i m b beginer of java...pls help me as u can...

Question

puppycrazy 0 Newbie Poster

12 Years Ago

my codes occur some problem inside.whn i answered 'N' to the last question(DO YOU WISH TO CONTINUE?(Y/N)), the program will repeat run again, but i cnt enter student name!anyone can help me solve the problem>?

import java.util.Scanner;

public class homewok{
	public static void main(String args[]){

		Scanner xx = new Scanner(System.in);

		boolean vv = true;
		do{

		System.out.println("----------------------------------------------");
		System.out.println("WELCOME TO  BOOKING SYSTEM");
		System.out.println("----------------------------------------------");
		System.out.println("\n---PLEASE ENTER YOUR PERSONAL INFORMATION---");

		System.out.print("\nPLEASE ENTER STUDENT NAME						: ");
		String name = xx.nextLine();

		boolean ag = true;
		do{
		System.out.print("\nPLEASE ENTER STUDENT ID NUMBER(eg : 1xxxxx) 	: TP ");
		int TP = xx.nextInt();

		if (TP < 100000){
			System.out.println("LOGIN FAIL!PLEASE ENTER AGAIN.");
			ag = true;}

		else if (TP> 999999){
			System.out.println("LOGIN FAIL!PLEASE ENTER AGAIN.");
			ag = true;}

		else if ((TP>=100000) && (TP<=999999)){
			System.out.println("LOGIN SUCCESSFUL!\nPLEASE CONTINUE TO FILL IN YOUR PERSONAL INFORMATION.");
			ag = false;}

		}while (ag == true);

		System.out.print("\nPLEASE ENTER YOUR SEX(M/F) : ");
		char s = xx.next().charAt(0);

		if (s=='m')
			System.out.println("WELCOME TO BOOKING SYSTEM, MR. "+(name));

		else if (s=='f')
			System.out.println("WELCOME TO BOOKING SYSTEM, MISS "+(name));

		System.out.print("DO YOU WISH TO CONTINUE?(Y/N) : ");
		char oo = xx.next().charAt(0);

		if (oo == 'y'){
			System.out.println("PLEASE CONTINUE.");
			vv = true;}

		else if (oo == 'n'){
			System.out.println("PLEASE ENTER YOUR PERSONAL INFORMATION AGAIN.");
			vv = false;}

		}while(vv == false);

	}
}

java

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Latest Post 12 Years Ago Latest Post by puppycrazy

zeroliken 79 Nearly a Posting Virtuoso

12 Years Ago

use .equals() when comparing strings rather than ==
try to read this link for more info
Java: ==,.equals()

Edited 12 Years Ago by zeroliken because: n/a

zeroliken 79 Nearly a Posting Virtuoso

12 Years Ago

change oo and s to string and use nextLine() for receiving input
e.g

String oo = xx.nextLine();

also just noticed now that the program continues when vv is equal to false and when you choose n the program continues to run cause it remains to be false

Edited 12 Years Ago by zeroliken because: n/a

zeroliken 79 Nearly a Posting Virtuoso

12 Years Ago

would you feel better if you don't use them...
like just use next() in lines 17, 39, 48

Edited 12 Years Ago by zeroliken because: n/a

puppycrazy commented: thx so much~problem sloved ad! +0

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puppycrazy 0 Newbie Poster · Answer 1 · 2011-12-29T14:45:05+00:00

use .equals() when comparing strings rather than ==
try to read this link for more info
Java: ==,.equals()

sry~i not really understand!i tried ad~if i change "if (s=='m')" to "if (s.equals("m"))"....it dosnt run...can u further explain to me?so sry for making trouble to u...><

JamesCherrill 4,733 Most Valuable Poster Team Colleague Featured Poster · Answer 2 · 2011-12-29T15:12:54+00:00

use .equals() when comparing strings rather than ==
try to read this link for more info
Java: ==,.equals()

Good advice, yes, but in this case the code is comparing chars, not Strings

char oo = xx.next().charAt(0);
if (oo == 'y')...{

so .equals cannot be used in this situation.

Maybe the problem here is that the y/n is read with a next(), which will leave a newline character unprocessed in the scanner's buffer, so the student name's nextLine() returns the remainder of the current line, ie "".
At line 49 try a nextLine() rather then a next() to read the y/n so that the newline character is processed normally.

zeroliken 79 Nearly a Posting Virtuoso · Answer 3 · 2011-12-29T15:19:39+00:00

Good advice, yes, but in this case the code is comparing chars, not Strings
char oo = xx.next().charAt(0);
if (oo == 'y')...{
so .equals cannot be used in this situation.

Maybe the problem here is that the y/n is read with a next(), which will leave a newline character unprocessed in the scanner's buffer, so the student name's nextLine() returns the remainder of the current line, ie "".
At line 49 try a nextLine() rather then a next() to read the y/n so that the newline character is processed normally.

Yes... sorry about that just noticed that char was used instead of string so the next step would have been my previous post... :$

It's correct but unnecessary... my bad

puppycrazy 0 Newbie Poster · Answer 4 · 2011-12-29T15:33:23+00:00

><.....ya~i changed ad!but the same problm still here....erm....if i change line 49 to "Line()", the program dosent ask user to enter Y/N....just display the question.

import java.util.Scanner;

public class homewok{
	public static void main(String args[]){

		Scanner xx = new Scanner(System.in);

		boolean vv = false;
		do{

		System.out.println("----------------------------------------------");
		System.out.println("WELCOME TO  BOOKING SYSTEM");
		System.out.println("----------------------------------------------");
		System.out.println("\n---PLEASE ENTER YOUR PERSONAL INFORMATION---");

		System.out.print("\nPLEASE ENTER STUDENT NAME						: ");
		String name = xx.nextLine();

		boolean ag = true;
		do{
		System.out.print("\nPLEASE ENTER STUDENT ID NUMBER(eg : 1xxxxx) 	: TP ");
		int TP = xx.nextInt();

		if (TP < 100000){
			System.out.println("LOGIN FAIL!PLEASE ENTER AGAIN.");
			ag = true;}

		else if (TP> 999999){
			System.out.println("LOGIN FAIL!PLEASE ENTER AGAIN.");
			ag = true;}

		else if ((TP>=100000) && (TP<=999999)){
			System.out.println("LOGIN SUCCESSFUL!\nPLEASE CONTINUE TO FILL IN YOUR PERSONAL INFORMATION.");
			ag = false;}

		}while (ag == true);

		System.out.print("PLEASE ENTER YOUR SEX(M/F) : ");
		String s = xx.next();

		if (s.equals("m"))
			System.out.println("WELCOME TO BOOKING SYSTEM, MR. "+(name));

		else if (s.equals("f"))
			System.out.println("WELCOME TO BOOKING SYSTEM, MISS "+(name));

		System.out.print("\nDO YOU WISH TO CONTINUE?(Y/N) : ");
		String oo = xx.next();

		if (oo.equals("y")){
			System.out.println("PLEASE CONTINUE.");
			vv = true;}

		else if (oo.equals("n")){
			System.out.println("PLEASE ENTER YOUR PERSONAL INFORMATION AGAIN.");
			vv = false;}

		}while(vv == false);

	}
}

JamesCherrill 4,733 Most Valuable Poster Team Colleague Featured Poster · Answer 5 · 2011-12-29T15:42:15+00:00

JamesCherrill 4,733 Most Valuable Poster

12 Years Ago

That's nextLine(), not Line().

zeroliken 79 Nearly a Posting Virtuoso · Answer 6 · 2011-12-29T15:42:56+00:00

Line 49 works fine without the ...Line() if you're using Strings

That's nextLine(), not Line().

Would he really just use Line() :) ;)

puppycrazy 0 Newbie Poster · Answer 7 · 2011-12-29T15:52:56+00:00

yaya~if i put "...Line()",the program can ask user to enter student name ad!but the program dosent ask user to enter Y/N(line 47) .....><....why?

Trentacle 112 Junior Poster in Training · Answer 8 · 2011-12-29T21:33:11+00:00

WELCOME TO FORUMS, puppycrazy
WOULD YOU LIKE TO ANNOY YOUR USERS?(Y/N) : y
IN THAT CASE PLEASE CONTINUE USING ALL CAPS

WOULD YOU LIKE TO GET GOOD ANSWERS?(Y/N) : y
IN THAT CASE PLEASE CONSIDER PROOFREADING YOUR POSTS TO ENSURE THEY MAKE SENSE. ALSO,PLEASE!AVOID>USING~WEIRD.......><....PUNCTUATION, SPACES EXIST FOR A REASON

puppycrazy 0 Newbie Poster · Answer 9 · 2011-12-30T09:39:42+00:00

thx so much to u all!sry for wasting you all so much time!my problem solved ad!by using 'next()' instead of 'nextLine()' in lines 17, 39, 48!!!thx thx thx^^