Question

这是我的web.xml

           <?xml version="1.0" encoding="UTF-8"?>
           <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org  /xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID"version="3.1">
          <display-name>Frontend</display-name>
          <welcome-file-list>
          <welcome-file>index.html</welcome-file>
          <welcome-file>index.htm</welcome-file>
          <welcome-file>index.jsp</welcome-file>
          <welcome-file>default.html</welcome-file>
          <welcome-file>default.htm</welcome-file>
          <welcome-file>default.jsp</welcome-file>
          </welcome-file-list>
          <servlet>
          <servlet-name>Jersey Webapp Application</servlet-name>
          <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>

          <init-param>
          <param-name>com.sun.jersey.config.property.packages</param-name>
          <param-value>com.restify.frontend;org.codehaus.jackson.jaxrs</param-value>
          </init-param>
          <init-param>
          <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
          <param-value>true</param-value>
          </init-param>
          <load-on-startup>1</load-on-startup>
          </servlet>
          <servlet-mapping>
          <servlet-name>Jersey Webapp Application</servlet-name>
          <url-pattern>/reporter/*</url-pattern>
          </servlet-mapping>
          </web-app>

这是我的第一个servlet：servlet的名称：响应

              @WebServlet({ "/Response", "/reportsto" })
              public class Response extends HttpServlet {
              private static final long serialVersionUID = 1L;
              protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
              .....

我的第二个servlet，即Response1：

              @WebServlet({ "/Response1", "/resp" })
              public class Response1 extends HttpServlet {
              private static final long serialVersionUID = 1L;
              protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        .....               
              Set<Map.Entry<Object,Object>> s2=map1.entrySet();
              PrintWriter out1=response.getWriter();
              out1.println("<html>"+
              "<form action=\"./reportsto\" method=\"get\">"+
              "<input type=\"cancel\" value=\"cancel\" name=\"cancel\">\n\">"+
        "</form>"+);
      So from the servlet Response1 I need to goto the previous page that is   servlet1 Response onclicking the cancel button.How can this be done please advice.

实际上我的servlet1响应将输出为：

                 Number User Id     Username    Count
                   1    A12354      Anagha R        0
                   2    M12345      Madhusudan S    1

我的第二个Servlet Response1显示：

                   1    CHARgska    Validating

在servlet2中，单击时需要有一个取消按钮，将我带回到上一页，即servlet1

Answer 1

单击Cancel按钮会将用户带到先前访问的页面。

out1.println("<html>"+
              "<form action=\"./reportsto\" method=\"get\">"+
              "<input onclick=\"window.history.go(-1); return false;\" type=\"button\" value=\"Cancel\" />"+
        "</form>"+);

单击取消按钮后如何转到上一页

1 个答案: