如何提取创建表的子节点

Question

我有以下 xml 文件。

<databaseChangeLog>
 <changeSet author="debdipta.h (generated)" id="1442927616936-399">
        <createTable tableName="SI_FORWARDING_EVENT_HISTORY_AU">
            <column name="FORWARDING_ID_O" type="NUMBER"/>
            <column name="FORWARDING_ID_N" type="NUMBER"/>
            <column name="ASSIGNMENT_ID_O" type="NUMBER"/>
            <column name="ASSIGNMENT_ID_N" type="NUMBER"/>
            <column name="PROCESS_INSTANCE_ID_O" type="NUMBER"/>
            <column name="PROCESS_INSTANCE_ID_N" type="NUMBER"/>
            <column name="ORIGINAL_USER_ID_O" type="NUMBER"/>
            <column name="ORIGINAL_USER_ID_N" type="NUMBER"/>
            <column name="FORWARD_FROM_USER_O" type="NUMBER"/>
            <column name="FORWARD_FROM_USER_N" type="NUMBER"/>
            <column name="FORWARD_TO_USER_O" type="NUMBER"/>
            <column name="FORWARD_TO_USER_N" type="NUMBER"/>
            <column name="COMMENTS_O" type="VARCHAR2(4000 CHAR)"/>
            <column name="COMMENTS_N" type="VARCHAR2(4000 CHAR)"/>
            <column name="XML_O" type="VARCHAR2(4000 CHAR)"/>
            <column name="XML_N" type="VARCHAR2(4000 CHAR)"/>
            <column name="CREATED_BY_O" type="NUMBER"/>
            <column name="CREATED_BY_N" type="NUMBER"/>
            <column name="CREATION_DATE_O" type="date"/>
            <column name="CREATION_DATE_N" type="date"/>
            <column name="AUDIT_TRAIL_REASON" type="VARCHAR2(1000 CHAR)"/>
            <column name="TRANSACTION_TYPE" type="VARCHAR2(100 CHAR)"/>
        </createTable>
    </changeSet>
    <changeSet author="deb.h" id="4588999797">
       </changeSet>
      </databaseChangeLog> 

我想显示元素 changeSet 的子节点，即

createTable
column
column
...
column  

但是我写的代码没有显示任何东西。

p

ackage com.xml.diff;

    import java.io.FileInputStream;


    import javax.xml.parsers.DocumentBuilder;
    import javax.xml.parsers.DocumentBuilderFactory;

    import org.w3c.dom.Document;
    import org.w3c.dom.Element;
    import org.w3c.dom.Node;
    import org.w3c.dom.NodeList;

    public class isolateXML {
    public static void main(String[] args)throws Exception{
        DocumentBuilderFactory dbf=DocumentBuilderFactory.newInstance();
        DocumentBuilder db=dbf.newDocumentBuilder();
        Document dom=db.parse(new FileInputStream("E:/1.xml"));
        Element ele=dom.getDocumentElement();
        ele.normalize();
        NodeList list=ele.getElementsByTagName("changeSet");
        for(int i=0;i<list.getLength();i++){
            Node r=list.item(i);
            NodeList rs=r.getChildNodes();
        for(int j=0;j<rs.getLength();j++){  
            Node k=rs.item(j);
            System.out.println(k.getNodeName());
        }

            }

          }




    }

我得到的唯一输出是

 #text
    createTable
    #text
    #text

谁能告诉我我做错了什么。

Answer 1

我会探索你的节点树，忽略空文本节点。

public static void exploreNode(Node node) {
    //Ignore empty text node
    if (node instanceof Text) {
        String value = node.getNodeValue().trim();
        if (value.equals("")) {
            return;
        }
    }
    System.out.println(node.getNodeName());
    if (node.hasChildNodes()) {
        NodeList childrens = node.getChildNodes();
        for (int i = 0; i < childrens.getLength(); i++) {
            exploreNode(childrens.item(i));
        }
    }
}