从所选节点中选择所有xml返回数据

Question

有XML设置

<?xml version="1.0" encoding="utf-8" ?>
    <TablesConfig>  
      <Logs>
        <Table name="T_1" key="id">
          <FieldsSearch>
            <Field>All</Field>        
          </FieldsSearch>
          <LinksDetail>
            <Link table="t_Det11" key ="id" srcKey="id_request" sortField ="id"/>
            <Link table="t_Det12" key ="id" srcKey="id_request" sortField ="id"/>
            <Link table="t_Det13" key ="id" srcKey="id_request" sortField ="id"/>      
          </LinksDetail>
        </Table>
        <Table name="T_2" key="id">
          <FieldsSearch>
            <Field>All</Field>        
          </FieldsSearch>

          <LinksDetail>
            <Link table="t_Det21"  key ="id" srcKey="operid" sortField ="id"/>
            <Link table="t_Det22" key ="id" srcKey="operid" sortField ="id"/>      
          <Link table="t_Det22"   key ="id" srcKey="operid" sortField ="id"/>
          </LinksDetail>
        </Table>
  </Logs>
</TablesConfig>

在我的TableSettings代码中，我选择了一个节点，在getFiltersList中我想从FieldsSearch选择节点获取列表，但node.SelectNodes（“// FieldsSearch / Field”）返回2行。为什么？我可以选择一个节点并使用它吗？

Class TableSettings{
    XmlNode node;
    public TableSettings(string TableName) {
                doc = new XmlDocument();
                tabSettingsPath = System.IO.Path.GetDirectoryName(Application.ExecutablePath) + "\\ExtFiles\\TablesConf.xml";
                doc.Load(tabSettingsPath);
                node = doc.SelectSingleNode("//Table[@name='"+TableName+"']"); 
            }
            public List<string> getFiltersList() {
                List<string> filtersList = new List<string>();
                foreach (XmlNode inNode in node.SelectNodes("//FieldsSearch/Field")) {
                    filtersList.Add(inNode.FirstChild.Value);
                }
                return filtersList;
            }
}

Answer 1

您没有检查所选节点的子节点，而是检查root的所有后代节点。您可以将其更正为：

public List<string> getFiltersList(XmlNode node)
{
    List<string> filtersList = new List<string>();
    foreach (XmlNode inNode in node.SelectNodes("FieldsSearch/Field"))
    {
        filtersList.Add(inNode.FirstChild.Value);
    }
    return filtersList;
}

Answer 2

路径//从根开始。对于使用currentnode的工作需要使用.// - 从当前节点“.//FieldsSearch / Field”开始或仅使用“FieldsSearch / Field”

public List<string> getFiltersList()
{
    List<string> filtersList = new List<string>();
    foreach (XmlNode inNode in node.SelectNodes(".//FieldsSearch/Field"))
    {
        filtersList.Add(inNode.FirstChild.Value);
    }
    return filtersList;
}