Develop Reference

How to use the Scanner class properly?

I'm doing an assignment for class. For some reason the program completely skips the part where the variable name is supposed to be typed in by the user. I can't think of any reason why it's behaving this way, since the rest of my code that is after the cardType part (which asks for things such as String and int types work fine and in order.
System.out.println("Enter the card information for wallet #" +
(n+1) + ":\n---\n");
System.out.println("Enter your name:");
String name = scan.nextLine();
name = capitalOf(name);
System.out.println("Enter card type");
String cardType = scan.nextLine();
cardType = capitalOf(cardType);

You probably need to consume the end of the last line you read prior to trying to get the user name :
scan.nextLine(); // add this
System.out.println("Enter the card information for wallet #" +
(n+1) + ":\n---\n");
System.out.println("Enter your name:");
String name = scan.nextLine();
name = capitalOf(name);
System.out.println("Enter card type");
String cardType = scan.nextLine();
cardType = capitalOf(cardType);

It is behaving this way because I am quite sure you used the same scanner object to scan for integers/double values before you used it to scan for name.
Having said that does not mean you have to create multiple scanner objects. (You should never do that).
One simple way to over come this problem is by scanning for strings even if you are expecting integers/doubles and cast it back.
Scanner scn = new Scanner(System.in);
int numbers = scn.nextInt(); //If you do this, and it skips the next input
scn.nextLine(); //do this to prevent skipping
//I prefer this way
int numbers = Integer.toString(scn.nextLine());
String str = scn.nextLine(); //No more problems

How to get the right input for the String variable using scanner class? [duplicate]

This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 8 years ago.
I am learning Java and I was trying an input program. When I tried to input an integer and string using instance to Scanner class , there is an error by which I can't input string. When I input string first and int after, it works fine. When I use a different object to Scanner class it also works fine. But what's the problem in this method when I try to input int first and string next using same instance to Scanner class?
import java.util.Scanner;
public class Program {
public static void main(String[] args) {
Scanner input=new Scanner(System.in);
//Scanner input2=new Scanner(System.in);
System.out.println("Enter your number :" );
int ip = input.nextInt();
System.out.println("Your Number is : " + ip);
System.out.println("Enter Your Name : ");
String name= input.nextLine();
System.out.println("Your Next is : " + name);
}
}

nextInt() doesn't wait for the end of the line - it waits for the end of the token, which is any whitespace, by default. So for example, if you type in "27 Jon" on the first line with your current code, you'll get a value of 27 for ip and Jon for name.
If you actually want to consumer a complete line, you might be best off calling input.nextLine() for the number input as well, and then use Integer.parseInt to parse the line. Aside from anything else, that represents what you actually want to do - enter two lines of text, and parse the first as a number.
Personally I'm not a big fan of Scanner - it has a lot of gotchas like this. I'm sure it's fine when it's being used in exactly the way the designers intended, but it's not always easy to tell what that is.

If you call input.nextInt(); the scanner reads the number from the input, but leaves the line separator there. That means, if you call input.nextLine(); next, it reads everything till the next line separator. And this is in this case only the line separator itself.
You can fix that in two ways.
Way 1:
int ip = Integer.parseInt(input.nextLine());
// output
String name= input.nextLine();
Ways 2:
int ip = input.nextInt();
// output
input.nextLine();
String name= input.nextLine();

This one working, Anyway if you want to save IP address it must be String.
import java.util.Scanner;
public class Program {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
//Scanner input2=new Scanner(System.in);
System.out.println("Enter your number :");
int ip = input.nextInt();
System.out.println("Your Number is : " + ip);
System.out.println("Enter Your Name : ");
input.nextLine();
String name = input.nextLine();
System.out.println("Your Next is : " + name);
}
}

Validation input to be string only and numbers only JAVA [duplicate]

This question already has answers here:
How to check if a String is numeric in Java
(41 answers)
Closed 8 years ago.
I am a student and i have a little problem with validation inputs.
String name;
System.out.println("Enter name:");
name = keyboard.nextLine();
//If name contain other than alphabetical character print an error
double number;
System.out.println("Enter number:");
name = keyboard.nextDouble();
//If number contain other than number print an error
What i have tried is test to parse the string to double but i could not.
I have no idea how to test if the double is only number.
Please give me a clue of what i should do.

You can use Regular expression to check if the input match your constraint as follow :
String name;
System.out.println("Enter name:");
name = keyboard.nextLine();
if (!name.matches("[a-zA-Z_]+")) {
System.out.println("Invalid name");
}
String number;
System.out.println("Enter number:");
number = keyboard.nextLine();
if (!number.matches("[0-9]+")) {
System.out.println("Invalid number");
}
Here is a good tutorial to learn regex .

You can loop through each character of the String and check if it's not alphabetic using Character.isAlphabetic(char):
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter name:");
String name = keyboard.nextLine();
for (char c : name.toCharArray()) {
if (!Character.isAlphabetic(c)){
System.out.println("INVALID");
break;
}
}
To only accept numbers, you can do something similar using the Character.isDigit(char) function, but note that you will have to read the input as a String not a double, or get the input as a double and the converting it to String using Double.toString(d).

double number = 0;
try {
number = Double.parseDouble(name)
} catch (NumberFormatException ex) {
System.out.println("Name is not a double.");
}
If number is not a double, you can catch a NumberFormatException.

It seems you are using scanner. If you are, you can use the Scanner class' hasNextDouble() to check if the input is a double before reading the double as shown below:
double number;
System.out.println("Enter number:");
if (keyboard.hasNextDouble()){
name = keyboard.nextDouble();
}
Have a look at the Scanner class docs for more information.

There is also the "fancier" regex solution.
Java Pattern Documentation
You can use this (untested code) given you used nextLine() for reading BOTH inputs:
boolean isWordOnly = Pattern.matches("\w*", name); //name is in your code
boolean isFloatOnly = Pattern.matches("\d*.?\d*", number); //number is in your code too
Now the too boolean values tell you if there is the desired format in your input. You can add it in a do - while loop or anything you want.
It is a good idea to start studying reg(ular) ex(presions) because they are useful in string formatting (Imagine that you have to test if the input is a valid email...). Also used to check for SQL injections and many key things in apps and programs generally.

How do I make Java register a string input with spaces?

Here is my code:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String question;
question = in.next();
if (question.equalsIgnoreCase("howdoyoulikeschool?") )
/* it seems strings do not allow for spaces */
System.out.println("CLOSED!!");
else
System.out.println("Que?");
When I try to write "how do you like school?" the answer is always "Que?" but it works fine as "howdoyoulikeschool?"
Should I define the input as something other than String?

in.next() will return space-delimited strings. Use in.nextLine() if you want to read the whole line. After reading the string, use question = question.replaceAll("\\s","") to remove spaces.

Since it's a long time and people keep suggesting to use Scanner#nextLine(), there's another chance that Scanner can take spaces included in input.
Class Scanner
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace.
You can use Scanner#useDelimiter() to change the delimiter of Scanner to another pattern such as a line feed or something else.
Scanner in = new Scanner(System.in);
in.useDelimiter("\n"); // use LF as the delimiter
String question;
System.out.println("Please input question:");
question = in.next();
// TODO do something with your input such as removing spaces...
if (question.equalsIgnoreCase("howdoyoulikeschool?") )
/* it seems strings do not allow for spaces */
System.out.println("CLOSED!!");
else
System.out.println("Que?");

I found a very weird thing in Java today, so it goes like -
If you are inputting more than 1 thing from the user, say
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();
double d = sc.nextDouble();
String s = sc.nextLine();
System.out.println(i);
System.out.println(d);
System.out.println(s);
So, it might look like if we run this program, it will ask for these 3 inputs and say our input values are 10, 2.5, "Welcome to java"
The program should print these 3 values as it is, as we have used nextLine() so it shouldn't ignore the text after spaces that we have entered in our variable s
But, the output that you will get is -
10
2.5
And that's it, it doesn't even prompt for the String input.
Now I was reading about it and to be very honest there are still some gaps in my understanding, all I could figure out was after taking the int input and then the double input when we press enter, it considers that as the prompt and ignores the nextLine().
So changing my code to something like this -
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();
double d = sc.nextDouble();
sc.nextLine();
String s = sc.nextLine();
System.out.println(i);
System.out.println(d);
System.out.println(s);
does the job perfectly, so it is related to something like "\n" being stored in the keyboard buffer in the previous example which we can bypass using this.
Please if anybody knows help me with an explanation for this.

Instead of
Scanner in = new Scanner(System.in);
String question;
question = in.next();
Type in
Scanner in = new Scanner(System.in);
String question;
question = in.nextLine();
This should be able to take spaces as input.

This is a sample implementation of taking input in java, I added some fault tolerance on just the salary field to show how it's done. If you notice, you also have to close the input stream .. Enjoy :-)
/* AUTHOR: MIKEQ
* DATE: 04/29/2016
* DESCRIPTION: Take input with Java using Scanner Class, Wow, stunningly fun. :-)
* Added example of error check on salary input.
* TESTED: Eclipse Java EE IDE for Web Developers. Version: Mars.2 Release (4.5.2)
*/
import java.util.Scanner;
public class userInputVersion1 {
public static void main(String[] args) {
System.out.println("** Taking in User input **");
Scanner input = new Scanner(System.in);
System.out.println("Please enter your name : ");
String s = input.nextLine(); // getting a String value (full line)
//String s = input.next(); // getting a String value (issues with spaces in line)
System.out.println("Please enter your age : ");
int i = input.nextInt(); // getting an integer
// version with Fault Tolerance:
System.out.println("Please enter your salary : ");
while (!input.hasNextDouble())
{
System.out.println("Invalid input\n Type the double-type number:");
input.next();
}
double d = input.nextDouble(); // need to check the data type?
System.out.printf("\nName %s" +
"\nAge: %d" +
"\nSalary: %f\n", s, i, d);
// close the scanner
System.out.println("Closing Scanner...");
input.close();
System.out.println("Scanner Closed.");
}
}

Simple Java Scanner code not working [duplicate]

This question already exists:
Scanner issue when using nextLine after nextXXX [duplicate]
Closed 9 years ago.
Here is the skeleton of some basic code I am writing to make a simple game:
Scanner in = new Scanner(System.in);
String name;
String playing;
int age;
do {
System.out.println("Enter your name");
name = in.nextLine();
System.out.println("Enter your age");
age = in.nextInt();
System.out.println("Play again?");
playing = in.nextLine();
} while (true);
The code does not work as expected, for example, here is the expected functioning of the code:
Enter your name
John
Enter your age
20
Play again?
Yes
Enter your name
Bill
...
However, there is an issue with reading the Play again line, this is the actual output:
Enter your name
John
Enter your age
20
Play again?
Enter your name
As you can see "Enter your name" is being displayed again before "Play again?" is able to accept input. When debugging the playing variable is set to "", so there is no input that I can see and I cannot figure out what is being consumed.
Any help would be appreciated, thanks!

nextInt() doesn't consume the end-of-line, even if the int is the only thing in there.
Add another nextLine() after reading the int, and either discard its value completely, or check that it is empty if you want to prevent people from entering anything but an int.

The problem is that after calling nextInt() there is still a '\n' in the buffer and so that is what is passed. use in.next() instead of in.nextLine().

Use next() method in Scanner class instead of nextLine() method.
Scanner in = new Scanner(System.in);
String name;
String playing;
int age;
do {
System.out.println("Enter your name");
name = in.next();
System.out.println("Enter your age");
age = in.nextInt();
System.out.println("Play again?");
playing = in.next();
} while (true);
for more information refer Java Documentation for Scanner class
Output :
Enter your name
Kanishka
Enter your age
23
Play again?
Yes
Enter your name
....

You should use the following code for the next line reading.
Scanner scanner = new Scanner(System.in).useDelimiter("\n");
And for reading the line you should write
scanner.next();