从选定的节点中选择所有XML返回数据

时间:2018-09-29 21:23:02

标签: c# xml xmlnode

有带有设置的XML

<?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TableSettings的代码中,我选择了一个节点,在getFiltersList中,我想从FieldsSearch选定的节点获取列表,但是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;
            }
}

2 个答案:

答案 0 :(得分:1)

您不是在检查所选节点的子节点,而是从根目录检查所有后代节点。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;
}

答案 1 :(得分:0)

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

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;
}