我正在尝试从我的注册表单中获取详细信息，使用servlet然后显示所有现有用户。 我知道我在servlet中出错了。无法真正找到答案。
这是我的reg.jsp
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Registration</title>
</head>
<body>
<form method="post" action="UserController.java">
<center>
<table border="1" width="30%" cellpadding="5">
<thead>
<tr>
<th colspan="2">Enter Information Here</th>
</tr>
</thead>
<tbody>
<tr>
<td>First Name</td>
<td><input type="text" name="fname" value="" /></td>
</tr>
<tr>
<td>Last Name</td>
<td><input type="text" name="lname" value="" /></td>
</tr>
<tr>
<td>Email</td>
<td><input type="text" name="email" value="" /></td>
</tr>
<tr>
<td>User Name</td>
<td><input type="text" name="uname" value="" /></td>
</tr>
<tr>
<td>Password</td>
<td><input type="password" name="pass" value="" /></td>
</tr>
<tr>
<td><input type="submit" value="Submit" /></td>
<td><input type="reset" value="Reset" /></td>
</tr>
<tr>
<td colspan="2">Already registered!! <a href="index.jsp">Login Here</a></td>
</tr>
</tbody>
</table>
</center>
</form>
</body>
</html>
这是我的servlet代码
import java.io.IOException;
import java.io.PrintWriter;
import java.sql.PreparedStatement;
import java.sql.SQLException;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.http.HttpSession;
import javax.swing.SwingUtilities;
import com.shreya.dao.UserDao;
import com.shreya.model.Details;
import com.shreya.model.User;
public class UserController extends HttpServlet {
private static final long serialVersionUID = 1L;
private static String INSERT_OR_EDIT = "/user.jsp";
private static String LIST_USER = "/listUser.jsp";
private UserDao dao;
public UserController() {
super();
dao = new UserDao();
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
String forward="";
String act = request.getParameter("act");
if (act != null && !act.equalsIgnoreCase("null") && act.equalsIgnoreCase("login")) {
forward= "/Login.jsp";
}
else if (act != null && !act.equalsIgnoreCase("null") && act.equalsIgnoreCase("delete"))
{
int userId = Integer.parseInt(request.getParameter("userId"));
dao.deleteUser(userId);
forward = LIST_USER;
request.setAttribute("users", dao.getAllUsers());
} else if (act!=null && !act.equalsIgnoreCase("null") && act.equalsIgnoreCase("edit")){
forward = INSERT_OR_EDIT;
int userId = Integer.parseInt(request.getParameter("userId"));
User user1 = dao.getUserById(userId);
request.setAttribute("user", user1);
} else if (act!=null && !act.equalsIgnoreCase("null") && act.equalsIgnoreCase("listUser")){
forward = LIST_USER;
request.setAttribute("users", dao.getAllUsers());
} else
forward = "/Login.jsp";
RequestDispatcher view = request.getRequestDispatcher(forward);
view.forward(request, response);
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
User user = new User();
Details details = new Details();
user.setFirstName(request.getParameter("firstName"));
user.setLastName(request.getParameter("lastName"));
details.setUsername(request.getParameter("username"));
details.setPassword(request.getParameter("password"));
// details.setId(request.getParameter("id"));
String str=request.getParameter("username");
String str1=request.getParameter("password");
if(str.equalsIgnoreCase("shreya")&&str1.equalsIgnoreCase("singh"))
System.out.println("Login!");
else
{
System.out.println("Login failed!");
request.setAttribute("users",dao.getAllUsers());
request.getRequestDispatcher("/listUser.jsp").forward(request, response);
}
}
}
编辑1：
else if (act!=null && !act.equalsIgnoreCase("null") && act.equalsIgnoreCase("register")){
forward = LIST_USER;
request.setAttribute("users", dao.getAllUsers());
传递了一个动作寄存器来保存doGet方法中的值。 我如何在dopost中进行相应的修改？
答案 0 :(得分：0)
改变这个：
details.setUsername(request.getParameter("username"));
details.setPassword(request.getParameter("password"));
到此：
details.setUsername(request.getParameter("uname"));
details.setPassword(request.getParameter("pass"));
输入的名称必须与您获得的请求参数的名称相匹配。
编辑：一些基本建议：
不直接访问请求参数进行存储，准备一个简单的实用函数来处理空值。请求参数在未发出时将为null，但如果为空，则为“”。
枚举请求参数名称的公共静态字符串常量，代码中包含魔术字符串可能会导致对其失去控制：
public static String PARAM_USERNAME = "uname";
public static String PARAM_PASSWORD = "pass";
如果参数是必需的，只需检查它是否具有实际值并处理错误/情况，在servlet中具有原始异常可能是以后调试的痛苦。
预见请求转储方法（保留在Utility类中）只是为了注销所有请求参数，属性和HTTP头，在调试时会为您节省大量的这些情况。
EDIT2：尝试调用此doPost方法并查看会发生什么（将参数的名称更改为正确的名称）：
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
PrintWriter out = response.getWriter();
try {
out.println("Dumping Request");
Enumeration<?> paramNames = request.getParameterNames();
while (paramNames.hasMoreElements()) {
String paramName = (String)paramNames.nextElement();
String paramValue = request.getParameter(paramName);
out.println(" " + paramName + ": " + paramValue);
}
// User user = new User();
// Details details = new Details();
// user.setFirstName(request.getParameter("firstName"));
// user.setLastName(request.getParameter("lastName"));
// details.setUsername(request.getParameter("username"));
// details.setPassword(request.getParameter("password"));
// details.setId(request.getParameter("id"));
// String str = request.getParameter("username");
// String str1 = request.getParameter("password");
// if (str.equalsIgnoreCase("shreya") && str1.equalsIgnoreCase("singh"))
// System.out.println("Login!");
// else
// {
// System.out.println("Login failed!");
// request.setAttribute("users", dao.getAllUsers());
// request.getRequestDispatcher("/listUser.jsp").forward(request, response);
// }
} catch (Throwable t) {
out.println("Exception in servlet");
t.printStackTrace(out);
}
}