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
Related
I'm making a java program that has to store data using classes and objects, My question is how do input characters like a name ( billy ) into my code.
Here is the class i did.
class bank
{
int AccountID;
int HolderName;
double AccountBalance;
}
And i'm assigning here
angel.AccountID = 7532;
angel.HolderName = 753; // angel
angel.AccountBalance = angelbalance;
I know that i can input integers using the following code
System.out.println("Set Balance for Angel: ");
int angelbalance = sc.nextInt();
My question is how do i input text/characters in a way (scanner) does with integer
Sorry for the bad explanation.
Do you mean getting the name input using the Scanner object? Because you've imported java.util.Scanner, you can do:
Scanner sc = new Scanner(System.in);
String name = sc.nextLine();
This will read the next line.
You can use (JOptionPane.showInputDialog) and it is easy to use if you want input from user
String name= JOptionPane.showInputDialog("Enter your name: "));
You need to either call the Scanner.nextLine() method or the Scanner.next() method to store String input such as a name like "Billy". Here is an example below:
//You can use the Scanner.next() method or the Scanner.nextLine() method to store Strings
Scanner scan = new Scanner(System.in);
System.out.println("Enter a name: ");
String billy = scan.next();
System.out.println("You entered: " + billy);
And here is your output:
Enter a name:
Billy
You entered: Billy
I'd check out the Scanner documentation here: https://docs.oracle.com/javase/8/docs/api/java/util/class-use/Scanner.html
im fairly sure the mistake I am making within this code is short sighted.
so this program starts by getting the first name and last name of the user and storing them as independent strings. the next part is for the program to manipulate that value into getting the first initial of the first name, which is where im having my problem (I have little experience with the CharArray function and have spent enough independent research time for me to opt to asking here lmao)
import java.util.Scanner; //Needed for the Scanner class
public class NumericTypes {
public static void main (String [] args) {
//TASK #2 Create a Scanner object here
//Reading from system.in
Scanner keyboard = new Scanner(System.in);
//prompt user for first name
System.out.println("Enter your first name: ");
//scans the next input as a double
String firstName = keyboard.nextLine();
//prompt user for last name
System.out.println("Enter your last name: ");
//scans the next input as a double
String lastName = keyboard.nextLine();
//concatenate the user's first and last names
String fullName = (firstName + " " + lastName);
//print out the user's full name
System.out.println(fullName);
//task 3 starts here
//get first initial from variable 'fullName'
String firstinitial = fullName.CharAt(0);
System.out.println("the first initial is: " + firstinitial);
}
}
my desired output is for the last set of lines to display the first initial of the first name (user input). any help would be greatly, greatly appreciated
This can be done in two ways -:
1.) Replace String firstinitial with char firstinitial
2.) Wrap fullName.charAt(0) with String.valueOf like this:
String firstinitial = String.valueOf(fullName.charAt(0));
Both will work just fine.
My code is below. Not sure how to eliminate this problem. I tried using next.Line(), but that did not work.
String demoEmpName;
int demoIdNum;
double demoPayRate;
int demoHoursWorked;
Scanner keyboard = new Scanner(System.in);
System.out.print("What is your name?");
demoEmpName = keyboard.nextString();
System.out.println("What is your ID number?");
demoIdNum = keyboard.nextInt();
System.out.println("What is your hourly pay rate?");
demoPayRate = keyboard.nextDouble();
System.out.println("How many hours did you work?");
demoHoursWorked = keyboard.nextInt();
Payroll pyrll = new Payroll(demoEmpName, demoIdNum, demoPayRate, demoHoursWorked);
Change keyboard.nextString() to keyboard.next().
There is no nextString() method in Scanner class.
Oh, and in case all this code you posted is not enclosed in some method, that would be another problem.
There is no existence of nextString() method in the Scanner class.If the employee's name consists of just a single word,you should use:-
demoEmpName=keyboard.next();
If the name consists of more than one words,use demoEmpName=keyboard.nextLine();
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);
}
}
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.");
}
}