是的,这是有效的。除了访问jsp或jsf页面外,您还可以访问Servlet。所以创建一个新的servlet。例如:
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* Servlet implementation class TestServlet
*/
public class TestServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public static void yourMethod() {
// do something useful
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
yourMethod();
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
}
}
然后在web.xml文件中创建一个新条目,以便将Servlet映射到/。
<servlet>
<display-name>TestServlet</display-name>
<servlet-name>TestServlet</servlet-name>
<servlet-class>your.packages.TestServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>TestServlet</servlet-name>
<url-pattern>/TestServlet</url-pattern>
</servlet-mapping>
在此之后,您应该可以调用localhost:8080/TestServlet
然后调用您的方法。
看到这个[相关问题](http://stackoverflow.com/q/5522702/620338) – 2012-04-23 07:25:24
@MattHandy你能给更多关于AjaxBehaviorEvent的信息?我必须重新加载我的页面几种类型才能看到应该显示的所有内容 – sameer 2012-04-23 08:16:17
您不需要此处。也许你需要使用属性'update =“@ all”'更新你的ajax调用的整个页面。 – 2012-04-23 08:24:06