Question

这是我的XML文件：

<applications>
<apps>
<appname>MYapp</appname>
<server>qwe</server>
<port>1234</port>
<uname>system</uname>
<passwd>security</passwd>
</apps>
 <apps>
<appname>tyas</appname>
<server>qwewe</server>
<port>1235</port>
<uname>asd</uname>
<passwd>wetry</passwd>
</apps>
</applications>

这是我的Java代码：

package trial;
import java.io.File;

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 parsing {

public static void main(String args[]) 
{
    try {

File applications = new File("appdetails.xml");
DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
Document doc = dBuilder.parse(applications);
doc.getDocumentElement().normalize();

System.out.println("application details:");
NodeList nodes = doc.getElementsByTagName("apps");
System.out.println("===================");

for (int i = 0; i < nodes.getLength(); i++) 
{
    Node node = nodes.item(i);

if (node.getNodeType() == Node.ELEMENT_NODE) 
{
    Element element = (Element) node;
    System.out.println("app name: " + getValue("appname", element));
    String uname = getValue("uname",element);
    String host = getValue("server",element);
    String passwd = getValue("passwd",element);
    String clu = getValue("clustername",element);
    String path = getValue("path",element);
    String cmd = getValue("cmd" ,element);
    String port = getValue("port" ,element);
    String user= getValue("user",element);
    System.out.println(user);
    String pswd= getValue("pswd" ,element);
    System.out.println(pswd);



}}} catch (Exception ex) {ex.printStackTrace();
    }
}
private static String getValue(String tag, Element element) 
{
NodeList nodes = element.getElementsByTagName(tag).item(0).getChildNodes();
Node node = (Node) nodes.item(0);
return node.getNodeValue();
}
}

我得到的输出遍历两个节点。如何修改此代码以仅显示我选择的节点的详细信息？如果我选择MYapp，它应该只显示Myapp的详细信息。

Answer 1

更改

if (node.getNodeType() == Node.ELEMENT_NODE)

到

if (node.getNodeType() == Node.ELEMENT_NODE && "MYapp".equals(getValue("appname", element)))

Answer 2

使用JAXB读取XML。它会让你的生活更轻松。

public class Parsing {

public static void main(String args[]) {
     String xml = "<applications>\n" +
                "<apps>\n" +
                "<appname>MYapp</appname>\n" +
                "<server>qwe</server>\n" +
                "<port>1234</port>\n" +
                "<uname>system</uname>\n" +
                "<passwd>security</passwd>\n" +
                "</apps>\n" +
                " <apps>\n" +
                "<appname>tyas</appname>\n" +
                "<server>qwewe</server>\n" +
                "<port>1235</port>\n" +
                "<uname>asd</uname>\n" +
                "<passwd>wetry</passwd>\n" +
                "</apps>\n" +
                "</applications>";
        Applications applications = JAXB.unmarshal(new StringReader(xml), Applications.class);
        Optional<Apps> appsOptional = applications.getApps().stream().filter(app -> app.getAppname().equals("MYapp")).findFirst();
        if(appsOptional.isPresent()) {
            Apps app = appsOptional.get();
            System.out.println("Got MyAPP object: "+ app.getUname());
        }
    }
}

class Applications {
    @XmlElement
    private List<Apps> apps;

    public List<Apps> getApps() {
        return apps;
    }
}
class Apps {
    @XmlElement
    private String appname;
    @XmlElement
    private String server;
    @XmlElement
    private String port;
    @XmlElement
    private String uname;
    @XmlElement
    private String passwd;

    public String getAppname() {
        return appname;
    }

    public String getServer() {
        return server;
    }

    public String getPort() {
        return port;
    }

    public String getUname() {
        return uname;
    }

    public String getPasswd() {
        return passwd;
    }
}

Answer 3

您可以使用XPath：

  XPath path = XPathFactory.newInstance().newXPath();
  NodeList nl=(NodeList)path.evaluate("applications/apps[appname = 'MYapp']", doc, XPathConstants.NODESET);
  for (int i = 0; i < nl.getLength(); i++) {
      System.out.println(nl.item(i).getNodeName());
  }

Answer 4

获取输入以阅读MyApp等应用。

阅读时如果appname => node.getValue().equals("MyApp")继续其他中断并查看下一步。如果XML结束显示错误消息。

如何从XML文件到Java代码一次解析一个节点？

4 个答案: