JAVA PROGRAMMING-LMS INTEG.MINDTAP
JAVA PROGRAMMING-LMS INTEG.MINDTAP
8th Edition
ISBN: 9781337091503
Author: FARRELL
Publisher: Cengage Learning
Expert Solution & Answer
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Chapter 8, Problem 9PE

Explanation of Solution

Program:

File name: “PhoneNumbers.java

//Import necessary header files

import javax.swing.*;

//Define a class named PhoneNumbers

class PhoneNumbers

{

    //Define a main method

    public static void main(String[] args)

    {

        /*Declare an array to store first names for 30 people*/

        String[] names = new String[30];

        //Assign names for 10 people

        names[0] = "Gina";    names[1] = "Marcia";

        names[2] = "Rita";    names[3] = "Jennifer";

        names[4] = "Fred";    names[5] = "Neil";

        names[6] = "Judy";    names[7] = "Arlene";

        names[8] = "LaWanda"; names[9] = "Deepak";

        /*Declare an array to store numbers for 30 people*/

        String numbers[] = new String[30];

        //Assign numbers for 10 people

        numbers[0] = "(847) 341-0912";

        numbers[1] = "(847) 341-2392";

        numbers[2] = "(847) 354-0654";

        numbers[3] = "(414) 234-0912";

        numbers[4] = "(414) 435-6567";

        numbers[5] = "(608) 123-0904";

        numbers[6] = "(608) 435-0434";

        numbers[7] = "(608) 123-0312";

        numbers[8] = "(920) 787-9813";

        numbers[9] = "(930) 412-0991";

        //Declare the variables and initialize the value

        String entryName, entryPhone;

        int x;

        boolean isFound = false;

        int highest = 10;

        //Prompt the user to enter a name

        entryName = JOptionPane.showInputDialog(null,

           "Enter name to look up. Type 'quit' to quit.");

        /*While the user enters a name until the arrays are full

        and until the user enters quit*/   

        while (highest < 30 && !entryName.equals("quit"))

        {

            //For loop to be executed until x exceeds highest

            for(x = 0; x < highest; ++ x)

                /*If the name is found in the list, display the

                corresponding phone number*/

                if(entryName...

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