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java.lang.NumberFormatException: For input string: “” [closed]

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Closed 1 year ago.
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The question is duplicate but I can't find a proper solution so please help me out with this,
I have to convert the credit card number(from edittext) to int.
Credit card number like : "3333 3333 3333 3333"
I removed white space using String removeWhiteSpace = cardNumEt.getText().toString().replace(" ", "");
Than converted to int like :
try
{
int nIntFromET = Integer.parseInt(removeWhiteSpace);
}
catch (NumberFormatException e)
{
Log.e("exptn",e.toString());
}
but unfortunately it's giving me exception :
java.lang.NumberFormatException: For input string: "3333333333333333"

The int type in Java can be used to represent any whole number from -2147483648 to 2147483647.
Your value is above the maximum positive integer.
But you can use instead long and BigInteger
class Scratch {
public static void main(String[] args) {
try
{
int nIntFromET = Integer.parseInt("2147483647");
}
catch (NumberFormatException e)
{
System.out.println(e);
}
}
}

Simple question about using names in else-if statements: [closed]

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I'm new and I need some guidance on how to use names in Java. I tried to use names to trigger the else if statements but I don't know how they work.
public class Else_If_Statements {
public static void main(String[] args){
int flavor = Chocolate;
if (flavor = Vanilla)
system.out.println("That will be 3$!");
else if (flavor = Caramel)
system.out.println("That will be 6$!");
else if (flavor = Chocolate)
system.out.println("That will be 5$!");
else
system.out.println("Sorry, that flavor isn't available!");
}
}

In Java, the if else works this way. The named value to be assigned to the String datatype and equals() method has to be used for the comparison.
public class Else_If_Statements {
public static void main(String[] args){
String flavor = "Chocolate";
if (flavor.equals("Vanilla"))
System.out.println("That will be 3$!");
else if (flavor.equals("Caramel"))
System.out.println("That will be 6$!");
else if (flavor.equals("Chocolate"))
System.out.println("That will be 5$!");
else
System.out.println("Sorry, that flavor isn't available!");
}
}

Simple Java Recursion Method [closed]

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Closed 5 years ago.
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What is the wrong of this code? need help.
public class TailRrecursion {
public static void tail(int i) {
if (i > 0) {
System.out.print(i + "");
tail(i - 1);
}
}
}

For a start, you are missing curly { braces
see
public class Tail-recursion{
public static void tail(int i) {
if(i>0){
System.out.print (i +"");
tail(i-1);
}
}
} // and here

Why isn't this recursion working in Java? [closed]

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Closed 6 years ago.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
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I am trying to calculate the factorial of the integer user inputs, but it does not return anything. Why?
Thanks.
import java.util.Scanner;
`class App {
public static int factorial(int n) {
if (n == 0) {
return 1;
} else {
int recurse = factorial(n - 1);
int result = n * recurse;
return result;
}
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter an integer to calculate its factorial: ");
int users = input.nextInt();
factorial(users);
}
}

The problem in your code is that you are have not giving a print statement for displaying factorial of the number entered. Just returning a value will not print it. If you are working in BlueJ environment, you can only use the code by directly executing the method factorial. Thank You.

The problem in your code is that , you are have not given a print statement for >displaying factorial of the number entered.

Cannot find i in the for loop [closed]

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Closed 9 years ago.
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I can't seem to get the last "i" in this code to fully function, it wont respond to the initial "i", just tells me that it "cannot find symbol". What am I doing wrong?
public void fillCB (JComboBox cb){
String sqlQueryCB = "select namn from ANSTALLD order by namn";
try {
ArrayList<String> listaAnstalld = idb.fetchColumn(sqlQueryCB);
for(int i = 0; i < listaAnstalld.size(); i++); {
cb.addItem(listaAnstalld.get(i));
}
}
catch(Exception e)
{
}

you have a semicolon at the end of your for loop, before the opening brace.
for(int i = 0; i < listaAnstalld.size(); i++); {
That semicolon closes the for loop. should remove it.
because of this, the contents of your block will not be in the scope of the for loop(and therefore not in the same scope that i is in).