java.lang.IllegalStateException: Cannot forward after response has been committed异常

最新推荐文章于 2020-04-16 16:08:08 发布

原创最新推荐文章于 2020-04-16 16:08:08 发布 · 508 阅读

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本文探讨了Java Web应用中登录阶段出现的IllegalStateException异常，该异常源于响应已提交后的页面转发操作。通过逐步分析代码并加入return语句解决了此问题。

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java.lang.IllegalStateException: Cannot forward after response has been committed

这个异常在我登入的时候就有而且页面还没有报错这个我真不懂

protected void doPost(HttpServletRequest req, HttpServletResponse resp)
			throws ServletException, IOException
			{	
		boolean isError=false;
		String username=req.getParameter(WebAttribute.ParameterOfUsername);
		String password=req.getParameter(WebAttribute.ParameterOfPssword);
		UserService service=new  SimpleUserService();
		try{
			User user=service.login(username, password);
			HttpSession session = req.getSession(false);
			if(session == null){
				req.getSession(true).setAttribute(WebAttribute.CURRENT_USER_SESSION_ATTRIBUTE, user);
			}else{
				session.setAttribute(WebAttribute.CURRENT_USER_SESSION_ATTRIBUTE, user);
			}	
			req.getRequestDispatcher("/WEB-INF/views/user/home.jsp").forward(req, resp);
			return;
		}catch(UsernameOrPasswordIsNullException e)
		{
			req.setAttribute(WebAttribute.VIEW_EXCEPTION_ATTRIBUTE,"账号或者密码不能为空");
			isError=true;
		}catch(UsernameNotFoundException e)
		{
			req.setAttribute(WebAttribute.VIEW_EXCEPTION_ATTRIBUTE,"密码错误");
			isError=true;
		}catch(UsernameMismatchPasswordException e)
		{
			req.setAttribute(WebAttribute.VIEW_EXCEPTION_ATTRIBUTE, "密码错误");
			isError=true;
		}
		catch(UserServiceException e)
		{
			req.setAttribute(WebAttribute.VIEW_EXCEPTION_ATTRIBUTE, "系统错误");
			isError=true;
		}finally
		{
			if(isError==true)
			{
				req.getRequestDispatcher("/index.jsp").forward(req, resp);
				return;
			}
		}

后来debug 后才知道转发是页面上执行的所以后面的代码加上return 就好了因为转发页面还在这，所以后面代码也会执行，系统就会报错。