-1
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletContext;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.http.HttpSession;
public class XmlServlet extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("text/html");
PrintWriter write = response.getWriter();
HttpSession session = request.getSession();
ServletContext context=request.getServletContext();
String userName = request.getParameter("userName");
if(userName!=""&&userName!=null){
session.setAttribute("savedUserName",userName);
context.setAttribute("savedUserName",userName);
}
write.println("Request parameter has username as "+userName);
write.println("Session parameter has username as "+session.getAttribute("savedUserName"));
write.println("Context parameter has username as "+context.getAttribute("savedUserName"));
write.println("Init parameter has default username as "+ getServletConfig().getInitParameter("defaultName"));
这是我的java servlet代码,我试图通过下面的web.xml文件打印servlet config参数值。Servlet配置返回空值的参数
<display-name>SimpleServletProject</display-name>
<servlet><servlet-name>xmlServlet</servlet-name>
<servlet-class>myservletpackage.XmlServlet</servlet-class>
<init-param>
<param-name>defaultName</param-name>
<param-value>Chris Jordan</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>xmlServlet</servlet-name>
<url-pattern>/xmlServletPath</url-pattern>
</servlet-mapping>
</web-app>
但运行配置参数值来null.Can任何人都可以告诉我做错了什么?
您在XML中拼写了默认错误。 – Zircon
我错误地粘贴了旧的代码。我已经更正了拼写,但它仍然不显示输出中的值。 – user6479280