Develop Reference

Java recursion method returns wrong value

I have a bit of a problem with this recursion method. I'm fairly new to Java.
This method checks if an input is either "exit" only or "start" followed by two times either "user" or "easy".
It works fine except for the return. If I enter a wrong input and then a right on it returns the previous wrong input with which I obviously can't continue working, why is that?
I've had this problem before but always somehow managed to avoid it.
You might notice that I print out a valid commadnd right when it's validated, this works fine and produces the result I need. But when printing out the return of the function on line 2 the above mentioned problem takes place. I've added numbers to the printed strings so I can recognize which is which.
I have tried returning immediately when there's a valid command but I still need that retrun at the end since the function gives me an error if return statements are exclusively in conditional statements so the problem persists.
Thanks for any help!
public static void main(String[] args) {
System.out.println(setup() + "3");
}
static String setup() {
System.out.print("Input command: ");
String command = input.nextLine();
String[] split = command.split(" ");
if(!(command.equals("exit") || split.length == 3)) {
System.out.println("Invalid parameters!");
setup();
}
else {
if(command.equals("exit")) {
System.out.println("Valid parameters! Exit");
System.out.println(command + "2");
}
else if(split[0].equals("start") && (split[1].equals("easy") || (split[1].equals("user")) && split[2].equals("easy") || split[2].equals("user"))) {
System.out.println("Valid parameters! Start");
System.out.println(command + "1");
}
else {
System.out.println("Invalid parameters!");
setup();
}
}
return command;
}

first of all I think that you meant to call the recusive call as
return setup()
second of all when using conditional operator (&&, ||) you should use () for make sure you get the logic condition you expect.
if you will update it to :
return setup instead of setup()
validate what you wrap the right part of condition with Parenthesis():
else if (split[0].equals("start") && ((split[1].equals("easy") || split[1].equals("user")) && (split[2].equals("easy") || split[2].equals("user")))) { System.out.println("Valid parameters! Start"); System.out.println(command + "1"); }

Handling exceptions in Java: How to reject int?

I'm having a problem handling exceptions. Honestly, I really don't understand how it works since I self study.
I'm working with a program where there would be a main menu with the following choices.
Odd/Even - asks an integer input from user and identify if it is an odd or even. Program would continuously ask for an integer input if the user keeps on giving character inputs. (I was able to do this but I keep on getting errors when I use br.readLine() in getting input. Pls see codes below. So I used the normal parsing. Since I didn't use Buffered Reader, I try to delete it but the Odd/Even program wouldn't handle the exception without it.)
Vowel/Consonant - asks the user for a character input and identify if it is a vowel or a consonant. Program should reject integer inputs. The program I made with the codes below doesn't reject integer inputs. I tried searching for answers but I can't find one.
Please ignore for now.
My problem/s involve/s the following questions.
1. Why doesn't the program Odd/Even handle the NumberFormat exception whenever I try to delete the BufferedReader line even though it wasn't used in the whole program?
How can I reject integer inputs for the Vowel/Consonant program?
Here is a video when I tried to run the program.
http://tinypic.com/r/24ou9kz/9
When I exit the program, the console shows this.
Exception in thread "main" java.lang.NumberFormatException: null at
java.lang.Integer.parseInt(Unknown Source) at
java.lang.Integer.parseInt(Unknown Source)
import javax.swing.JOptionPane;
import java.io.*;
import java.util.InputMismatchException;
public class DoWhileIf {
public static void main(String[] args) throws Exception {
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String input;
int choice, num = 0;
char again = 0;
boolean err = true;
do {
input = JOptionPane.showInputDialog("Menu\n[1] Odd/Even\n[2] Vowel/Consonant\n[3] CQM\n[4] Fuel Efficiency\n[5] Scholarship\n[6] Exit program.\n\nEnter Choice.");
choice = Integer.parseInt(input);
if (choice == 1) {
do {
do {
try {
input = JOptionPane.showInputDialog("Input an integer : ");
num = Integer.parseInt(input);
err = false;
} catch (NumberFormatException o) {
JOptionPane.showMessageDialog(null,"Error!");
err = true;
}
} while (err);
if (num % 2 == 0) {
JOptionPane.showMessageDialog(null,"Even.");
}
else {
JOptionPane.showMessageDialog(null,"Odd.");
}
do {
input = JOptionPane.showInputDialog("Try again? Press Y for yes or N to go back to main menu.");
again = input.charAt(0);
} while (again != 'Y' && again != 'y' && again !='N' && again !='n');
} while (again == 'Y' || again == 'y');
}
if (choice == 2) {
char letter = 0;
do {
do {
try {
input = JOptionPane.showInputDialog("Character : ");
letter = input.charAt(0);
err = false;
} catch (InputMismatchException a) {
JOptionPane.showMessageDialog(null,"Error!");
err = true;
}
} while (err);
if (letter == 'a' || letter == 'A' || letter == 'e' || letter == 'E' || letter == 'i' || letter == 'I' || letter == 'o' || letter == 'O' || letter == 'u' || letter == 'U') {
JOptionPane.showMessageDialog(null,"Vowel");
}
else {
JOptionPane.showMessageDialog(null,"Consonant");
}
do {
input = JOptionPane.showInputDialog("Try again? Press Y for yes or N to go back to main menu.");
again = input.charAt(0);
} while (again != 'Y' && again != 'y' && again !='N' && again !='n');
} while (again == 'Y' || again == 'y');
}
} while (choice <= 0 || choice > 6 || again == 'N' || again == 'n');
}

Why doesn't the program Odd/Even handle the NumberFormat exception whenever I try to delete the BufferedReader line even though it wasn't
used in the whole program?
I am not able to duplicate this problem. I removed the BufferedReader and option #1 works the same as it did before. I entered integer values, special characters, letters, spaces and it works fine.
How can I reject integer inputs for the Vowel/Consonant program?
You could modify your else condition from this:
else {
JOptionPane.showMessageDialog(null,"Consonant");
}
to this:
else if(Character.isLetter(letter)){
JOptionPane.showMessageDialog(null,"Consonant");
}
else{
JOptionPane.showMessageDialog(null,"Error! You must enter a valid letter.");
}
When I exit the program, the console shows this.
Exception in thread "main" java.lang.NumberFormatException: null at ...
Regarding the NumberFormatException you're seeing, I'm guessing you're pressing the Cancel button on the dialog. When you press cancel the variable input receives the value null. When you try to parse null as an integer it fails and throws the exception:
Exception in thread "main" java.lang.NumberFormatException: null
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at exception.DoWhileIf.main(DoWhileIf.java:18)
Line 18 is this line: choice = Integer.parseInt(input);
Notice how the exception told us - java.lang.NumberFormatException: null which tells us that the parameter being passed to the parseInt method is null.
Lastly some additional thoughts for you to consider:
Whenever you get input from the user you must account for all the possibilities somehow. For example when you have code like this:
letter = input.charAt(0);
you're not accounting for the possibility that the input could be null or empty in which case this logic will throw an exception.
A concrete example is when the user clicks Cancel on the dialog that asks whether they want to try again:
input = JOptionPane.showInputDialog("Try again? Press Y for yes or N to go back to main menu.");
When the user clicks Cancel on this dialog the same thing happens that I described above regarding the NumberFormatException - input becomes null. If you try to use input like this:
again = input.charAt(0);
it will fail with the exception:
Exception in thread "main" java.lang.NullPointerException
because you can't invoke a method on a null.
Another example is when the user enters nothing at the main menu but simply presses OK. The result is this exception:
Exception in thread "main" java.lang.NumberFormatException: For input string: ""
This happens because input was set to an empty string and parseInt does not know how to convert empty string into an integer value.
Another item I want to bring up is that you're using this same piece of code over and over again. Whenever you have code you want to reuse you should not copy and paste it but instead create a method, object, or other construct so that you can refer to it.
do {
input = JOptionPane.showInputDialog("Try again? Press Y for yes or N to go back to main menu.");
again = input.charAt(0);
} while (again != 'Y' && again != 'y' && again !='N' && again !='n');
Breaking up your logic into smaller more manageable pieces will help you to debug, test, and maintain your code more easily.
Another point I want to touch on regarding this same block of logic is that you're using the same kind of dialog to ask for many different kinds of input. Since you're using a GUI dialog, you could use a dialog that is better suited to your task such as one that asks the user to press either a Yes button or No button.
You can learn more about different kinds of dialogs by reading the How to Make Dialogs Tutorial
Here is an example of how you could create a more friendly dialog:
/**
* Asks the user if they want to try something again and
* returns a boolean representing the user's response.
* #return true if the user answers Yes, false otherwise.
*/
private static boolean promptToRepeatSelectedOption(){
int n = JOptionPane.showOptionDialog(null,
"Try again?",
"Repeat Selection",
JOptionPane.YES_NO_OPTION,
JOptionPane.QUESTION_MESSAGE,
null,
null,
null);
return n == JOptionPane.YES_OPTION;
}
The above method, when invoked, will create and display a dialog with two buttons - Yes and No - and the user will have to select one of them or close the dialog. The logic simply looks for if the user said Yes (by clicking the Yes button) and returns true when that is the case. If the user closes the dialog or chooses the No option the method returns false simply because either of those two scenarios will cause the n == JOptionPane.YES_OPTION comparison to result in a value of false.
You can replace your entire loop with a call to this method like this:
First, define a variable to hold the user's response.
boolean repeat = false;
Then invoke the method and set the variable to its result:
repeat = promptToRepeatSelectedOption();
Now replace the outer loop condition
while (again == 'Y' || again == 'y');
with this: while (repeat);
and finally replace part of the outermost loop condition
again == 'N' || again == 'n'
with this: !repeat
One final thought is that you're using very general error messages when the user enters something incorrect or invalid:
JOptionPane.showMessageDialog(null, "Error!");
It's always better to explain to the user a little bit about what they did wrong so that they know how to avoid the error next time. You should probably consider adding more detail to your error messages.
Hope this helps!

How to validate phone number

I am trying to validate phone numbers in java. In my country, phone numbers either begin with 9 or 8 and only have 8 numbers. I have done
try {
Integer.parseInt(phoneNo);
}
catch (NumberFormatException e) {
msg += "Plese enter amount in Integers.\n";
}
if (phoneNo.length() == 0)
msg += "Please Enter Phone Number.\n";
if (phoneNo.length() != 8)
msg += "Invalid Phone Number.\n";
However I need to validate when the first digit of the number isn't 9 or 8. I am not entirely sure of how I am supposed to do that. Please explain how your code works as I am a student and I am trying to learn.

Just in case you are looking for a regular expression solution.
You can do use the following pattern ^(?=(?:[8-9]){1})(?=[0-9]{8}).* to perform the check.
Essentially what it does is;
From first character position ^
Look ahead and see if the first character is a 8 or 9 (?=(?:[8-9]){1})
Then see if there are a total of 8 digits (?=[0-9]{8})
If the above conditions is a match then mark this as matched .*
public static void main(String[] args) {
String telephoneNr = "88765432";
if (telephoneNr.matches("^(?=(?:[8-9]){1})(?=[0-9]{8}).*")) {
System.out.println("Valid phone number!");
}
else {
System.out.println("Invalid!");
}
}
Output:
Valid phone number!

The methods that you need to put this together are readily available on the String and Character classes.
Here is an example program that does what you are looking for:
public class Foo {
public static void main(String[] args) {
// First try null and the empty string
System.out.println(isValidPhoneNumber(null));
System.out.println(isValidPhoneNumber(""));
// Now try an otherwise valid string that doesn't have the right first character
System.out.println(isValidPhoneNumber("01234567"));
// Now try an invalid string
System.out.println(isValidPhoneNumber("9a934581"));
// Finally a valid number
System.out.println(isValidPhoneNumber("94934581"));
}
static boolean isValidPhoneNumber(String phoneNo) {
// First validate that the phone number is not null and has a length of 8
if (null == phoneNo || phoneNo.length() != 8) {
return false;
}
// Next check the first character of the string to make sure it's an 8 or 9
if (phoneNo.charAt(0) != '8' && phoneNo.charAt(0) != '9') {
return false;
}
// Now verify that each character of the string is a digit
for (char c : phoneNo.toCharArray()) {
if (!Character.isDigit(c)) {
// One of the characters is not a digit (e.g. 0-9)
return false;
}
}
// At this point you know it is valid
return true;
}
}
The output it produces is:
false
false
false
false
true
The final for-each loop could avoid re-checking the first character by using a for loop with an explicit counter, but the performance gain of not checking a single int doesn't outweigh the cleaner code and better readability of the for each construct.
Edit: also please note that I removed the validation error messages from the original question for better readability as the OP asked to explain what the code was doing.

Instead of taking the phone number in a Integer variable take it in a String variable.
Then check whether the 1st number is 9, 8 or not by using stringVariable.charAt(0)
and for length of the phone number use int len=stringVariable.length();

You can check the first character of the phoneNo:
if (phoneNo.charAt(0) != '9' && phoneNo.charAt(0) != '8') {
// the first character is not a 9 or an 8
}
Documentation for charAt from Oracle.

(Java) use one code for multiple responses?

If in a code, I have two options - choosing 1 or 2 -- in '2' you go straight to another code, but with '1' you must do a few things before and then do to that code, is this possible with one code? This is an example:
Scanner scanner = new Scanner(System.in);
reader = scanner.nextInt();
if (reader == 1) {
System.out.println("Not yet, pick another one");
int number = scanner.nextInt();
}
} else if (reader == 2) {
System.out.println("Okay, you have picked 2!");
}
Is it possible to go directly from the first to the second? Provided the first one picks '2' in the second try, do I have to retype the code in that area or can I move it directly to the else if(reader == 2){....} code after this?
Thanks!

Sure, just modify you're conditions a bit:
int someVar = ...;
if(someVar == 1){
...
//do stuff before other code
}
...
//do stuff you do for 1 and 2

You can't use else-if in this scenario, since the else-if part will never be evaluated if the first condition is true. You need two separate conditions :
int reader = scanner.nextInt();
if (reader == 1){
System.out.println("Not yet, pick another one");
reader = scanner.nextInt();
}
if (reader == 2) {
System.out.println("Okay, you have picked 2!");
}
Note that you also have to read both integers into the same variable.

While not (or equivalent) java

I'm making a program with Java that needs to involve some error checking. I can stop users from entering bad numerical inputs like this (assume the input scanner has already been created):
while (n == 0){
System.out.println("Can't use 0 as a denominator! Please enter a real, nonzero number");
n = input.nextInt();
}
But how do I stop users from entering an invalid string? I can't use !=, because strings can only be compared with the string.equals() method, right? So, is there a while not loop? ie:
while !(string.equals("y") || string.equals("n")){
//here have code
}
Or something of that nature?

While there is no such thing as a while-not loop, you can always invert the condition:
while (!(string.equals("y") || string.equals("n"))){
This is read, "while the string is not equal to "y" or "n"".
You could also apply DeMorgan's identity to rewrite this as:
while (!(string.equals("y")) && !(string.equals("n"))){
which is a bit clearer as "While the string isn't equal to "y" and isn't equal to "n"".

There isn't a while-not instruction, but you can simply negate the condition in a normal while loop. Try this:
while (!string.equals("y") && !string.equals("n"))
Or even better, to guard against the case where the string is null and/or it's in a different case:
while (!"y".equalsIgnoreCase(string) && !"n".equalsIgnoreCase(string))

You almost get it, just change where you position your !
like this:
while (!(string.equals("y") || string.equals("n")))

Why not try regex?
Scanner sc = new Scanner(System.in);
String string = sc.nextLine();
while (!string.matches("(?i)^(?:y|n|yes|no)$"))
{
System.out.println("Invalid input...");
string = sc.nextLine();
}
boolean answer = string.matches("(?i)^(?:y|yes)$");