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Is there a way to keep track of the elements and to loop the menu?


How does the Java 'for each' loop work?What is the best way to iterate over a dictionary?What's the simplest way to print a Java array?Accessing the index in 'for' loops?How do I loop through or enumerate a JavaScript object?JavaScript closure inside loops – simple practical exampleHow do I break out of nested loops in Java?Looping through the content of a file in BashLoop through an array in JavaScriptWould it make any difference giving arguments using scanner class instead of command line arguments?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








0















The program needs to be able to enter people into a Queue and keep track of them. the user has 3 options "A" to enter a new person(int) into the queue, "N" to just let the queue be processed, and "Q" to quit the queue and to then display how many people are in the queue. I can't quite figure out how to loop and keep track.



package pkg3650queue;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner; // Import the Scanner class

public class Main
/**
* @param args the command line arguments
*/
public static void main(String[] args)
// TODO code application logic here
Queue<Integer> line = new LinkedList<Integer>();
Scanner input = new Scanner(System.in);
Scanner addperson = new Scanner(System.in);

String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
System.out.println("Enter a number to add a person to the line: ");
int addtoLine = addperson.nextInt();
line.add(addtoLine);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");
else if (option.equalsIgnoreCase("N"))
if(line.isEmpty())
System.out.println("There are no elements in the line to be processed");
System.exit(0);

else
int requestsProccessed = line.remove();
System.out.println(requestsProccessed);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");



while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());











share|improve this question


























  • Not related to your problem, but to simplify your code, instead of using option.equals("q") || option.equals("Q") you can do option.equalsIgnoreCase("q"). See the corresponding doc

    – vincrichaud
    Mar 27 at 15:34











  • You may just put all your code (from the displying of the question, to the end) inside a loop. Use a boolean flag as the condition of your loop. When the user enter q set the flag to false.

    – vincrichaud
    Mar 27 at 15:36

















0















The program needs to be able to enter people into a Queue and keep track of them. the user has 3 options "A" to enter a new person(int) into the queue, "N" to just let the queue be processed, and "Q" to quit the queue and to then display how many people are in the queue. I can't quite figure out how to loop and keep track.



package pkg3650queue;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner; // Import the Scanner class

public class Main
/**
* @param args the command line arguments
*/
public static void main(String[] args)
// TODO code application logic here
Queue<Integer> line = new LinkedList<Integer>();
Scanner input = new Scanner(System.in);
Scanner addperson = new Scanner(System.in);

String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
System.out.println("Enter a number to add a person to the line: ");
int addtoLine = addperson.nextInt();
line.add(addtoLine);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");
else if (option.equalsIgnoreCase("N"))
if(line.isEmpty())
System.out.println("There are no elements in the line to be processed");
System.exit(0);

else
int requestsProccessed = line.remove();
System.out.println(requestsProccessed);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");



while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());











share|improve this question


























  • Not related to your problem, but to simplify your code, instead of using option.equals("q") || option.equals("Q") you can do option.equalsIgnoreCase("q"). See the corresponding doc

    – vincrichaud
    Mar 27 at 15:34











  • You may just put all your code (from the displying of the question, to the end) inside a loop. Use a boolean flag as the condition of your loop. When the user enter q set the flag to false.

    – vincrichaud
    Mar 27 at 15:36













0












0








0








The program needs to be able to enter people into a Queue and keep track of them. the user has 3 options "A" to enter a new person(int) into the queue, "N" to just let the queue be processed, and "Q" to quit the queue and to then display how many people are in the queue. I can't quite figure out how to loop and keep track.



package pkg3650queue;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner; // Import the Scanner class

public class Main
/**
* @param args the command line arguments
*/
public static void main(String[] args)
// TODO code application logic here
Queue<Integer> line = new LinkedList<Integer>();
Scanner input = new Scanner(System.in);
Scanner addperson = new Scanner(System.in);

String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
System.out.println("Enter a number to add a person to the line: ");
int addtoLine = addperson.nextInt();
line.add(addtoLine);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");
else if (option.equalsIgnoreCase("N"))
if(line.isEmpty())
System.out.println("There are no elements in the line to be processed");
System.exit(0);

else
int requestsProccessed = line.remove();
System.out.println(requestsProccessed);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");



while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());











share|improve this question
















The program needs to be able to enter people into a Queue and keep track of them. the user has 3 options "A" to enter a new person(int) into the queue, "N" to just let the queue be processed, and "Q" to quit the queue and to then display how many people are in the queue. I can't quite figure out how to loop and keep track.



package pkg3650queue;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner; // Import the Scanner class

public class Main
/**
* @param args the command line arguments
*/
public static void main(String[] args)
// TODO code application logic here
Queue<Integer> line = new LinkedList<Integer>();
Scanner input = new Scanner(System.in);
Scanner addperson = new Scanner(System.in);

String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
System.out.println("Enter a number to add a person to the line: ");
int addtoLine = addperson.nextInt();
line.add(addtoLine);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");
else if (option.equalsIgnoreCase("N"))
if(line.isEmpty())
System.out.println("There are no elements in the line to be processed");
System.exit(0);

else
int requestsProccessed = line.remove();
System.out.println(requestsProccessed);
System.out.println(line);
System.out.println("There are " + line.size() + " people in the queue");



while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());








java loops queue






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 27 at 21:15







Osama

















asked Mar 27 at 15:26









OsamaOsama

83 bronze badges




83 bronze badges















  • Not related to your problem, but to simplify your code, instead of using option.equals("q") || option.equals("Q") you can do option.equalsIgnoreCase("q"). See the corresponding doc

    – vincrichaud
    Mar 27 at 15:34











  • You may just put all your code (from the displying of the question, to the end) inside a loop. Use a boolean flag as the condition of your loop. When the user enter q set the flag to false.

    – vincrichaud
    Mar 27 at 15:36

















  • Not related to your problem, but to simplify your code, instead of using option.equals("q") || option.equals("Q") you can do option.equalsIgnoreCase("q"). See the corresponding doc

    – vincrichaud
    Mar 27 at 15:34











  • You may just put all your code (from the displying of the question, to the end) inside a loop. Use a boolean flag as the condition of your loop. When the user enter q set the flag to false.

    – vincrichaud
    Mar 27 at 15:36
















Not related to your problem, but to simplify your code, instead of using option.equals("q") || option.equals("Q") you can do option.equalsIgnoreCase("q"). See the corresponding doc

– vincrichaud
Mar 27 at 15:34





Not related to your problem, but to simplify your code, instead of using option.equals("q") || option.equals("Q") you can do option.equalsIgnoreCase("q"). See the corresponding doc

– vincrichaud
Mar 27 at 15:34













You may just put all your code (from the displying of the question, to the end) inside a loop. Use a boolean flag as the condition of your loop. When the user enter q set the flag to false.

– vincrichaud
Mar 27 at 15:36





You may just put all your code (from the displying of the question, to the end) inside a loop. Use a boolean flag as the condition of your loop. When the user enter q set the flag to false.

– vincrichaud
Mar 27 at 15:36












1 Answer
1






active

oldest

votes


















0














You mean how to loop the user input? You can use a do-while:



 String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
// do something
else if (option.equalsIgnoreCase("N"))
// do something


// notice we don't need an if for 'Q' here. This loop only determines how many
// times we want to keep going. If it's 'Q', it'll exit the while loop, where
// we then print the size of the list.

while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());


Note I didn't test this code, but it should get you going on the right track.



Also notice that we do not need a System.exit(0) in this case, as the program will just naturally end. Though there are exceptions, you normally do not want to use System.exit(0), and would rather find ways for code to 'finish itself'.






share|improve this answer



























  • Thank you it works. But I noticed another problem while trying to run it, whenever i try to use "n" and let the queue process itself it only subtracts 1 from the first int ie [5,3,2,1] -> n -> 4(only displays the first int in queue) and if I add another int (6) its going to display [5,3,2,1,6] . any advise?

    – Osama
    Mar 27 at 20:42











  • @Osama I'm not sure what you mean. What would you like it to do when the user selects n?

    – Tiberiu
    Mar 27 at 20:49











  • Ok, so let's think like programmers! You have a LinkedList and you'd like to remove the first int in the queue. So if we go here: docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html and check which method allows us to do that (hint, it's pop()). So you would do: int requestsProccessed = line.pop();

    – Tiberiu
    Mar 27 at 20:53












  • Also, I would appreciate if you accepted my answer if you thought it was helpful :)

    – Tiberiu
    Mar 27 at 20:54











  • Sorry, pop() will remove the last element that was put in. Since you're dealing with a Queue, you will want to use the removeFirst() method.

    – Tiberiu
    Mar 27 at 20:59










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














You mean how to loop the user input? You can use a do-while:



 String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
// do something
else if (option.equalsIgnoreCase("N"))
// do something


// notice we don't need an if for 'Q' here. This loop only determines how many
// times we want to keep going. If it's 'Q', it'll exit the while loop, where
// we then print the size of the list.

while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());


Note I didn't test this code, but it should get you going on the right track.



Also notice that we do not need a System.exit(0) in this case, as the program will just naturally end. Though there are exceptions, you normally do not want to use System.exit(0), and would rather find ways for code to 'finish itself'.






share|improve this answer



























  • Thank you it works. But I noticed another problem while trying to run it, whenever i try to use "n" and let the queue process itself it only subtracts 1 from the first int ie [5,3,2,1] -> n -> 4(only displays the first int in queue) and if I add another int (6) its going to display [5,3,2,1,6] . any advise?

    – Osama
    Mar 27 at 20:42











  • @Osama I'm not sure what you mean. What would you like it to do when the user selects n?

    – Tiberiu
    Mar 27 at 20:49











  • Ok, so let's think like programmers! You have a LinkedList and you'd like to remove the first int in the queue. So if we go here: docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html and check which method allows us to do that (hint, it's pop()). So you would do: int requestsProccessed = line.pop();

    – Tiberiu
    Mar 27 at 20:53












  • Also, I would appreciate if you accepted my answer if you thought it was helpful :)

    – Tiberiu
    Mar 27 at 20:54











  • Sorry, pop() will remove the last element that was put in. Since you're dealing with a Queue, you will want to use the removeFirst() method.

    – Tiberiu
    Mar 27 at 20:59















0














You mean how to loop the user input? You can use a do-while:



 String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
// do something
else if (option.equalsIgnoreCase("N"))
// do something


// notice we don't need an if for 'Q' here. This loop only determines how many
// times we want to keep going. If it's 'Q', it'll exit the while loop, where
// we then print the size of the list.

while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());


Note I didn't test this code, but it should get you going on the right track.



Also notice that we do not need a System.exit(0) in this case, as the program will just naturally end. Though there are exceptions, you normally do not want to use System.exit(0), and would rather find ways for code to 'finish itself'.






share|improve this answer



























  • Thank you it works. But I noticed another problem while trying to run it, whenever i try to use "n" and let the queue process itself it only subtracts 1 from the first int ie [5,3,2,1] -> n -> 4(only displays the first int in queue) and if I add another int (6) its going to display [5,3,2,1,6] . any advise?

    – Osama
    Mar 27 at 20:42











  • @Osama I'm not sure what you mean. What would you like it to do when the user selects n?

    – Tiberiu
    Mar 27 at 20:49











  • Ok, so let's think like programmers! You have a LinkedList and you'd like to remove the first int in the queue. So if we go here: docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html and check which method allows us to do that (hint, it's pop()). So you would do: int requestsProccessed = line.pop();

    – Tiberiu
    Mar 27 at 20:53












  • Also, I would appreciate if you accepted my answer if you thought it was helpful :)

    – Tiberiu
    Mar 27 at 20:54











  • Sorry, pop() will remove the last element that was put in. Since you're dealing with a Queue, you will want to use the removeFirst() method.

    – Tiberiu
    Mar 27 at 20:59













0












0








0







You mean how to loop the user input? You can use a do-while:



 String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
// do something
else if (option.equalsIgnoreCase("N"))
// do something


// notice we don't need an if for 'Q' here. This loop only determines how many
// times we want to keep going. If it's 'Q', it'll exit the while loop, where
// we then print the size of the list.

while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());


Note I didn't test this code, but it should get you going on the right track.



Also notice that we do not need a System.exit(0) in this case, as the program will just naturally end. Though there are exceptions, you normally do not want to use System.exit(0), and would rather find ways for code to 'finish itself'.






share|improve this answer















You mean how to loop the user input? You can use a do-while:



 String option;
do
System.out.println("Type A to add a person to the line (# of requests)n"
+ "Type N to do nothing and allow the line to be processedn"
+ "Type Q to quit the applicationn");
option = input.nextLine();

if(option.equalsIgnoreCase("A"))
// do something
else if (option.equalsIgnoreCase("N"))
// do something


// notice we don't need an if for 'Q' here. This loop only determines how many
// times we want to keep going. If it's 'Q', it'll exit the while loop, where
// we then print the size of the list.

while (!option.equalsIgnoreCase("Q"));

System.out.println("Q was chosen. The number of ppl in this queue are " + line.size());


Note I didn't test this code, but it should get you going on the right track.



Also notice that we do not need a System.exit(0) in this case, as the program will just naturally end. Though there are exceptions, you normally do not want to use System.exit(0), and would rather find ways for code to 'finish itself'.







share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 27 at 15:46

























answered Mar 27 at 15:40









TiberiuTiberiu

3764 silver badges20 bronze badges




3764 silver badges20 bronze badges















  • Thank you it works. But I noticed another problem while trying to run it, whenever i try to use "n" and let the queue process itself it only subtracts 1 from the first int ie [5,3,2,1] -> n -> 4(only displays the first int in queue) and if I add another int (6) its going to display [5,3,2,1,6] . any advise?

    – Osama
    Mar 27 at 20:42











  • @Osama I'm not sure what you mean. What would you like it to do when the user selects n?

    – Tiberiu
    Mar 27 at 20:49











  • Ok, so let's think like programmers! You have a LinkedList and you'd like to remove the first int in the queue. So if we go here: docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html and check which method allows us to do that (hint, it's pop()). So you would do: int requestsProccessed = line.pop();

    – Tiberiu
    Mar 27 at 20:53












  • Also, I would appreciate if you accepted my answer if you thought it was helpful :)

    – Tiberiu
    Mar 27 at 20:54











  • Sorry, pop() will remove the last element that was put in. Since you're dealing with a Queue, you will want to use the removeFirst() method.

    – Tiberiu
    Mar 27 at 20:59

















  • Thank you it works. But I noticed another problem while trying to run it, whenever i try to use "n" and let the queue process itself it only subtracts 1 from the first int ie [5,3,2,1] -> n -> 4(only displays the first int in queue) and if I add another int (6) its going to display [5,3,2,1,6] . any advise?

    – Osama
    Mar 27 at 20:42











  • @Osama I'm not sure what you mean. What would you like it to do when the user selects n?

    – Tiberiu
    Mar 27 at 20:49











  • Ok, so let's think like programmers! You have a LinkedList and you'd like to remove the first int in the queue. So if we go here: docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html and check which method allows us to do that (hint, it's pop()). So you would do: int requestsProccessed = line.pop();

    – Tiberiu
    Mar 27 at 20:53












  • Also, I would appreciate if you accepted my answer if you thought it was helpful :)

    – Tiberiu
    Mar 27 at 20:54











  • Sorry, pop() will remove the last element that was put in. Since you're dealing with a Queue, you will want to use the removeFirst() method.

    – Tiberiu
    Mar 27 at 20:59
















Thank you it works. But I noticed another problem while trying to run it, whenever i try to use "n" and let the queue process itself it only subtracts 1 from the first int ie [5,3,2,1] -> n -> 4(only displays the first int in queue) and if I add another int (6) its going to display [5,3,2,1,6] . any advise?

– Osama
Mar 27 at 20:42





Thank you it works. But I noticed another problem while trying to run it, whenever i try to use "n" and let the queue process itself it only subtracts 1 from the first int ie [5,3,2,1] -> n -> 4(only displays the first int in queue) and if I add another int (6) its going to display [5,3,2,1,6] . any advise?

– Osama
Mar 27 at 20:42













@Osama I'm not sure what you mean. What would you like it to do when the user selects n?

– Tiberiu
Mar 27 at 20:49





@Osama I'm not sure what you mean. What would you like it to do when the user selects n?

– Tiberiu
Mar 27 at 20:49













Ok, so let's think like programmers! You have a LinkedList and you'd like to remove the first int in the queue. So if we go here: docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html and check which method allows us to do that (hint, it's pop()). So you would do: int requestsProccessed = line.pop();

– Tiberiu
Mar 27 at 20:53






Ok, so let's think like programmers! You have a LinkedList and you'd like to remove the first int in the queue. So if we go here: docs.oracle.com/javase/7/docs/api/java/util/LinkedList.html and check which method allows us to do that (hint, it's pop()). So you would do: int requestsProccessed = line.pop();

– Tiberiu
Mar 27 at 20:53














Also, I would appreciate if you accepted my answer if you thought it was helpful :)

– Tiberiu
Mar 27 at 20:54





Also, I would appreciate if you accepted my answer if you thought it was helpful :)

– Tiberiu
Mar 27 at 20:54













Sorry, pop() will remove the last element that was put in. Since you're dealing with a Queue, you will want to use the removeFirst() method.

– Tiberiu
Mar 27 at 20:59





Sorry, pop() will remove the last element that was put in. Since you're dealing with a Queue, you will want to use the removeFirst() method.

– Tiberiu
Mar 27 at 20:59








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