Issues with nextLine();

jl90 picture jl90 · Jan 22, 2013 · Viewed 41.2k times · Source

Possible Duplicate:
Scanner issue when using nextLine after nextInt

I am trying create a program where it lets the user inputs values into an array using scanner.

However, when the program asks for the student's next of kin, it doesn't let the user to input anything and straight away ends the program.

Below is the code that I have done:

if(index!=-1)
    {
        Function.print("Enter full name: ");
        stdName = input.nextLine();

        Function.print("Enter student no.: ");
        stdNo = input.nextLine();

        Function.print("Enter age: ");
        stdAge = input.nextInt();

        Function.print("Enter next of kin: ");
        stdKin = input.nextLine();

        Student newStd = new Student(stdName, stdNo, stdAge, stdKin);
        stdDetails[index] = newStd;
    }

I have tried using next(); but it will only just take the first word of the user input which is not what I wanted. Is there anyway to solve this problem?

Answer

Swapnil picture Swapnil · Jan 22, 2013

The problem occurs as you hit the enter key, which is a newline \n character. nextInt() consumes only the integer, but it skips the newline \n. To get around this problem, you may need to add an additional input.nextLine() after you read the int, which can consume the \n.

Function.print("Enter age: ");
stdAge = input.nextInt();
input.nextLine();.

// rest of the code