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# How to create a java program that generates and prints a random phone number in the form xxx-xxx-xxxx

Posted on 2004-10-12
251 Views
Thats basically what i need to know in a nutshell.....
these are the parameters
-The first digit can't be a zero
-The first 3 digits can't contain an 8 or 9
-the second set of 3 digits cant begin with a zero
-the second set of 3 digits cant be greater than 610
-The first digit of the last set of digits cant be a zero

(ultimately i need it written in java form)
0
Question by:vsands
• 6

LVL 37

Expert Comment

ID: 12286275
0

LVL 37

Expert Comment

ID: 12286292
We're not here to write a complete programs for you. (Lots of other sites exist for that)
0

LVL 37

Expert Comment

ID: 12286313
Use java.util.Random.nextInt() to generate the parts of your telephone number
0

LVL 37

Accepted Solution

zzynx earned 50 total points
ID: 12286387
This can give you an idea:

String digits[] = { "0", "1", "2", "3", "4", "5", "6", "7", "8", "9" };
StringBuffer telephoneNr = new StringBuffer();
Random rd = new Random();

// Generate first digit (which shouldn't be 0, 8 or 9)
String tmp;
do {
tmp = digits[ rd.nextInt(8) ];
} while ( !tmp.equals("0") );
telephoneNr.append(tmp);

for (int i=1; i<=2; i++) { // Generate 2nd/3rd digit (which shouldn't be 8 or 9)
do {
tmp = digits[ rd.nextInt(10) ];
} while ( !tmp.equals("0") );
telephoneNr.append(tmp);
}

...
0

LVL 37

Expert Comment

ID: 12286400
Typo:

tmp = digits[ rd.nextInt(10) ];

should be

tmp = digits[ rd.nextInt(8) ];   // since 8 and 9 shouldn't be "picked"
0

LVL 37

Expert Comment

ID: 12286503
Or easier:

StringBuffer telephoneNr = new StringBuffer();
Random rd = new Random();

// Generate first part
String tmp;
do {
tmp = "" + (100 + rd.nextInt(900));
} while ( tmp.indexOf("8")!=-1 || tmp.indexOf("9")!=-1 );
telephoneNr.append(tmp);

telephoneNr.append("-");

// Generate 2nd part
tmp = "" + (100 + rd.nextInt(511));
telephoneNr.append(tmp);

telephoneNr.append("-");

// Generate 3rd part
tmp = "" + (1000 + rd.nextInt(9000));
telephoneNr.append(tmp);

System.out.println( telephoneNr.toString() );
0

LVL 2

Expert Comment

ID: 12321679
Try this code.. Compile & run the program.

public class Application1
{

String finalString = null;
public void generateNumber()
{
int p1 = 0, p2 = 0, p3 = 0;
String s1 = null, s2 = null, s3 = null;

while(true)
{
p1 = (int) (Math.random() * 1000);
String temp = "" + p1;
int tempFlag = 0;
for(int i = 0; i < temp.length();i ++)
{
if(temp.charAt(i) == '8' || temp.charAt(i) == '9')
{
tempFlag = 1;
}
}
if(p1 > 100 && tempFlag == 0)
break;
}

finalString = "" + p1;

while(true)
{
p2 = (int) (Math.random() * 1000);
if(p2 > 100 && p2 < 610)
break;
}

finalString = finalString + "-" + p2;

while(true)
{
p3 = (int) (Math.random() * 10000);
if(p3 > 1000)
break;
}

finalString = finalString + "-" + p3;

System.out.println(finalString);

}

public static void main(String args[])
{
Application1 app = new Application1();
for(int i = 0; i < 1000; i++) // this will generate 1000 numbers..
{
app.generateNumber();
}
}
}
0

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