如上,提交表单时,会提示用户错误信息,并将他刚刚输入的用户名和电子邮件回传,这样用户只用重新输入密码这些就可!
怎么做呢?
//把回显信息,保存在Request域中request.setAttribute("msg", "用户名已经存在!");request.setAttribute("username", username);request.setAttribute("email", email);
//把回显信息,保存在Request域中request.setAttribute("msg", "验证码错误!");request.setAttribute("username", username);request.setAttribute("email", email);
完整是这样的:👇👇(不能只看片段)
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {//获取请求的参数String username = Parameter("username");String password = Parameter("password");String email = Parameter("email");String code = Parameter("code");//检查验证码是否正确(这里先写死,要求验证码为abcde)if ("abcde".equalsIgnoreCase(code)) {if (istsUsername(username)) {System.out.println("用户名["+username+"]已经存在!");//把回显信息,保存在Request域中request.setAttribute("msg", "用户名已经存在!");request.setAttribute("username", username);request.setAttribute("email", email);RequestDispatcher("/pages/user/regist.jsp").forward(request, response);}else {istUser(new User(null,username,password,email));RequestDispatcher("/pages/user/regist_success.jsp").forward(request, response);}}else {//把回显信息,保存在Request域中request.setAttribute("msg", "验证码错误!");request.setAttribute("username", username);request.setAttribute("email", email);System.out.println("验证码["+code+"]错误");RequestDispatcher("/pages/user/regist.jsp").forward(request, response);//错误的话跳转到注册页面}}}
<span class="errorMsg"><%Attribute("msg")==null?"":Attribute("msg") %>
</span>
2⃣️在用户提交后回显的内容这样写:主要是value部分:
<label>用户名称:</label>
<input class="itxt" type="text" value="<%Attribute("username")==null?"":Attribute("username") %>" name="username" id="username" />
<label>电子邮件:</label>
<input class="itxt" type="text" value="<%Attribute("email")==null?"":Attribute("email") %>" name="email" id="email" />
以上,完成。
本文发布于:2024-01-31 12:37:36,感谢您对本站的认可!
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