Need help in Java!! AGAIN!!

J

j.kaii

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An electronics shop has announced the following seasonal discounts on the purchase of certain
items.
Purchase Amount Discount on Laptop Discount on Desktop PC
0 – 250 0.0% 5.0%
251 – 570 5.0% 7.6%
571 – 1000 7.5% 10.0%
More than 1000 10.0% 15.0%

Write a program based on the above criteria to input name, address, amount of purchase and
type of purchase (L for Laptop and D for Desktop) by a customer. Compute and print the net
amount to be paid by a customer along with his name and address.
discount = (discount rate/100)* amount of purchase
Net amount = amount of purchase – discount


package Assignment;
import java.util.Scanner;
public class Assignment
{
public static void main(String[] args) throws Exception
{
String name;
String address;
int amount;
char type;
int net = 0;
Scanner keyboard = new Scanner(System.in);

System.out.print("Enter Name : ");
name = keyboard.nextLine();
System.out.print("Enter Address : ");
address = keyboard.nextLine();
System.out.print("Enter Amount of purchase : ");
amount = keyboard.nextInt();
System.out.print("Enter type of purchase : ");
type = (char)System.in.read();
System.in.read();

while(type != 'D' && type!= 'L')
{
System.out.println("Invaild type of purchase");
System.out.print("Enter type of purchase : ");
type = (char)System.in.read();
System.in.read();
}
if(amount<=250)
{
if(type == 'D')
{
net = ((95/100)*amount);
System.out.println("net price : " +net);
}
else if(type == 'L')
{
net = amount;
}
}
{
System.out.println("Name : " +name);
System.out.println("Address : " +address);
System.out.println("Net price : " +amount);
System.out.println("Type of purchase : " +type);

}
}
}

output
run:
Enter Name : ffdhfjf
Enter Address : gstgdgd 435454
Enter Amount of purchase : 200
Enter type of purchase : D
net price : 0
Name : ffdhfjf
Address : gstgdgd 435454
Net price : 200
Type of purchase : D
BUILD SUCCESSFUL (total time: 24 seconds)

Why is my IF part not being able to calculate if my type is D?
 
S

Swiftbladez

Master Member
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Nov 24, 2006
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It should be an 'or' operator instead.


EDIT: I obviously don't know what I'm saying lol.
 
Last edited:
davidktw

davidktw

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j.kaii said:
An electronics shop has announced the following seasonal discounts on the purchase of certain
items.
Purchase Amount Discount on Laptop Discount on Desktop PC
0 – 250 0.0% 5.0%
251 – 570 5.0% 7.6%
571 – 1000 7.5% 10.0%
More than 1000 10.0% 15.0%

Write a program based on the above criteria to input name, address, amount of purchase and
type of purchase (L for Laptop and D for Desktop) by a customer. Compute and print the net
amount to be paid by a customer along with his name and address.
discount = (discount rate/100)* amount of purchase
Net amount = amount of purchase – discount

ORIGINALLY BADLY FORMATTED CODES

output
run:
Enter Name : ffdhfjf
Enter Address : gstgdgd 435454
Enter Amount of purchase : 200
Enter type of purchase : D
net price : 0
Name : ffdhfjf
Address : gstgdgd 435454
Net price : 200
Type of purchase : D
BUILD SUCCESSFUL (total time: 24 seconds)

Why is my IF part not being able to calculate if my type is D?
Click to expand...

Do us a favour and also do yourself a favour at the same time. Please properly format your code for good readability. Your code is already flawed, you don't need to add on the complexity having a human Java syntax parser :)

PHP: 
package Assignment;
import java.util.Scanner;
public class Assignment
{
    public static void main(String[] args) throws Exception
    {
        String name;
        String address;
        int amount;
        char type;
        int net = 0;
        Scanner keyboard = new Scanner(System.in);
        
        System.out.print("Enter Name : ");
        name = keyboard.nextLine();
        System.out.print("Enter Address : ");
        address = keyboard.nextLine();
        System.out.print("Enter Amount of purchase : ");
        amount = keyboard.nextInt();
        System.out.print("Enter type of purchase : ");
        type = (char)System.in.read();
        System.in.read();
        
        while(type != 'D' && type!= 'L')
        {
            System.out.println("Invaild type of purchase");
            System.out.print("Enter type of purchase : ");
            type = (char)System.in.read();
            System.in.read();
        }
        if(amount<=250)
        {
            if(type == 'D')
            {
                net = ((95/100)*amount);
                System.out.println("net price : " +net);
            }
            else if(type == 'L')
            {
                net = amount;
            }
        }
        {
            System.out.println("Name : " +name);
            System.out.println("Address : " +address);
            System.out.println("Net price : " +amount);
            System.out.println("Type of purchase : " +type);
            
        }
    }
}
 
davidktw

davidktw

Arch-Supremacy Member
Joined
Apr 15, 2010
Messages
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ykgoh has more or less pointed out your issue.

I will just give you a bit more glimpse into how you can code better
PHP: 
package Assignment;
import java.util.Scanner;
public class Assignment
{
    public static void main(String[] args) throws Exception
    {
        String name;
        String address;
        int amount;
        char type;
        int net = 0;
        Scanner keyboard = new Scanner(System.in);
        
        System.out.print("Enter Name : ");
        name = keyboard.nextLine();
        System.out.print("Enter Address : ");
        address = keyboard.nextLine();
        System.out.print("Enter Amount of purchase : ");
        amount = keyboard.nextInt();
        
        //////////////////////////////////////////////////////////////
        // I PURPOSELY ISOLATED THIS FRAGMENT OF CODES TO GUIDE YOU
        // ON BETTER LOGIC LOOPING CONCEPT
        /* ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
           set input condition to be false, so while loop will enter
           while (input condition false) {
             (1) read input from user
             if (input is true) {
               optionally perform your necessary post processing here
               and break out of the while loop
             }
             // reaching here means while loop condition has to be re-evaluated
           }
           ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ */
        System.out.print("Enter type of purchase : ");
        type = (char)System.in.read();
        System.in.read(); // <-- Why is there an extra read in here ?
                          //     Would your user knows the need of an
                          //     extra RETURN keypress ?
        
        while(type != 'D' && type!= 'L')
        {
            System.out.println("Invaild type of purchase");
            System.out.print("Enter type of purchase : ");
            type = (char)System.in.read();
            System.in.read();
        }
        ///////////////////////////////////////////////////////////////
        
        
        if(amount<=250)
        {
            if(type == 'D')
            {
                net = ((95/100)*amount);
                System.out.println("net price : " +net);
            }
            else if(type == 'L') // <-- is there any possibility for type to be
                                 //     other than L and D ?
            /*
            You can take this approach for such cases
            else // if (type == 'L')
            it serves as comment and also save evaluation.
             */
            {
                net = amount;
            }
        }
        // <-- is there something missing here ?
        // Why an extra block for the codes below ?
        {
            System.out.println("Name : " +name);
            System.out.println("Address : " +address);
            System.out.println("Net price : " +amount);
            System.out.println("Type of purchase : " +type);
            
        }
    }
}
 
Y

ykgoh

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Joined
Jan 1, 2000
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davidktw said:
You might want to point out why is it zero ? :)
Click to expand...

Source: Beginning Java - Unit 2 Math Operators

Confusing DIVISIONS

Be careful when performing integer division. When dividing an integer by an integer, the answer will be an integer (not rounded).

Compare these divisions: (5 is an integer while 5.0 is a double)

Integer division 8 / 5 = 1
Double division 8.0 / 5.0 = 1.6
Mixed division 8.0 / 5 = 1.6
Click to expand...

TS do you see the issue now?
 
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