我正在尝试使用XStream将XML文件解析为对象,但我得到了这个异常:
线程“main”中的异常 com.thoughtworks.xstream.mapper.CannotResolveClassException:servers 在 com.thoughtworks.xstream.mapper.DefaultMapper.realClass(DefaultMapper.java:56) 在 com.thoughtworks.xstream.mapper.MapperWrapper.realClass(MapperWrapper.java:30) 在 com.thoughtworks.xstream.mapper.DynamicProxyMapper.realClass(DynamicProxyMapper.java:55) [...]
这是我的XML:
<servers>
<server>
<ip>10.196.113.27</ip>
</server>
<server>
<ip>10.196.113.31</ip>
</server>
</servers>
这是我的代码:
public class ServerIP {
private String ip;
public String getIp() {
return ip;
}
public void setIp(String ip) {
this.ip = ip;
}
}
public class ServerHandler {
private String fileName = "servers.xml";
private String path = "J:\\workspace\\LOG730\\src\\Q3\\";
private XStream xstream = new XStream(new DomDriver());
public void readFromXML() {
try {
FileInputStream fis = new FileInputStream(path + fileName);
ServerIP server = (ServerIP) xstream.fromXML(fis, new ServerIP());
System.out.println("Host: " + server.getIp());
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
由此触发异常:
ServerHandler serverHandler = new ServerHandler();
serverHandler.readFromXML();
答案 0 :(得分:6)
尝试添加类Servers来保存您的ServerIP实例并添加以下行:
xstream.alias("servers", Servers.class);
xstream.alias("server", ServerIP.class);
在这里,您可以找到有关别名的简单教程:http://x-stream.github.io/alias-tutorial.html
答案 1 :(得分:3)
@XStreamAlias("server")
public class ServerIP {
private String ip;
public String getIp() {
return ip;
}
public void setIp(String ip) {
this.ip = ip;
}
}
对不完整的回答感到抱歉,在完成之前我分心并发布了。与此同时,@ Teg指出了这个方向。