Java

nextLine() after nextInt() or nextDouble gives trouble

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Join Date: Oct 2008

Posts: 4

Reputation: Isky is an unknown quantity at this point

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Isky

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Newbie Poster

nextLine() after nextInt() or nextDouble gives trouble

Oct 21st, 2009

Hi,
I'm new to this Java world, and just can't figure out this problem. After I use nextInt or nextDouble, and try to use nextLine, it won't work unless I put nextLine twice. What's the problem there, and what would be the good way to handle this?

Just gave a code to illustrate a simple example I can understand.

 Help with Code Tags 
 java Syntax (Toggle Plain Text) 
 
import java.util.Scanner;
 
public class nextLineProblem {
 
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		System.out.println("Type in something for line 1");
		String line1 = sc.nextLine();
 
		System.out.println("Type an integer for line 2");
		int line2 = sc.nextInt();
 
		System.out.println("Type in something for line 3");
		String line3 = sc.nextLine();
 
		System.out.println("Line1: " + line1 + "\n" + "Line2: " + line2 + "\n" + "Line3:" + line3);
 
	}
 
}
import java.util.Scanner;

public class nextLineProblem {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		System.out.println("Type in something for line 1");
		String line1 = sc.nextLine();

		System.out.println("Type an integer for line 2");
		int line2 = sc.nextInt();
		
		System.out.println("Type in something for line 3");
		String line3 = sc.nextLine();
		
		System.out.println("Line1: " + line1 + "\n" + "Line2: " + line2 + "\n" + "Line3:" + line3);
		
	}

}

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Join Date: Sep 2009

Posts: 23

Reputation: jmaat7 is an unknown quantity at this point

Solved Threads: 3

jmaat7

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Newbie Poster

Oct 21st, 2009

I had this problem before also.

this is what I came across:

http://www.robocommunity.com/blog/en...Scanner-Class/

For my assignment a while back , I just used the .next() to read the next input. But my assignment only required a single word , not separated by spaces (so i could just use tokens).

looks like you're just going to have to add the extra line as far as I can tell.

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Join Date: Sep 2008

Posts: 1,658

Reputation: BestJewSinceJC is a splendid one to behold

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BestJewSinceJC BestJewSinceJC is offline

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Posting Virtuoso

Oct 21st, 2009

The reason it gives you trouble is because when the user enters an integer then hits enter, two things have just been entered - the integer and a "newline" which is \n. The method you are calling, "nextInt", only reads in the integer, which leaves the newline in the input stream. But calling nextLine() does read in newlines, which is why you had to call nextLine() before your code would work. You could have also called next(), which would also have read in the newline.

Out.

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Join Date: Oct 2008

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Reputation: cgeier is an unknown quantity at this point

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cgeier

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Oct 21st, 2009

Just change the delimiter.

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
Scanner in = new Scanner(System.in);
in.useDelimiter("\n");
Scanner in = new Scanner(System.in);
in.useDelimiter("\n");

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Join Date: Oct 2008

Posts: 4

Reputation: Isky is an unknown quantity at this point

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Isky

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Newbie Poster

Thanks

Oct 22nd, 2009

Thanks, now I understand what's going on now and how to deal with it when using the Scanner class.
Just left pondering if it's a bug or a feature

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Join Date: Oct 2008

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Reputation: Isky is an unknown quantity at this point

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Isky

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New problem evolves

Oct 22nd, 2009

A new problem came to mind today.
It's easy to make a way around this problem if I know I'm going to have it, but what if it's unknown that I'll have a nextLine() after nextInt() , how would I be able to deal with the problem then?
For instance if there's a random function involved. Would I just have to scrap the nextLine() and go with something else?

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Join Date: Oct 2008

Posts: 103

Reputation: cgeier is an unknown quantity at this point

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cgeier

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Junior Poster

Oct 24th, 2009

Check out the documentation for the scanner class.
http://java.sun.com/j2se/1.5.0/docs/...l/Scanner.html

It will throw "NoSuchElementException" if no line was found. If you don't haven't learned about exception handling yet, I would use "hasNext".

 Help with Code Tags 
 Java Syntax (Toggle Plain Text) 
int myInt = 0;
String myString = "";
 
Scanner in = new Scanner(System.in);
in.useDelimiter("\n");
 
if (in.hasNextInt()){
     //read in int value
     myInt = in.nextInt();
}//if
else if (in.hasNext()){
     //read in String value
     myString = in.next();
}//else if
int myInt = 0;
String myString = "";

Scanner in = new Scanner(System.in);
in.useDelimiter("\n");

if (in.hasNextInt()){
     //read in int value
     myInt = in.nextInt();
}//if
else if (in.hasNext()){
     //read in String value
     myString = in.next();
}//else if