This question already exists:
Scanner issue when using nextLine after nextXXX [duplicate]
Closed 9 years ago.
I am using a do-while loop that loops based on the input of the user as a 'Y' or 'N'. If the user chooses to continue by pressing 'Y' or 1 (for simplicity), on the second iteration it skips the first input in the same loop.
do{
System.out.println("Enter the location: ");
cartItem.purchaseLocation = input.nextLine(); // This input is skipped on second iteration
System.out.println("Enter the product description: ");
cartItem.productDescription = input.nextLine();
System.out.println("Enter the price of the item: ");
cartItem.productPrice = input.nextDouble();
System.out.println("Enter the quantity: ");
cartItem.quantity = input.nextInt();
cartList.add(cartItem);
System.out.println("Do you want to add more items ? y = 1 / n = 0: ");
exitVar = input.nextInt();
}while(exitVar!=0); // Repeat till no more items need to be added in the cart
So when the user presses input 0 to repeat the procedure, the line
cartItem.purchaseLocation = input.nextLine();
which is the very first input in the loop will be skipped. Any suggestions what might be wrong here ?
same question as Skipping nextLine() after use nextInt()
input.nextInt() does not read a line, when your input is "1\n", it read "1" and leaves "\n" to the next read, which is
cartItem.purchaseLocation = input.nextLine();
Instead of using input.nextLine() use input.next() assuming input is a Scanner object.
If it is not, please clarify what input is.
Related
enter image description here
}
Why does the code on line 6 have the same effect on line 8? Is there no distinction between the starting point of this while loop?
Also, you should follow this tips to use the Scanner properly:
Mixing any nextXXX method with nextLine from the Scanner class for user input, will not ask you for input again but instead result in an empty line read by nextLine.
To prevent this, when reading user input, always only use nextLine. If you need an int, do
int value = Integer.parseInt(scanner.nextLine());
instead of using nextInt.
Assume the following:
Scanner sc = new Scanner(System.in);
System.out.println("Enter your age:");
int age = sc.nextInt();
System.out.println("Enter your name:");
String name = sc.nextLine();
System.out.println("Hello " + name + ", you are " + age + " years old");
When executing this code, you will be asked to enter an age, suppose you enter 20.
However, the code will not ask you to actually input a name and the output will be:
Hello , you are 20 years old.
The reason why is that when you hit the enter button, your actual input is
20\n
and not just 20. A call to nextInt will now consume the 20 and leave the newline symbol \n in the internal input buffer of System.in. The call to nextLine will now not lead to a new input, since there is still unread input left in System.in. So it will read the \n, leading to an empty input.
So every user input is not only a number, but a full line. As such, it makes much more sense to also use nextLine(), even if reading just an age. The corrected code which works as intended is:
Scanner sc = new Scanner(System.in);
System.out.println("Enter your age:");
// Now nextLine, not nextInt anymore
int age = Integer.parseInt(sc.nextLine());
System.out.println("Enter your name:");
String name = sc.nextLine();
System.out.println("Hello " + name + ", you are " + age + " years old");
The nextXXX methods, such as nextInt can be useful when reading multi-input from a single line. For example when you enter 20 John in a single line.
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 7 years ago.
I have this piece of code that is supposed to add String values to an arrayList through the Scanner's next() or nextLine() method. The problem with next() is that it ignores everything after the first white space so I get that I should use the nextLine() method instead. The problem with nextLine() is that that it doesn't record the input, instead it stores a couple of blank spaces into the arrayList. Here's the code:
System.out.println("\nWhat is your idea? ");
String i = in.nextLine();
in.nextLine();
meals.add(i);
System.out.println("\n" + i + " has been entered into the idea pool. \n");
System.in.read();
I added the extra "in.nextLine()" after the initial "String i = in.nextLine()" because that's the only fix I found when I researched this problem but it doesn't work for me, it just still just stores a couple of blank spaces. Also, the System.in.read(); at the end there is only there so that it doesn't just jump forward after taking the input.
Here is the code where the above sample fits into:
ArrayList<String> meals = new ArrayList<String>();
String select = "";
while(!select.equals("")){
System.out.println("What would you like to do?");
System.out.println("1. <Irrelevant>");
System.out.println("2. Enter an idea");
System.out.println("3. <Irrelevant>");
System.out.println("4. <Irrelevant>");
System.out.println("Q. <Irrelevant>");
select = in.next();
switch(select){
case "1":
//Some stuff here.
case "2":
//Here's where the above problem fits into.
case "3":
//More stuff here
//and so on...
}
}
The reason why you are facing such issue is because you are using firstly next() method to read the input, and the further inputs you are taking using nextLine().
next() accepts the input, and the input pointer stays on the same line as of the current input.
So, as soon as you enter your choice and hit Enter, the choice is saved into the select variable, but, the input pointer is still on the same line. You should use a nextLine() to move the pointer to a new line.
Then you should use any number of nextLine()'s to receive multiple lines.
Also, remove an extra nextLine() method call from the case 2 statement. Remove the System.in.read() too, as your problem would have been solved.
ArrayList<String> meals = new ArrayList<String>();
String select = "";
while(!select.equals("")){
System.out.println("What would you like to do?");
System.out.println("1. <Irrelevant>");
System.out.println("2. Enter an idea");
System.out.println("3. <Irrelevant>");
System.out.println("4. <Irrelevant>");
System.out.println("Q. <Irrelevant>");
select = in.next();
in.nextLine(); // add this extra line in your code
switch(select){
case "1":
//Some stuff here.
case "2":
System.out.println("\nWhat is your idea? ");
String i = in.nextLine();
meals.add(i);
System.out.println("\n" + i + " has been entered into the idea pool. \n");
case "3":
//More stuff here
//and so on...
}
I am brand new at Java and this one is throwing me. Using the below code it loops through for the first question until I enter anything but an integer but after finishing that loop it does not stop for the remaining question.
Through a bit of research and reading I have found that I need to use the in.nextLine() to eat the newline character after the input. However no matter where I place the nextLine() it doesn't work? I thought it would be after the first int input = in.nextInt(); line but that did not work. Any help on where it would go and why?
System.out.print("How many CUs per course are remaining in your degree program? Enter any letter to quit: ");
while (in.hasNextInt()) { // Verify input is an integer
int input = in.nextInt();
if (input <= 0) // Verify that input is not negative or zero
{
System.out.println("Please enter a positive number or any letter to quit");
System.out.print("Add another course or any letter to quit: ");
} else {
courseCuList.add(input);
System.out.print("Add another course or any letter to quit: ");
}
}
System.out.print("How many CUs do you plan to take per term?");
while (in.hasNextInt()) {
int input = in.nextInt();
// in.nextLine(); This line consumes the \n
if (input <= 0) {
System.out.println("Please enter a whole positive number.");
System.out.println("How many CUs do you plan to take per term?");
} else {
cuPerTerm = in.nextInt();
}
}
Your problem is that in while (in.hasNextInt()) each call of hasNextInt needs to wait for user input, and then test if it is integer or not.
So each time user give integer, condition will be evaluated to ture, loop will execute and condition will need to be checked again, and if it is integer loop will execute again. This will go again and again until hasNextInt will be able to return false, for instance when user will give non-integer - like letter. But in this case condition in next loop will also return false because this non-integer value was not consumed after first loop. To let second loop work you would need to invoke nextLine two times
to consume line separator after previously put correct integer
to consume actual non-integer value
But this may also fail if user will not put any integer before non-integer value because there will be no line separator to consume.
So consider changing your logic to something similar to
boolean iterateAgain = true;
System.out.print("give me positive number: ");
while (iterateAgain) {
// this inner loop will move on only after getting integer
while (!in.hasNextInt()) {//here program waits for user input
in.nextLine();// consume non-integer values
System.out.print("that wasn't positive number, try again: ");
}
int number = in.nextInt();// now there must be number here
in.nextLine();// consume line separator
if (number > 0) {
System.out.println("you gave " + number);
// do what you want with this number
iterateAgain = false;// we can leave loop
} else
System.out.print("that wasn't positive number, try again: ");
}
If you want to execute next loop then all you need is reset iterateAgain value to true.
You need to read twice.
The exit condition on your while loop is hasNextInt() - checking to see if the next token is an integer doesn't actually clear that token, which means that the next nextLine() is going to read the token, and the subsequent nextLine() will read the newline character.
To demonstrate this, place the following between the loops:
System.out.println(in.nextLine() + " | " + in.nextLine());
For the input 4, 4, A, you will see the output:
How many CUs per course are remaining in your degree program? Enter any letter to quit: 4
Add another course or any letter to quit: 4
Add another course or any letter to quit: A
| A
How many CUs do you plan to take per term?
There are two tokens that need to be cleared from the buffer, and neither of them are integers. Because of this, no matter where you put nextLine(), it will fail - you need to insert it twice. If you only insert it once, the next token won't be an integer, and hasNextInt() will fail when the program tries to enter the second loop.
In order to get your program to work, simply insert:
in.nextLine(); in.nextLine();
before the second loop. (Note that you shouldn't put both this and the print-out in, as this will read four times.)
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 6 years ago.
I am using Java's Scanner to read user input. If I use nextLine only once, it works OK. With two nextLine first one doesnt wait for user to enter the string(second does).
Output:
X: Y: (wait for input)
My code
System.out.print("X: ");
x = scanner.nextLine();
System.out.print("Y: ");
y = scanner.nextLine();
Any ideas why could this happen? Thanks
It's possible that you are calling a method like nextInt() before. Thus a program like this:
Scanner scanner = new Scanner(System.in);
int pos = scanner.nextInt();
System.out.print("X: ");
String x = scanner.nextLine();
System.out.print("Y: ");
String y = scanner.nextLine();
demonstatres the behavior you're seeing.
The problem is that nextInt() does not consume the '\n', so the next call to nextLine() consumes it and then it's waiting to read the input for y.
You need to consume the '\n' before calling nextLine().
System.out.print("X: ");
scanner.nextLine(); //throw away the \n not consumed by nextInt()
x = scanner.nextLine();
System.out.print("Y: ");
y = scanner.nextLine();
(actually a better way would be to call directly nextLine() after nextInt()).
This question already has answers here:
Scanner issue when using nextLine after nextXXX [duplicate]
user input string and integers in Java [duplicate]
(3 answers)
Closed 8 years ago.
int id;
float grade;
String name;
Scanner z= new Scanner(System.in);
System.out.println("Give the id:\n");
id=z.nextInt();
System.out.println("your id is :"+id+"\n");
System.out.println("Give the name:");
name=z.nextLine();
System.out.println("your name is :"+name);
System.out.println("Give the grade:\n");
grade=z.nextFloat();
The problem goes like this.It inputs the integer but when it comes to the String, it prints "Give the name" but it doesn't waits until I type something, it skips to the next instruction.
Why's that?
You have used name=z.nextLine(), hence such behavior, Replace it with name=z.next(). Below is the edited code:
int id;
float grade;
String name;
Scanner z= new Scanner(System.in);
System.out.println("Give the id:\n");
id=z.nextInt();
System.out.println("your id is :"+id+"\n");
System.out.println("Give the name:");
name=z.next();
System.out.println("your name is :"+name);
System.out.println("Give the grade:\n");
grade=z.nextFloat();
When you read int value using nextInt, it reads only the int value, it skips the new line character. The latter will be read in the next nextLine causing it to skip the "real" input.
You can fix this by adding another nextLine before the "real" nextLine, it'll swallow the '\n' that you don't want to read.
Important note: Don't use int to store ID value! Use String instead!
The problem is with the input.nextInt() command it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine()
id = z.nextInt();
System.out.println ("your id is :"+id+"\n");
z.nextLine ();// add this line between the next line read
System.out.println("Give the name:");
or
id = Integer.parseInt(z.nextLine());
System.out.println ("your id is :"+id+"\n");
System.out.println("Give the name:");