forward一问? 在dispatcher后面加个return,okgetServletContext().getRequestDispatcher(address1.jsp).forward(request,response);return; 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 if(.....)//条件1 getServletContext().getRequestDispatcher(address1.jsp).forward(request,response);System.out.println("1");getServletContext().getRequestDispatcher(address2.jsp).forward(request,response);上面说的对,你的if条件里怎么不return,或者不是if(){ 条件1 }else if() {条件2 }.....就可以不用return了。不过我现在都是这么做的:public class SaveServlet extends javax.servlet.http.HttpServlet { private String[] strCmdList = {"UPDATE","INSERT","RETN"}; public void doPost( javax.servlet.http.HttpServletRequest request, javax.servlet.http.HttpServletResponse response) throws IOException { try { response.setContentType("text/html;charset=GB2312"); String strCmd = null; int intI; if (request.getParameter("CMD") != null) { strCmd = request.getParameter("CMD"); } for (intI = 0; intI < strCmdList.length; ++intI) { if (strCmd.equals(strCmdList[intI])) break; } switch (intI) { case 0 : update(request, response);// update break; case 1 : insert(request, response);// insert break; case 2 : return(request, response); // return break; } } catch (Exception e) { System.out.println("SaveServlet.doPost:" + e); } }public void insertEtcode( HttpServletRequest request, HttpServletResponse response){..........}} 难道forward不是直接跳转到jsp页面吗,为什么还需要return?我看相关资料也没有具体说明,请问servlet是怎么个执行过程。请大侠赐教,多谢了! 免费下载:JSP版网页视频聊天软件系统 一行字符串,“asfas你好按时打发sfasfa”,我用什么方法可以判断中间是否有“你好”这两个字符呢 页面跳转问题 怎么样得到servlet的返回值 求一个提交到本页面的问题? 数据库连接池问题 再次提问,为何出现以下错误信息! 请会的看看 用Tomcat如何新建一个工作目录 spring mvs ${}获取不到变量 网站程序在执行过程中Tomcat窗口经常会连续打印大量的null,已全部屏蔽system.out.println()。请求帮助! 引用问题?
getServletContext().getRequestDispatcher(address1.jsp).forward(request,response);System.out.println("1");getServletContext().getRequestDispatcher(address2.jsp).forward(request,response);上面说的对,你的if条件里怎么不return,
或者不是
if()
{ 条件1 }
else if() {条件2 }
.....
就可以不用return了。
不过我现在都是这么做的:
public class SaveServlet extends javax.servlet.http.HttpServlet {
private String[] strCmdList = {"UPDATE","INSERT","RETN"};
public void doPost(
javax.servlet.http.HttpServletRequest request,
javax.servlet.http.HttpServletResponse response)
throws IOException { try { response.setContentType("text/html;charset=GB2312");
String strCmd = null;
int intI;
if (request.getParameter("CMD") != null) {
strCmd = request.getParameter("CMD");
} for (intI = 0; intI < strCmdList.length; ++intI) {
if (strCmd.equals(strCmdList[intI]))
break;
}
switch (intI) {
case 0 :
update(request, response);// update
break;
case 1 :
insert(request, response);// insert
break;
case 2 :
return(request, response); // return
break;
} } catch (Exception e) {
System.out.println("SaveServlet.doPost:" + e);
}
}
public void insertEtcode(
HttpServletRequest request,
HttpServletResponse response){
..........}}
我看相关资料也没有具体说明,请问servlet是怎么个执行过程。
请大侠赐教,多谢了!