Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 194
  • Last Modified:

Prompt the user to select one of four patterns

I asked to prompt the user to select one of 4 patterns and display the pattern selected. How do I prompt the user which one to choose from, (either Pattern I or Pattern II)?

<code>


public class Pattern {
  public static void main(String[] args) {
    final int NUM_OF_LINES = 6;
   
    for (int row = 1; row <= NUM_OF_LINES; row++) {
   
      for (int num = 2; num <= row; num++)
        System.out.print(num);

        for (int column = 1; column <= NUM_OF_LINES - row; column++)
        System.out.print(" ");
       
     
      System.out.println();
    }
   
   
    for (int row = 1; row <= NUM_OF_LINES; row++) {
   
      for (int column = 1; column <= NUM_OF_LINES - row; column++)
        System.out.print(" ");

     
      for (int num = 2; num <= row; num++)
        System.out.print(num);

       
     
      System.out.println();
    }
   
  }
}
0
biloonline
Asked:
biloonline
  • 11
  • 5
  • 2
1 Solution
 
biloonlineAuthor Commented:
for (int num = 2; num <= row; num++) should be (int num = 1; .... )
0
 
expertmbCommented:
use

System.in.read()
0
 
biloonlineAuthor Commented:
I know how to input an answer from the user, in my case( Pattern I, Pattern II), but how would the program know what to display, whether Pattern I or II?
0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
biloonlineAuthor Commented:
Also how can I construct a for loop to display the following pattern:
         1
       21
     321
   4321
  54321
654321


Thanks
0
 
CI-Ia0sCommented:
You could also use JOptionPane.

At the top of your class file put "import javax.swing.JOptionPane;"

Code:

String inString = JOptionPane.showInputDialog("Some text to display telling the user what to enter.");

/*At this point you can just check if they entered the right string. In the "Some text to display..." area you should tell them what the possible inputs are.*/

if (inString.equals("Pattern 1")){//Do Pattern 1
}
else if (inString.equals("Pattern 2")){//Do Pattern 2
}
else if (inString.equals("Pattern 3")){//Do Pattern 3
}
else if (inString.equals("Pattern 4")){//Do Pattern 4
}
else{JOptionPane.showMessageDialog(null, "Bad input.");}
0
 
biloonlineAuthor Commented:
Disregard the last question, I got it.
0
 
biloonlineAuthor Commented:
CI-Ia0s,

I am asked to use:

BufferedReader stdin  = new BufferedReader (new InputStreamReader(System.in));

Thanks
0
 
expertmbCommented:
int number = 0;      
try{
          number = System.in.read();
      }catch(Exception ex){
      }

if((char)number == '1'){
// first pattern
}else if((char)number == '2'){
//pattern 2
}
0
 
expertmbCommented:
public class Pattern1 {
  public static void main(String[] args) {
    System.out.println("Select Pattern 1 or 2, or 3 to exit the application");

    System.out.println("Read pattern ");

      int number = 0;
      try{
          number = System.in.read();
      }catch(Exception ex){
      }

    System.out.println("Pattern: " + (char)number);

    final int NUM_OF_LINES = 6;

          if((char)number == '1'){
            for (int row = 1; row <= NUM_OF_LINES; row++) {

              for (int num = 2; num <= row; num++)
                  System.out.print(num);

               for (int column = 1; column <= NUM_OF_LINES - row; column++)
                  System.out.print(" ");


              System.out.println();
            }
      }else if ((char)number == '2'){

            for (int row = 1; row <= NUM_OF_LINES; row++) {

              for (int column = 1; column <= NUM_OF_LINES - row; column++)
                  System.out.print(" ");


              for (int num = 2; num <= row; num++)
                  System.out.print(num);



              System.out.println();
            }
    }else if((char)number == '3'){
          System.exit(0);
    }else {
            System.out.println("Enter 1 or 2 or 3to exit");
      }

  }
}
0
 
CI-Ia0sCommented:
First question: Use an if statement.

Second question:

//Declare numDigits to be the number of digits you want
int wholeNumber = 0;
int digitHolder = 1;
for (int c = 1; c < numDigits; c++){

wholeNumber = wholeNumber + (c * digitHolder);
digitHolder = digitHolder * 10;
}
0
 
biloonlineAuthor Commented:
<code>

import java.io.*;


public class Pattern4 {
  public static void main(String[] args) throws IOException {
    final int NUM_OF_LINES = 6;
    int number;
   
    BufferedReader stdin  = new BufferedReader (new InputStreamReader(System.in));
    System.out.println("Select Pattern 1 or 2 or 3 or 4 or 9 to exit the application");
   
    number = stdin.readLine();
 
   
    if (number == '1')
    {
 


                for (int row = 1; row <= NUM_OF_LINES; row++) {

            for (int num = 1; num <= row; num++)
              System.out.print(num);
      
         for (int column = 1; column <= NUM_OF_LINES - row; column++)
                 System.out.print(" ");


            System.out.println();
          }
    }
   
      if (number == '2')
      {
      
      
          for (int row = 1; row <= NUM_OF_LINES; row++) {

            for (int column = 1; column <= NUM_OF_LINES - row; column++)
              System.out.print(" ");


            for (int num = row; num >= 1; num--)
              System.out.print(num);



            System.out.println();
            }
    }

      if (number == '3')
      {
      
             for (int row = 1; row <= NUM_OF_LINES; row++) {

             for (int num = 1; num < row; num++)
               System.out.print(" ");

           for (int column = 1; column <= NUM_OF_LINES - row + 1; column++)
                System.out.print(column);
            
                System.out.print(" ");
                }
                
    if (number == '4')
       {
       
           for (int row = 1; row <= NUM_OF_LINES; row++) {

                    for (int num = 1; num < row; num++)
               System.out.print(" ");

            for (int column = 1; column <= NUM_OF_LINES - row + 1; column++)
            System.out.print(column);


             System.out.println();
             }
     }
  }
}
}


</code>

This is what I have gotten so far, I am getting 1 error
E:\Program Files\Xinox Software\JCreatorV3 LE\MyProjects\Hw3Pr1\Pattern4.java:12: incompatible types
found   : java.lang.String
required: int
    number = stdin.readLine();
                           ^
1 error

Process completed.

Please advise,
Thanks
0
 
biloonlineAuthor Commented:
if (number == '1')  <-- I guess that would not be right, since in this case '1' is considered as a CHAR.

How do I convert from String to Char?

0
 
biloonlineAuthor Commented:
Here is an updated version, It compiled with no errors, I will run it now ...

import java.io.*;


public class Pattern4 {
  public static void main(String[] args) throws IOException {
    final int NUM_OF_LINES = 6;
   
    BufferedReader stdin  = new BufferedReader (new InputStreamReader(System.in));
    System.out.println("Select Pattern 1 or 2 or 3 or 4 or 9 to exit the application");
   
    String number = stdin.readLine();
   
    int numb = Integer.parseInt(number);
 
   
    if (numb == 1)
    {
 


                for (int row = 1; row <= NUM_OF_LINES; row++) {

            for (int num = 1; num <= row; num++)
              System.out.print(num);
      
         for (int column = 1; column <= NUM_OF_LINES - row; column++)
                 System.out.print(" ");


            System.out.println();
          }
    }
   
      if (numb == 2)
      {
      
      
          for (int row = 1; row <= NUM_OF_LINES; row++) {

            for (int column = 1; column <= NUM_OF_LINES - row; column++)
              System.out.print(" ");


            for (int num = row; num >= 1; num--)
              System.out.print(num);



            System.out.println();
            }
    }

      if (numb == 3)
      {
      
             for (int row = 1; row <= NUM_OF_LINES; row++) {

             for (int num = 1; num < row; num++)
               System.out.print(" ");

           for (int column = 1; column <= NUM_OF_LINES - row + 1; column++)
                System.out.print(column);
            
                System.out.print(" ");
                }
                
    if (numb == 4)
       {
       
           for (int row = 1; row <= NUM_OF_LINES; row++) {

                    for (int num = 1; num < row; num++)
               System.out.print(" ");

            for (int column = 1; column <= NUM_OF_LINES - row + 1; column++)
            System.out.print(column);


             System.out.println();
             }
     }
  }
}
}
0
 
biloonlineAuthor Commented:
and added this at the end:

       if (numb == 9)
             return;
             

Looks like it's working fine.

Thanks guys.
0
 
expertmbCommented:
if (number == '1')
should be
if ((char)number == '1')
0
 
biloonlineAuthor Commented:
But I am working with Int and not Char, the program compiles and works fine.
0
 
expertmbCommented:
from api
read
public abstract int read()
                  throws IOExceptionReads the next byte of data from the input stream. The value byte is returned as an int in the range 0 to 255. If no byte is available because the end of the stream has been reached, the value -1 is returned. This method blocks until input data is available, the end of the stream is detected, or an exception is thrown.
A subclass must provide an implementation of this method.


0
 
biloonlineAuthor Commented:
Thanks again.
0

Featured Post

Become an Android App Developer

Ready to kick start your career in 2018? Learn how to build an Android app in January’s Course of the Month and open the door to new opportunities.

  • 11
  • 5
  • 2
Tackle projects and never again get stuck behind a technical roadblock.
Join Now