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luofuxiang
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直接将要跳转的servlet的url-pattern作为跳转链接就行,假设我有一个servlet如下所示(注意url-pattern):
import java.io.IOException; import javax.servlet.ServletException; import javax.servlet.annotation.WebServlet; import javax.servlet.http.HttpServlet; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; @WebServlet("/CourseInfoManage") //注意这里,/CourseInfoManage为servlet的url-pattern public class CourseInfoManage extends HttpServlet { public CourseInfoManage() { super(); } protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { ...... } protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { doGet(request, response); ...... } }
在jsp页面的超链接中:
<a href="CourseInfoManage">课程信息管理</a>
java代码中:
response.sendRedirect("CourseInfoManage");
当然,有时候可能你的jsp页面和servlet不在同一目录下,可能需要用绝对路径。
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