java.lang.NumberFormatException: null

This exception throws when you want to parse a non-numeric character whole, or a null , which is your case: java.lang.NumberFormatException: null java.lang.Integer .parseInt (Integer.java:542)

Make sure that on your html or jsp page you are using the name of your variable Cedula well, since evidently your servlet is not receiving any value.

You have problems when paging from String to Int. There is no try{}catch and therefore if there is an exception such as the value is a non-numeric character or a value null , this finally return an error.

Therefore, what you must do is to add a try{} catch(){} so that it can be executed without problems.

Here's the code below:

protected void doPost(HttpServletRequest request, HttpServletResponse response)
       throws ServletException, IOException {  
    try{

        response.setContentType("text/html");  
        PrintWriter out = response.getWriter();  

        int n= Integer.parseInt(request.getParameter("Cedula")); 
        String p=request.getParameter("Contraseña");  


        if(metodos.Validate.checkUser(n, p)){  
            RequestDispatcher rd=request.getRequestDispatcher("Welcome");  
            rd.forward(request,response);  
        }else{  
            out.print("Sorry username or password error");  
            RequestDispatcher rd=request.getRequestDispatcher("index.html");  
            rd.include(request,response);  
        }  

    out.close();  
    } catch (Exception e) {
        System.out.println("Hubo un error");
    }
}  

I hope that with this I clarify the doubts and you can understand where your error comes from. Greetings!

NOTE:

Another way for you to avoid this type of errors in case you are working with web pages, is that before sending the form, corroborate a javascript information to send, or, said input is not type "text", but "number".

Ex: <input type="number" id="Cedula" name="Cedula">

Success!

  

NumberFormatException Launched to indicate that the application has tried to convert a   string to one of the numeric types, but the string does not have the   appropriate format.

The NumberFormatException exception is generated in this line of code at try to convert a string that does not have a numeric format using Integer.parseInt() :

int n = Integer.parseInt(request.getParameter("Cedula"));

You have to ensure the value of request.getParameter("Cedula") is numeric to avoid the problem.

Examples:

Launch NumberFormatException :

Integer.parseInt(null);
Integer.parseInt("");
Integer.parseInt("a97");
Integer.parseInt("@#$");

Correct conversion:

Integer.parseInt("12");
Integer.parseInt("0");
Integer.parseInt("9245");

The Exception shown NumberFormatException is clear, the method parseInt (String value) of the class Integer wait for a string with a numeric value, if this value is not numerical, throw the Exception

Integer.parseInt("") /* Lanza Exception*/
Integer.parseInt("A") /* Lanza Exception*/
Integer.parseInt("123") /* No Lanza Exception*/

The possible error is that the parameter celula of request has a null or non-numeric value. To solve this, you would have to validate that the entry is a numeric value. One option would be to use Regular Expression. ^[1-9]\d*$ .

if(request.getParameter("Cedula")!=null){
  if(request.getParameter("Cedula").matches("^[1-9]\d*$")){
    System.out.println("Entrada Correcta");
  }
  else
    System.out.println("Entrada Incorrecta");
}
else
    System.out.println("Verificar los Datos");