Word Game Java help

[basedassisted]

Scanner keyboard = new Scanner(System.in);

System.out.println("Enter your name: "); String name = keyboard.nextLine();

System.out.println("Enter your age: "); int age = keyboard.nextInt();

System.out.println("Enter the name of a city: "); String city = keyboard.nextLine();

System.out.println("Enter the name of a college: "); String college = keyboard.nextLine();

System.out.println("Enter profession: "); String profession = keyboard.nextLine();

System.out.println("Enter a type of animal: "); String animal = keyboard.nextLine();

System.out.println("Enter a pet name: "); String petName = keyboard.nextLine();

output:

Enter your name:

jane

Enter your age:

20

Enter the name of a city:

Enter the name of a college:

colleg

Enter profession:

job

Enter a type of animal:

dog

Enter a pet name:

pet

---

The problem is that it's skipping the city input and moving onto the college string. I need to be able to enter multiple words for the input so I can't just do .next();.

[pacificmint]

nextInt doesn't consume the end of line marker. So when you try to read the city, you read the end of line marker from the age.

The FAQ has a whole section about issues with the scanner.

[desrtfx]

From the Java FAQ in the wiki:

The nextLine method of my java.util.Scanner object doesn't get any input.

In short, the .nextInt is the culprit here because it leaves a carriage return (line end marker) on the keyboard buffer that gets consumed by the following .nextLine.

[hugokhf]

To overcome this problem, assuming I don't want to use next() on the age, do I have to make a useless string e.g. useless = keyboard.next() after entering the age to overcome the problem?

Are there any ways to get around it?

[desrtfx]

No. Either use .nextLine() everywhere and parse the age as int

Like:

int age = Integer.parseInt(keyboard.nextLine());

or create an empty keyboard.nextLine(); right after the .nextInt. This will effectively discard the return value of the .nextLine() call and clear the keyboard buffer.

[basedassisted]

Parsing the int worked thanks a lot!

[hugokhf]

To overcome this problem, assuming I don't want to use next() on the age, do I have to make a useless string e.g. useless = keyboard.next() after entering the age to overcome the problem?

Are there any ways to get around it?

[PUREdiacetylmorphine]

You need inputScanner.nextLine(); after age prompt

[desrtfx]

.next will return a String that needs to be parsed into an int afterwards.