Why is the first user input ignored when entering string input?

Question

if((type.toLowerCase()).equals("phone number")) {                                   
    System.out.println("Enter phone number: ");
    kb.nextLine();
    individual.setPhoneNumber(kb.nextLine());
}

Basically if the user enters a phone number the first time, it is ignored. The user must then re-enter the information for the string and it will work the second time.

For an identical piece of code in which an integer is enterred, however, it works on the first input.

Maybe http://stackoverflow.com/questions/13102045/skipping-nextline-after-use-nextint — Sotirios Delimanolis, Nov 07 '14 at 23:16
I think we need some more info. For example, what type is kd variable? Other than that, Why do you call kd.nextLine() once first (ignoring the return value) before setting the value from the second call? — Lennart Rolland, Nov 07 '14 at 23:16
How did you format this? By having your dog/cat sit on the keyboard?? — Boris the Spider, Nov 07 '14 at 23:17
Terrible code formatting -- fixed. Please next time fix it before posting it. You're asking volunteers to help, so you should strive to make it easy for them to do so. — Hovercraft Full Of Eels, Nov 07 '14 at 23:17
You're asking the user for input, ignore the entered data and then you're asking for more input, which you're using then. And know you don't know why the user have to input the data twice? — Tom, Nov 07 '14 at 23:23

score 2 · Answer 1 · answered Nov 07 '14 at 23:17

Did you mean to call nextLine() again as a parameter?

if((type.toLowerCase()).equals("phone number")){                                    
    System.out.println("Enter phone number: ");
    String number = kb.nextLine();
    individual.setPhoneNumber(number);
}

Why is the first user input ignored when entering string input?

1 Answers1