FunctionName>FName2</FunctionName>
<ActType>Update</ActType>
<RequestNumber>5</RequestNumber>
<FrequencyNumber>2</FrequencyNumber>
<DelayNumber>3</DelayNumber>
</User>
</Case>
<Case CaseID="Case002">
<User UserID="User001">
<FunctionName>FName21</FunctionName>
<ActType>Create</ActType>
<RequestNumber>3</RequestNumber>
<FrequencyNumber>2</FrequencyNumber>
<DelayNumber>1</DelayNumber>
</User>
<User UserID="User002">
<FunctionName>FName22</FunctionName>
<ActType>Update</ActType>
<RequestNumber>5</RequestNumber>
<FrequencyNumber>2</FrequencyNumber>
<DelayNumber>3</DelayNumber>
</User>
</Case>
</TestCase>
現要取得CaseID="Case001"且UserID="User002"的各元素的值該如何做？
------解决方案--------------------
XmlDocument doc = new XmlDocument();
doc.Load(@"D:\Project\C#Test\WindowsApplication1\WindowsApplication1\XMLFile1.xml");
XmlNodeList nodeList = doc.SelectNodes("//TestCase//Case[@CaseID='Case001']//User[@UserID='User002']//*");
List<string> values = new List<string>();
foreach (XmlElement element in nodeList)
{
if (!values.Contains(element.InnerText))
{
Debug.WriteLine(string.Format("ElementName:{0} Value:{1}",element.Name ,element.InnerText));
values.Add(element.InnerText);
}
}
------解决方案--------------------
- JScript code
<html>
<head>
<script type="text/javascript" src="loadxmldoc.js">
</script>
</head>
<body>
<script type="text/javascript">
xmlDoc = loadXMLDoc("Xml1.xml");
var nodeList = xmlDoc.selectNodes("//Case[@CaseID='Case001']//User[@UserID='User002']//*");
for(i=0;i<nodeList.length;i++)
{
alert(nodeList.item(i).nodeName+"-"+nodeList.item(i).childNodes[0].nodeValue);
}
</script>
</body>
</html>
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