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How do I make my code loop back to the beginning in Java?

I've read that I would need to put "continue" after an "if" statement, but every I tried this with my if statement and it states that "continue cannot be used outside of a loop."

Set it in the loop. EX:
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
while (true){
System.out.print("Enter a password: The password must have at least eight characters, only letters and digits, and at least two digits. ");
String s = input.nextLine();
if (thepassword(s)) {
System.out.println("Valid Password");
break;
} else {
System.out.println("Invalid Password");
}
}
}
See more : Java Break and Continute

Use an outer infinite loop with a lable, then break the loop using break loop_lable. Check until a valid input is made.
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter a password: The password must have at least eight characters, only letters and digits, and at least two digits. ");
loop:for(;;)
{
String s = input.nextLine();
if (thepassword(s))
{
System.out.println("Valid Password");
break loop;
}
else
{
System.out.println("Invalid Password");
continue loop;
}
}
input.close();
}
public static boolean thepassword(String password) {
boolean thepassword = true;
if (password.length() < 8) {
thepassword = false;
} else {
int totaldigits = 0;
for (int n = 0; n < password.length(); n++) {
if (thedigit(password.charAt(n)) || theletter(password.charAt(n))) {
if (thedigit(password.charAt(n))) {
totaldigits++;
}
} else {
thepassword = false;
break;
}
}
if (totaldigits < 2) {
thepassword = false;
}
}
return thepassword;
}
public static boolean theletter(char c) {
return ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z'));
}
public static boolean thedigit(char c) {
return (c >= '0' && c <= '9');
}
}

Need to prompt user to enter integers until binary number is entered

This is the program that checks if the input integer is binary or not, now I need to create a loop that will prompt the user to renter integers until binary number is entered.
import java.util.Scanner;
public class BinaryNumbers {
public static void main(String[] args) {
int value, userValue;
int binaryDigit = 0, notBinaryDigit = 0;
Scanner scan = new Scanner(System.in);
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0) {
if ((userValue % 10 == 0) || (userValue % 10 == 1)) {
binaryDigit++;
} else {
notBinaryDigit++;
}
userValue = userValue / 10;
}
if (notBinaryDigit == 0) {
System.out.println(value + " is a Binary Number.");
} else {
System.out.println(value + " is not a Binary Number.");
}
}
}

Why not using Regular Expressions?
All those checking on user inputs by assuming them as numbers (by calling Scanner#nextInt or Scanner#nextFloat or ...) are very breakable. How can be so sure user won't enter anything wrong?
It's better to hold the user input in a String variable and check it against being binary integer (which has to be 32 bit at most as defined in many languages such as java) using Regex is more safe:
import java.util.Scanner;
public class CheckBinaryInteger {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
boolean isValid = false;
String userInput = "";
do {
System.out.print("Please Enter a binary integer: ");
userInput = sc.next();
isValid = userInput != null && !userInput.trim().isEmpty()
&& userInput.matches("[01]{1,32}");//assume every digit as one bit
if(!isValid)
System.out.println("invalid binary integer entered! ");
}while(!isValid);
System.out.println("Valid input: "+userInput);
}
}
Hope this helps.

import java.util.Scanner;
public class BinaryNumbers {
public static void main(String[] args) {
int value, userValue;
Scanner scan = new Scanner(System.in);
while(true){
int binaryDigit = 0, notBinaryDigit = 0;
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0) {
if ((userValue % 10 == 0) || (userValue % 10 == 1)) {
binaryDigit++;
} else {
notBinaryDigit++;
}
userValue = userValue / 10;
}
if (notBinaryDigit == 0) {
System.out.println(value + " is a Binary Number.");
return; //does the trick ;)
} else {
System.out.println(value + " is not a Binary Number.");
}
}
}
}
A simple return can end the program then and there :)

import java.util.Scanner;
class calculatePremium
{
public static void main(String[] args) {
int value, userValue;
int binaryDigit = 0, notBinaryDigit = 0;
Scanner scan = new Scanner(System.in);
while(true) /*use a while loop to iterate till you get the binary number*/
{
binaryDigit = 0; notBinaryDigit = 0;
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0) {
if ((userValue % 10 == 0) || (userValue % 10 == 1)) {
binaryDigit++;
} else {
notBinaryDigit++;
}
userValue = userValue / 10;
}
if (notBinaryDigit == 0) {
System.out.println(value + " is a Binary Number.");
break; /* breaks out of loop when gets the correct input */
} else {
System.out.println(value + " is not a Binary Number.\n");
}
}
}
}
You just need to use a loop till you get a binary number.
Hope it helps.

import java.util.Scanner;
public class BinaryNumbers {
public static void main(String[] args) {
int value, userValue;
int binaryDigit = 0, notBinaryDigit = 0;
Scanner scan = new Scanner(System.in);
while(true)
{
System.out.println("Please enter positive integers: ");
userValue = scan.nextInt();
value = userValue;
while (userValue > 0)
{
if ((userValue % 10 == 0) || (userValue % 10 == 1))
binaryDigit++;
else
notBinaryDigit++;
userValue = userValue / 10;
}
if (notBinaryDigit == 0)
{
System.out.println(value + " is a Binary Number.");
break;
}
else
System.out.println(value + " is not a Binary Number.");
}
}
}

A loop for checking prime number

package pureTest;
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class test3 {
public static void main(String[] args) {
/* Enter your code here. */
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for (int i = 2; i< n; i++){
if( n <= 3){
System.out.println("Prime");
break;
}else if( n%i ==0){
System.out.println("Not Prime");
break;
}else{
System.out.println("Prime");
}
}
}
}
the input of 7; the out put is repetitions of Prime:
7
Prime
Prime
Prime
Prime
Prime
Just wondering why the if condition doesn't work out here.

your code will print prime until it finds divisor !
for (int i = 2; i< n; i++){
if( n <= 3){
System.out.println("Prime");
break;
}else if( n%i ==0){
System.out.println("Not Prime");
break;
}else{
System.out.println("Prime"); --> this line will be printed every time in your loop!
}
}
Also you don't need to iterate till n, as after n/2 there would be no number which can divide n :-)
Check this code...
private static boolean checkPrime(int n) {
int i = 2;
while(i<=n/2){
if(n%i++ == 0){
return false;
}
}
return true;
}

Your else clause is wrong. It prints "Prime" each time n is not divisible by i. It will even print prime for non prime inputs (for example, it will print "Prime" for 21 before printing "Not Prime", since 21%2 != 0).
Change your loop to something like :
for (int i = 2; i < n; i++) {
if( n <= 3){
System.out.println("Prime");
return;
} else if(n%i == 0){
System.out.println("Not Prime");
return;
}
}
System.out.println("Prime");

public static void main(String[] args) {
/* Enter your code here. */
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
boolean isPrime = true;
for (int i = 2; i< n; i++){
if( n <= 3){
break;
}else if( n%i ==0){
System.out.println("Not Prime");
break;
}
}
if (isPrime) {
System.out.println("Prime");
} else {
System.out.println("Not prime");
}
}
If you want to simply check in the main function the above is ok. If you want to do that well try the following:
public static void main() {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
if (isPrime(n)) {
System.out.println("Prime");
} else {
System.out.println("Not prime");
}
}
boolean isPrime(int n) {
if (n < 2) {
return false;
if (n == 2)
return true;
if (n%2 == 0)
return false;
for (int i = 3; i*i<=n; i+=2){
if (n%i == 0)
return false
return true;
}

boolean isPrime = true;
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for (int i = 2; i< n; i++){
if(n%i==0) {
System.out.println("Not Prime");
isPrime=false;
break;
}
}
if(isPrime) {
System.out.println("Prime");
}

Java: checking if a phone number fits a valid format

I am trying to find if a phone number fits the format (xxx)xxx-xxxx where x is a digit. First, I have the program for the length and the '(',')', and '-'. When I type in something that doesn't work, I get the logical output. However, when I type in a properly formatted number, my program doesn't return anything.
import java.util.Scanner;
public class Program04 {
public static void main(String args[])
{
Scanner stdIn = new Scanner(System.in);
String pN;
System.out.print("Please enter a phone number (ddd)ddd-dddd :");
pN = stdIn.nextLine();
char a = pN.charAt(1);
char b = pN.charAt(2);
char c = pN.charAt(3);
char d = pN.charAt(5);
char e = pN.charAt(6);
char f = pN.charAt(7);
char g = pN.charAt(9);
char h = pN.charAt(10);
char i = pN.charAt(11);
char j = pN.charAt(12);
if (pN.length() == 13 && pN.charAt(0)=='(' && pN.charAt(4)== ')' && pN.charAt(8)=='-')
{
if (a>=0 && a<=9)
{
if (b>=0 && b<=9)
{
if (c>=0 && c<=9)
{
if (d>=0 && d<=9)
{
if (e>=0 && e<=9)
{
if (f>=0 && f<=9)
{
if (g>=0 && g<=9)
{
if (h>=0 && h<=9)
{
if (i>=0 && i<=9)
{
if (j>=0 && j<=9)
{
System.out.print("This is a valid phone number!");
}
}
}
}
}
}
}
}
}
}
}
else System.out.println("Not a vaid phone number.");
}
}

It's easier to use pattern-matching (regex) for validation:
...
pN = stdIn.nextLine();
System.out.println(pN.matches("\\(\\d{3}\\)\\d{3}-\\d{4}"));
Even if you want to have it check if each character is a digit, using so many nested if's is not the way to go. Instead, define a simple method that applies the check, say:
private static boolean isDigit(char x) {
return x>=0 && x<=9;
}
and then do:
if ( isDigit(a) && isDigit(b) && ...) {
return true;
}
else {
return false;
}

If you're not allowed to use RegEx or if it is too difficult to understand, try simplifying your nested if's by a simple switch inside a loop, it is much more readable and maintenance is the easiest :
public static void main(String[] args) {
String pn = scan.nextLine();
boolean valid = true;
if (pn.length() == 13){
for (int i = 0 ; i < 13 ; i++){
switch(i){
case 0 : valid = pn.charAt(0) == '(';break;
case 4 : valid = pn.charAt(4) == ')';break;
case 8 : valid = pn.charAt(8) == '-';break;
default : valid = Character.getNumericValue(pn.charAt(i)) >= 0 && Character.getNumericValue(pn.charAt(i))<= 9 ; break;
}
if (!valid) break;
}
if (!valid) System.out.println("Invalid number");
}else{
valid = false;
System.out.println("Invalid length");
}
if (valid){
System.out.println(pn + " is a valid telephone number");
}
}
Also, to avoid working with the ASCII value of a char, try using the Character.getNumericValue(Char c) method. It returns a numeric that you can use for your tests, like above.

It is better to use regex in this case:
You can use following :
String pattern = "(\\(\\d{3}\\)\\d{3}-\\d{4})";
Pattern r = Pattern.compile(pattern);
pN = stdIn.nextLine();
Matcher m = r.matcher(pN);
if (m.find( )) {
System.out.println("Found value: " + m.group(0) );
} else {
System.out.println("NO MATCH");
}

How to check if a character in a string is a digit or letter

I have the user entering a single character into the program and it is stored as a string. I would like to know how I could check to see if the character that was entered is a letter or a digit. I have an if statement, so if its a letter its prints that it's a letter, and the same for a digit. The code I have so far doesn't work but I feel like I'm close. Any help you can offer is appreciated.
System.out.println("Please enter a single character: ");
String character = in.next();
System.out.println(character);
if (character.isLetter()){
System.out.println("The character entered is a letter.");
}
else (character.isDigit()){
Syste.out.println("The character entered is a digit.");

You could use:
if (Character.isLetter(character.charAt(0))){
....

You could use the existing methods from the Character class. Take a look at the docs:
http://download.java.net/jdk7/archive/b123/docs/api/java/lang/Character.html#isDigit(char)
So, you could do something like this...
String character = in.next();
char c = character.charAt(0);
...
if (Character.isDigit(c)) {
...
} else if (Character.isLetter(c)) {
...
}
...
If you ever want to know exactly how this is implemented, you could always look at the Java source code.

Ummm, you guys are forgetting the Character.isLetterOrDigit method:
boolean x;
String character = in.next();
char c = character.charAt(0);
if(Character.isLetterOrDigit(charAt(c)))
{
x = true;
}

This is a little tricky, the value you enter at keyboard, is a String value, so you have to pitch the first character with method line.chartAt(0) where, 0 is the index of the first character, and store this value in a char variable as in char c= line.charAt(0)
now with the use of method isDigit() and isLetter() from class Character you can differentiate between a Digit and Letter.
here is a code for your program:
import java.util.Scanner;
class Practice
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
System.out.println("Input a letter");
String line = in.nextLine();
char c = line.charAt(0);
if( Character.isDigit(c))
System.out.println(c +" Is a digit");
else if (Character.isLetter(c))
System.out.println(c +" Is a Letter");
}
}

By using regular expressions:
boolean isChar = character.matches("[a-zA-z]{1}");
boolean isDigit = character.matches("\\d{1}");

char charInt=character.charAt(0);
if(charInt>=48 && charInt<=57){
System.out.println("not character");
}
else
System.out.println("Character");
Look for ASCII table to see how the int value are hardcoded .

This is the way how to check whether a given character is alphabet or not
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
char c = sc.next().charAt(0);
if((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z'))
System.out.println(c + " is an alphabet.");
else
System.out.println(c + " is not an alphabet.");
}

char temp = yourString.charAt(0);
if(Character.isDigit(temp))
{
..........
}else if (Character.isLetter(temp))
{
......
}else
{
....
}

import java.util.*;
public class String_char
{
public static void main(String arg[]){
Scanner in = new Scanner(System.in);
System.out.println("Enter the value");
String data;
data = in.next();
int len = data.length();
for (int i = 0 ; i < len ; i++){
char ch = data.charAt(i);
if ((ch >= '0' && ch <= '9')){
System.out.println("Number ");
}
else if((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')){
System.out.println("Character");
}
else{
System.out.println("Symbol");
}
}
}
}

You need to convert your string into character..
String character = in.next();
char myChar = character.charAt(0);
if (Character.isDigit(myChar)) {
// print true
}
Check Character for other methods..

You could do this by Regular Expression as follows
you could use this code
EditText et = (EditText) findViewById(R.id.editText);
String NumberPattern = "[0-9]+";
String Number = et.getText().toString();
if (Number.matches(NumberPattern) && s.length() > 0)
{
//code for number
}
else
{
//code for incorrect number pattern
}

I have coded a sample program that checks if a string contains a number in it! I guess it will serve for this purpose as well.
public class test {
public static void main(String[] args) {
String c;
boolean b;
System.out.println("Enter the value");
Scanner s = new Scanner(System.in);
c = s.next();
b = containsNumber(c);
try {
if (b == true) {
throw new CharacterFormatException();
} else {
System.out.println("Valid String \t" + c);
}
} catch (CharacterFormatException ex) {
System.out.println("Exception Raised-Contains Number");
}
}
static boolean containsNumber(String c) {
char[] ch = new char[10];
ch = c.toCharArray();
for (int i = 0; i < ch.length; i++) {
if ((ch[i] >= 48) && (ch[i] <= 57)) {
return true;
}
}
return false;
}
}
CharacterFormatException is a user defined Exception. Suggest me if any changes can be made.