C＃遍歷xml以獲取特定子節點的內部文本值

Question

我的XML文檔如下所示：

<Runs>
 <Run>
  <LotInfo>
   <Column name="Entity">HST123,</Column> 
   <Column name="Product">XXX123</Column> 
   <Column name="WSOp">1234</Column> 
   <Column name="Route">V234</Column> 
   <Column name="Recipe" /> 
   <Column name="LotNumber">K898722</Column> 
   <Column name="RunStartTime">2014-05-20T17:43:11.8872105</Column> 
  </LotInfo>
  <Operations>
   <Operation type="INTRODUCTION">
   <Column name="Size">1490</Column> 
   <Column name="TimeStamp">2014-05-20T17:43:11.8872105</Column> 
   <Column name="Operator">nismail9</Column> 
   <Column name="Description">
    <![CDATA[ Unknown ]]> 
   </Column>
  <Column name="Status">Success</Column> 
  </Operation>
  <Operation type="RNUCHECK">
   <Column name="Size">1490</Column> 
   <Column name="TimeStamp">2014-05-20T17:43:15.3091731</Column> 
   <Column name="Operator">nismail9</Column> 
   <Column name="Description">
    <![CDATA[ ]]> 
   </Column>
   <Column name="Status">True</Column> 
  </Operation>
  <Operations>
   <Operation type="INTRODUCTION">
   <Column name="Size">1490</Column> 
   <Column name="TimeStamp">2014-05-20T17:58:47.0830259</Column> 
   <Column name="Operator">nismail9</Column> 
   <Column name="Description">
    <![CDATA[ Unknown ]]> 
   </Column>
   <Column name="Status">Success</Column> 
   </Operation>
  </Operations>
 </Run>
</Runs>

我想遍歷整個XML以獲取內部文本值。 我需要獲取節點<Operation type="INTRODUCTION">時間戳。 每當我得到<Operation type="INTRODUCTION"> ，我都會去獲取此節點下的時間戳： <Column name="TimeStamp">2014-05-20T17:43:11.8872105</Column> ，即2014-05-20T17:43:11.8872105 。

我有一個代碼來獲取值，但是我不知道如何遍歷整個文本來獲取所有值。 我只能得到一個。

到目前為止，我的代碼：

XmlDocument readDoc = new XmlDocument();
            readDoc.Load(fileName);
            int count = readDoc.SelectNodes("/Runs/Run/Operations/Operation[@type='INTRODUCTION']").Count;
            MessageBox.Show(count.ToString());

                        var node = readDoc.SelectSingleNode("/Runs/Run/Operations/Operation[@type='INTRODUCTION']/Column[@name='TimeStamp']");
MessageBox.Show(node.InnerText);

Answer 1

您可以使用SelectNodes獲取所有操作節點的內部文本，然后對其進行迭代以獲取內部文本。

var nodes = readDoc.SelectNodes("/Runs/Run/Operations/Operation/
                                    Column[@name='TimeStamp']");
var innerTexts = nodes.OfType<XmlNode>().Select(n => n.InnerText);

確保添加System.Linq命名空間以使用可枚舉的擴展方法。

Answer 2

使用XmlDocument.SelectNodes代替XmlDocument.SelectSingleNode獲取所有匹配的節點。 您不需要節點數，但需要它們的列表。 獲得節點后，使用常規的foreach循環構造遍歷它們：

    var xmlFile = "c:\\input.xml";
    XmlDocument readDoc = new XmlDocument();
    readDoc.Load(xmlFile);
    var nodes = readDoc.SelectNodes("/Runs/Run/Operations/Operation[@type='INTRODUCTION']/Column[@name='TimeStamp']");
    foreach (XmlElement node in nodes)
    {
        Console.WriteLine(node.InnerText);
    }

輸出為：

2014-05-20T17:43:11.8872105
2014-05-20T17:58:47.0830259

另一種可能性是使用LINQ2XML以獲得所需的節點：

try
{           
    var xmlFile = "c:\\input.xml";
    // load the xml file
    var xml = XDocument.Load(xmlFile);
    // find all operations nodes
    var operations = xml.Root.Descendants("Operations").ToList();
    // iterate over all nodes
    foreach (var operation in operations)
    {
        // find the operation node with type INTRODUCTION
        var op = operation.Elements().Where (o => o.Name == "Operation" && (o.Attribute("type").Value.ToUpper() == "INTRODUCTION")).FirstOrDefault();
        if (op != null)
        {
            // find the timestamp node
            var timestamp = op.Elements().Where (o => o.Name == "Column" && (o.Attribute("name").Value.ToUpper() == "TIMESTAMP")).FirstOrDefault();
            if (timestamp != null)
            {
                // and get the value
                Console.WriteLine(timestamp.Value);
            }
        }
    }
}
catch (Exception exception)
{
    Console.WriteLine(exception.Message);
}

輸出與上面相同：

2014-05-20T17:43:11.8872105
2014-05-20T17:58:47.0830259

Answer 3

使用LINQ to XML

（ using System.Xml.Linq; ）

var query = XDocument.Load(xmlPath)
    .Descendants("Operation")
    .Where(w => (string)w.Attribute("type") == "INTRODUCTION")
    .SelectMany(s => s.Elements("Column")
        .Where(w => (string)w.Attribute("name") == "TimeStamp")
        .Select(x => (string)x));