Я создаю простую систему crud в сервлете Java. Я новичок в Servlet. Я заполнил форму, нажал кнопку отправки и получил ошибку - HTTP Status 404 - Not Found. Я прикрепил снимок экрана ниже вместе с кодом.
структура папок
Форма
<form method = post action = "employee/employee.java">
<table class="table table-borederd">
<tr>
<td>Enter Employee ID:</td>
<td ><input class="form-control" type = "text" name = "txtEmpId"/></br></td>
</tr>
<tr>
<td>Enter Employee FirstName:</td>
<td ><input class="form-control" type = "text" name = "txtFName"/></td>
</tr>
<tr>
<td>Enter Employee LastName:</td>
<td ><input class="form-control" type = "text" name = "txtLName"/></td>
</tr>
<td><input type = "submit" value = "submit"/></td>
</form>
employee.java
public class employee extends HttpServlet
{
Connection con;
int row;
public void doPost(HttpServletRequest req, HttpServletResponse rsp) throws IOException, ServletException
{
try
{
Class.forName("com.mysql.jdbc.Driver");
con = DriverManager.getConnection("jdbc:mysql://localhost/empp","root","");
System.out.println("Connection Established");
}
catch(Exception e)
{
System.out.println("Sorry...........");
}
rsp.setContentType("text/html");
PrintWriter out = rsp.getWriter();
String empId = req.getParameter("txtEmpId");
String empFName = req.getParameter("txtFName");
String empLName = req.getParameter("txtLName");
try
{
PreparedStatement stat = con.prepareStatement("insert into record (id,fname,lname) Values(?,?,?)");
stat.setString(1,empId);
stat.setString(2,empFName);
stat.setString(3,empLName);
row = stat.executeUpdate();
}
catch(Exception e)
{
}
out.println("<html>");
out.println("<font color = 'red'>Sucessfully registered!</font>");
out.println("</body>");
out.println("<h2>");
out.println(row);
out.println("</h2>");
out.println("</body>");
}
}