Unique Random numbers

How to get unique random numbers every time in AS3
I have searched but not able to get satisfactory results

var rand :Number;
rand = Number(movXml.movie.length());
for(var n:uint = 0 ; n<5 ; n++)
num1 = Number(Math.round(Math.random()* ((rand-1) - 0) + 0 ) );
if(tempArray.indexOf(num1) == -1)
tempArray.push(num1);
//trace(tempArray);
num2 = Number (tempArray[0]);
//trace("num1:"+num2);
trace(tempArray);
some time this produce more than 5 numbers
and some time less than 3
I want 4 random numbers , but different
the above loop runs 4 times , so if number is repeated , it will not add to Array , hence i will not able to get 4 Required numbers

Similar Messages

Array of unique random numbers - help?

basically, i'm trying to create a BINGO card.
i stared learning Java a week ago and am finding it easy to learn and use....at least so far.
my instructor gave me the assignment to created a BINGO card. so i used the system.out.format command to make 5 tab fields, 8 spaces each ("%8s %8s %8s %8s %8s", "B", "I", "N", "G", "O"). although the assignment only wanted me to think numbers out of thin air, just to use as examples, i went above and beyond with the idea to make a BINGO card generator that could actually function.
then i started the random number sequences, using the Math.random() command. all told, there are 24 of these in the program, one for each number on a bingo card. the field in the middle is the FREE space, so it has no Math.random command.
in BINGO, each letter (one of my five tab fields) can have a number in ranges of 15. B is 1 to 15, I is 16 to 30, etc. it looks similar to this:
B I N G O
9 19 39 57 66
3 28 32 51 74
3 29 FREE 46 70
14 28 43 55 67
9 24 35 59 62
as you can tell, i'm having trouble with actually making unique random numbers so that none repeat.
is there a command or string or something to help me accomplish this?

The best way I've come up with is to use an object to store the numbers that implements Collection--like ArrayList...
Then you load it with the range of numbers that you want and call, shuffle() on the object, that will randomize your range of numbers and then you can choose the quantity you want from the storage object. So let's say you need 25 number in the range of 1 to 100:
Add the numbers (you have to use classes so I would do Integers) 1 to 100;
call shuffle()
pull back the first 25 numbers
This will guarantee that each number is distinct and in a random order.
If you need multiple sheets populated, you can just call shuffle() between population of each sheet.
package Junk;
import java.util.ArrayList;
import java.util.Collections;
class Junk{
private ArrayList<Integer> l;
Junk(){
    l = new ArrayList<Integer>();
public void loadList(int s, int e){
    for(int i=s; i<=e; i++){
      l.add(new Integer(i));
public void randomizeList(){
    Collections.shuffle(l);
public void printList5(){
    for(int i=0; i<5; i++){
      System.out.println(l.get(i));
public static void main(String[] args){
    Junk j = new Junk();
    j.loadList(10,99);
    j.randomizeList();
    j.printList5();
    System.out.println();
    j.randomizeList();
    j.printList5();
}

How to generate unique random numbers

Hi All,
I am wondering whether there is already a random library or built-in function available in Java to produce some random numbers between certain ranges that are not repetitive. Let's look at a common examples as follows:
Random diceRoller = new Random();
for (int i = 0; i < 10; i++) {
int roll = diceRoller.nextInt(6) + 1;
System.out.println(roll);
}My understanding from this approach is that it allows the same number to be repeated over and over again. However, I would like to find out how to continue generating random numbers from remaining ones that haven't been generated earlier.
Using the above example to illustrate my intention:
1st random number generated - possibility of 1 - 6 showed up. Say 5 is picked.
2nd random number generated - possibility of 1, 2, 3, 4, 6 only. Say 2 is picked.
3rd random number generated - possibility of 1, 3, 4, 6 available. Say 1 is picked.
4th random number generated - possibility of 3, 4, 6 left. Say 6 is picked.
5th random number generated - possibility of 3, 4 remains. Say 4 is picked.
Any assistance would be much appreciated.
Many thanks,
Jack

htran_888 wrote:
Hi All,
I am wondering whether there is already a random library or built-in function available in Java to produce some random numbers between certain ranges that are not repetitive. Let's look at a common examples as follows:
Random diceRoller = new Random();
for (int i = 0; i < 10; i++) {
int roll = diceRoller.nextInt(6) + 1;
System.out.println(roll);
}My understanding from this approach is that it allows the same number to be repeated over and over again. However, I would like to find out how to continue generating random numbers from remaining ones that haven't been generated earlier.
Using the above example to illustrate my intention:
1st random number generated - possibility of 1 - 6 showed up. Say 5 is picked.
2nd random number generated - possibility of 1, 2, 3, 4, 6 only. Say 2 is picked.
3rd random number generated - possibility of 1, 3, 4, 6 available. Say 1 is picked.
4th random number generated - possibility of 3, 4, 6 left. Say 6 is picked.
5th random number generated - possibility of 3, 4 remains. Say 4 is picked.
Any assistance would be much appreciated.If it is your school assignment then you have the answer above (List & the lists length).
(You might want to look at Collections)

How can I generate multiple unique random numbers?

Hello,
I am trying to generate multiple random numbers between a given set of numbers (1-52) and not have the same number generated twice within that set. I can compare the last and next numbers with the compare function but how would I go about comparing all of the generated numbers without using a huge list of shift registers...
Any help/ideas are welcome and appreciated.
Jason
Solved!
Go to Solution.

Here is an implementation of Jason's solution.
steve
Help the forum when you get help. Click the "Solution?" icon on the reply that answers your
question. Give "Kudos" to replies that help.
Attachments:
random.vi ‏10 KB

Random numbers

I'm doing a java program using random numbers.
I would like to avoid that each generated number, besides unique, is different from its index in array - for instance, the first number in array is not number 1, the second is not number 2, the third not number 3, and so one.
Can someone help ?
That�s my relevant code :
     boolean isRepeated = false;
        int [] rNums = new int [t1];
        for (int x = 0 ; x < t1 ; x++)
        do
        isRepeated = false;
       rNums [x] = (int) ((Math.random () * t1) + 1);
       for (int y = 0 ; y < x ; y++)
           if (rNums [x] == rNums [y])
       isRepeated = true;
       break;
       while (isRepeated);

Any suggestion ?Not really.
The not reapeating part by itself is easy. Either populate the array with numbers that match the indices and then shuffle, or just keep track of which numbers you've used to far and get the next number in a loop until you get one you haven't used up yet.
The no-number-at-its index (and by the way, array indices start with 0, not 1) is fairly easy. Say your array is 10 elements, 0..9. So you generate a random number in the range 0..8 (9 possible values). Say we're trying to populate index 3. If the number generated is < 3, then use it as-is. If it's >= 3, then add 1.
Putting them together is harder. And you could end up with a situation where 0-8 are populated, and the only number that's not used and the only index that's not used are both 9.
Why are you imposing these restrictions?

7 digit unique random number

I would like to generate 7 digit unique random number. I have it something like this :
Random generator = new Random();
          generator.setSeed(System.currentTimeMillis());
          //generates seven digit random number.
          return generator.nextInt(9999999) + 999999;
But it generates 8 digit numbers sometimes. I am giving the Max limit in the braces and min limit after the "+" sign. Could anybody help me figure out what is wrong.

This is not a horrible solution but an intelligent &
easy solution ;)Actually this would be the easy and intellegent solution to the problem "give a random 7 digit integer."
Random generator = new Random();
//generates seven digit random number.
return generator.nextInt(10000000);Though the orginal post does not specify this as part of the problem others have assumed (probably correctly) that in fact the poster wants a number between 1000000 and 9999999 (inclusive) which can easily be produced with the following modification to the above.
Random generator = new Random();
//generates seven digit random number.
return generator.nextInt(9000000) + 1000000;I am confused by your response. Are you saying the becasue the original question was asking why flawed code doesn't work, that makes your solution good?

How to generate random numbers that doesnt contain characters?

How do we generate random numbers in ESB ROuting Service/ XSL transformation that does not contain characters. I have been using "orcl:generate-guid()" , but it contains some characters, so, I substring it to get only numbers. I dont want to take this risk and in future, my substring itself may contain characters.
Has anybdy tried this before?
Thanks,

If the goal is to have a pseudo random number then consider using the translate function to replace the occurances of the A - E characters with another digit. For example:
translate(orcl:generate-guid(),'0123456789ABCDEF','0123456789123456')
This will produce a 1 for A, 2 for B, etc. While not a valid hex to decimal conversion of the GUID value, it does get around the hex issue and might be sufficient for your purposes.
I would perform a large sample test to see just how unique the values end up being.
As for why hex is used, well GUID's are most commonly written in text as a sequence of hexadecimal digits such as: 3F2504E0-4F89-11D3-9A0C-0305E82C3301. i.e. 128 bits represented in 32 characters formatted into 5 sections.
Hope this helps,
Peter

Working with random numbers and probabilities

Hi again,
i am working with random numbers at the moment.
in the first step i do create a random number between 3 and 8 which is stored to a variable.
set myVAR1 to random number from 3 to 8 as integer
Lets assume myVAR1 is 5
Now i want to select 5 numbers out of number pool from 1 to 8. Each number should be pickable only once.
How would i realize that ?
I guess i need some kind of pool, array and then select 5 out of this array, but i am not sure about the syntax.
In best case i would like to add probabilities to those 8 numbers in the pool.
i.e.
1 (5%),2(5%),3(20%),4(10%),5(20%),6(10%),7(15%),8(15%)
any help is heavily appreciated
best regards
fidel

Hello fidel,
For selecting unique numbers from given pool, try something like this.
--SCRIPT 1
(* select unique numbers from pool *)
set pool to {1, 2, 3, 4, 5, 6, 7, 8} -- number pool
set n to random number from 3 to 8 -- selection count
return {n, random(n, pool)}
on random(n, pool)
set kk to {}
repeat with i from 1 to count pool
set end of kk to i
end repeat
set rr to {}
repeat n times
set k to kk's some integer
set kk's item k to false
set end of rr to pool's item k
end repeat
return rr
end random
--END OF SCRIPT 1
For introducing selection weight of each number in the pool, you may try the below.
--SCRIPT 2
(* select unique numbers from pool with stochastic weights *)
set pool to {1, 2, 3, 4, 5, 6, 7, 8} -- number pool
set weights to {5, 5, 20, 10, 20, 10, 15, 15} -- selection weight
set n to random number from 3 to 8 -- selection count
return {n, random(n, pool, weights)}
on random(n, pool, weights)
script o
property ww : weights
property pp : {}
property kk : {}
property rr : {}
repeat with i from 1 to count pool
repeat my ww's item i times
set end of my pp to pool's item i
end repeat
end repeat
repeat with i from 1 to count my pp
set end of my kk to i
end repeat
repeat n times
set k to my kk's some integer
set end of my rr to my pp's item k
set j to 0
set i to 1
repeat with w in my ww
set j to j + w
if k > j then
set i to j + 1
else
repeat with h from i to j
set my kk's item h to false
end repeat
set i to 1
exit repeat
end if
end repeat
end repeat
return my rr's contents
end script
tell o to run
end random
--END OF SCRIPT 2
Hope this may help,
H
Message was edited by: Hiroto

Unique random alpha num string

Hi Guys,
I have a unique question.
I need to generate a unique random Alpha Num string.
I cannot use 0,5,8 and 0,S,B in the generated output string.
The string generated should be no more than length 7 max
Pls help.

Following solution generates random unique strings combinations
from any given base set. Here is a sample output.
$ java UniqueStringGenerator 7 100 10
FCT2RYT
2M31BU8
VV8FU4Z
BON2MKE
OKUJZT0
68E8X8S
YTNLZAY
CPEKPFP
C176M24
GQB9QE9The idea is quite simple.
Pre-work:
1. The base set is first mixed to increase entropy.
2. All possible combinations are divide up into random buckets.
Work for every unique string:
1. Pick a bucket at random.
2. Get next number in bucket.
3. Convert the number to a unique string token.
It is possible to adjust how much entropy you want in your
sequence of generated combinations by choosing how many buckets
the generator contains. However, every bucket will take up some
memory.
import java.util.*;
import java.math.*;
public class UniqueStringGenerator {
    public static String[] ALPHA_NUM =
     {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P",
      "Q","R","S","T","U","V","X","Y","Z",
      "0","1","2","3","4","5","6","7","8","9"};
    private List _parts = null;
    private List _buckets = null;
    private int _length;
    private BigInteger _range = null;
    public UniqueStringGenerator(String[] parts, int length,
                     int numberOfBuckets) {
     _length = length;
     // mixing the base set to increase entropy
     _parts = new Vector();
     List all = new Vector();
     for (int i = 0;i < parts.length;i++) {
         all.add(parts);
     while (all.size()>0) {
     int p = (int)(((double)all.size())*Math.random());
     _parts.add(all.get(p));
     all.remove(p);
     // dividing all possible numbers into given number of buckets
     BigInteger numParts = BigInteger.valueOf(parts.length);
     _range = BigInteger.ONE;
     for (int i = 0;i<_length;i++) {
     range = range.multiply(numParts);
     TreeSet starts = new TreeSet();
     Random rand = new Random();
     while (starts.size()<numberOfBuckets-1) {
     BigInteger bi = new BigInteger(_range.bitLength(), rand);
     if (bi.compareTo(_range)<=0) {
          starts.add(bi);
     _buckets = new Vector();
     Bucket last = new Bucket(BigInteger.ZERO);
     _buckets.add(last);
     for (Iterator i = starts.iterator();i.hasNext();) {
     Bucket next = new Bucket((BigInteger)i.next());
     _buckets.add(next);
     last.setTo(next.getStart());
     last = next;
     last.setTo(_range);
private class Bucket {
     private BigInteger _start;
     private BigInteger _next;
     private BigInteger _to;
     public Bucket(BigInteger start) {
     _start = start;
     _next = start;
     public void setTo(BigInteger to) {
     _to = to;
     public BigInteger getStart() {
     return _start;
     public boolean hasNext() {
     return next.compareTo(to)<0;
     public BigInteger next() {
     BigInteger ret = _next;
     next = next.add(BigInteger.ONE);
     return ret;
     public String toString() {
     return "["+start+", "+next+", "+_to+"]:"+hasNext();
public String nextUnique() {
     // getting next big integer
     BigInteger left = null;
     while (left==null) {
     Bucket b = (Bucket)
          buckets.get((int)(Math.random()*buckets.size()));
     if (b.hasNext()) {
          left = b.next();
     // converting it to a combination
     StringBuffer unique = new StringBuffer();
     BigInteger base = BigInteger.valueOf(_parts.size());
     for (int i = 0;i<_length;i++) {
     BigInteger[] resRem = left.divideAndRemainder(base);
     left = resRem[0];
     unique.append(_parts.get(resRem[1].intValue()));
     return unique.toString();
public static void main(String[] args) {
     if (args.length!=3) {
     System.out.println("Usage: UniqueStringGenerator "+
               "[string size] [# buckets] [# prints]");
     return;
     int length = Integer.parseInt(args[0]);
     int numBuckets = Integer.parseInt(args[1]);
     int num = Integer.parseInt(args[2]);
     UniqueStringGenerator usg = new UniqueStringGenerator
     (UniqueStringGenerator.ALPHA_NUM, length, numBuckets);
     for (int i=0;i<num;i++) {
     System.out.println(usg.nextUnique());

I am trying to generate an array of 30 random numbers. after every 5 readings a new vi should open and tell the user that 5 readings has been completed. and again carry on with the generation of array.

since i do not have a transducer now, i am currently generating an array of 30 random numbers. after every 5 readings a warning should be given to the user that 5 readngs are complete. this cycle should repeat. the array size is 30.
please help me out, waiting for reply asap.
once i have the transducer, i will be taking 30 analog samples and then after every 5 smaples that wraning will be displaye din a new VI
Solved!
Go to Solution.

Use a while loop with a time delay representing your sampling period.
Use the count terminal to check if equals 4, so 4th iteration=5th sample.
Use a case structure. The true case will only be executed on the 4th iteration.
In the true case place a subVI with your desired message in the front panel. Go to the VI properties window and set "open front panel when called".
The closing condition of the warnign is not giving in your description.
Consider that rather than usign a subvi for this, you could use the "One/Two/Three button dialog" or "display message" vis at the "dialog and user interface" pallete.
Please give it a try and send your own VI. Do not expect us to provide a working solution.
Regards,

How can you generate Multiple random numbers from 1 to 49?

I am very new at programming with the iPhone SDK and I need some help with a project I am working on. What I want to do is set up a window with 6 labels and 1 button. When the user clicks the button, the program will populate 6 randomized numbers each ranging from 1 to 49, and then place each of 6 numbers in a label. But each time the program ends and starts again the numbers cannot be the same, and also when the user clicks the button, no label can have the same number twice, so for example. If label 1 had the number 10, the other 5 labels cannot have that number until the button is clicked again. I know how to set up the interface, I just need the code. I would so greatly appreciate someone's help in this matter. I have been trying to do this for days and I cannot figure it out. Possibly someone who knows tons about Objective C programming can help me!
Thank-you so very much!!

I see that you're writing a lottery number generator. Perhaps the easiest way to do it is to emulate a real lottery: fill an array (NSMutableArray, probably) with the numbers between 1 and 49, pick one at random, remove that number from the array, and repeat. (You can think of the numbers as being the balls and the array as being the machine that pops them out.)
One simple way to get the random indices needed is to extract six random bytes from Randomization Services using SecRandomCopyBytes, then loop over them, using the modulo operator (%) to select an index within the size of the array.
And no, I'm not going to write your code for you. If I was going to do that, I could package and sell the app myself.

I need help with this program ( Calculating Pi using random numbers)

hi
please understand that I am not trying to ask anymore to do this hw for me. I am new to java and working on the assignment. below is the specification of this program:
Calculate PI using Random Numbers
In geometry the ratio of the circumference of a circle to its diameter is known as �. The value of � can be estimated from an infinite series of the form:
� / 4 = 1 - (1/3) + (1/5) - (1/7) + (1/9) - (1/11) + ...
There is another novel approach to calculate �. Imagine that you have a dart board that is 2 units square. It inscribes a circle of unit radius. The center of the circle coincides with the center of the square. Now imagine that you throw darts at that dart board randomly. Then the ratio of the number of darts that fall within the circle to the total number of darts thrown is the same as the ratio of the area of the circle to the area of the square dart board. The area of a circle with unit radius is just � square unit. The area of the dart board is 4 square units. The ratio of the area of the circle to the area of the square is � / 4.
To simuluate the throwing of darts we will use a random number generator. The Math class has a random() method that can be used. This method returns random numbers between 0.0 (inclusive) to 1.0 (exclusive). There is an even better random number generator that is provided the Random class. We will first create a Random object called randomGen. This random number generator needs a seed to get started. We will read the time from the System clock and use that as our seed.
Random randomGen = new Random ( System.currentTimeMillis() );
Imagine that the square dart board has a coordinate system attached to it. The upper right corner has coordinates ( 1.0, 1.0) and the lower left corner has coordinates ( -1.0, -1.0 ). It has sides that are 2 units long and its center (as well as the center of the inscribed circle) is at the origin.
A random point inside the dart board can be specified by its x and y coordinates. These values are generated using the random number generator. There is a method nextDouble() that will return a double between 0.0 (inclusive) and 1.0 (exclusive). But we need random numbers between -1.0 and +1.0. The way we achieve that is:
double xPos = (randomGen.nextDouble()) * 2 - 1.0;
double yPos = (randomGen.nextDouble()) * 2 - 1.0;
To determine if a point is inside the circle its distance from the center of the circle must be less than the radius of the circle. The distance of a point with coordinates ( xPos, yPos ) from the center is Math.sqrt ( xPos * xPos + yPos * yPos ). The radius of the circle is 1 unit.
The class that you will be writing will be called CalculatePI. It will have the following structure:
import java.util.*;
public class CalculatePI
public static boolean isInside ( double xPos, double yPos )
public static double computePI ( int numThrows )
public static void main ( String[] args )
In your method main() you want to experiment and see if the accuracy of PI increases with the number of throws on the dartboard. You will compare your result with the value given by Math.PI. The quantity Difference in the output is your calculated value of PI minus Math.PI. Use the following number of throws to run your experiment - 100, 1000, 10,000, and 100,000. You will call the method computePI() with these numbers as input parameters. Your output will be of the following form:
Computation of PI using Random Numbers
Number of throws = 100, Computed PI = ..., Difference = ...
Number of throws = 1000, Computed PI = ..., Difference = ...
Number of throws = 10000, Computed PI = ..., Difference = ...
Number of throws = 100000, Computed PI = ..., Difference = ...
* Difference = Computed PI - Math.PI
In the method computePI() you will simulate the throw of a dart by generating random numbers for the x and y coordinates. You will call the method isInside() to determine if the point is inside the circle or not. This you will do as many times as specified by the number of throws. You will keep a count of the number of times a dart landed inside the circle. That figure divided by the total number of throws is the ratio � / 4. The method computePI() will return the computed value of PI.
and below is what i have so far:
import java.util.*;
public class CalculatePI
public static boolean isInside ( double xPos, double yPos )
     double distance = Math.sqrt( xPos * xPos + yPos * yPos );
public static double computePI ( int numThrows )
    Random randomGen = new Random ( System.currentTimeMillis() );
    double xPos = (randomGen.nextDouble()) * 2 - 1.0;
    double yPos = (randomGen.nextDouble()) * 2 - 1.0;
    int hits = 0;
    int darts = 0;
    int i = 0;
    int areaSquare = 4 ;
    while (i <= numThrows)
        if (distance< 1)
            hits = hits + 1;
        if (distance <= areaSquare)
            darts = darts + 1;
        double PI = 4 * ( hits / darts );
        i = i+1;
public static void main ( String[] args )
    Scanner sc = new Scanner (System.in);
    System.out.print ("Enter number of throws:");
    int numThrows = sc.nextInt();
    double Difference = PI - Math.PI;
    System.out.println ("Number of throws = " + numThrows + ", Computed PI = " + PI + ", Difference = " + difference );
}when I tried to compile it says "cannot find variable 'distance' " in the while loop. but i thought i already declare that variable in the above method. Please give me some ideas to solve this problem and please check my program to see if there is any other mistakes.
Thanks a lot.

You've declared a local variable, distance, in the method isInside(). The scope of this variable is limited to the method in which it is declared. There is no declaration for distance in computePI() and that is why the compiler gives you an error.
I won't check your entire program but I did notice that isInside() is declared to be a boolean method but doesn't return anything, let alone a boolean value. In fact, it doesn't even compute a boolean value.

Problem with Purchase Order creation with Random numbers .

Hi Experts
Currently i am facing an issue with Bapi BAPI_PO_CREATE1 to create Purchase order with random numbers for example items 1, 3,5.
Please let me know the settings .
Thanks in Advance

Hi Neha,
A reset of the release strategy only takes place if
- the changeability of the release indicator is set to '4' in
   case of a purchase requisition and '4' or '6' in case of
   another purchasing document (purchase order, request for
   quotation, contract, scheduling agreement),
- the document is still subject to the previous release
   strategy,
- the new TOTAL NET ORDER VALUE is higher than the old one.
The total net order value is linked to CEKKO-GNETW for a purchase order,
CEBAN-GSWRT for a purchase requisition item-wise release and CEBAN-GFWRT
for a purchase requisition overall release.
If you have maintained a Tolerance for value changes during release
(V_161S-TLFAE), the release strategy is reset when the value of the
document is higher than the percentage you have specified and if the
document is still subject to the previous release strategy.
Review the SAP note 365604 and 493900 for more
information about Release strategy.
Regards,
Ashwini.

Program to create random numbers in plsql

How do we Write a program to create random numbers.
Thanks

No need to - we have DBMS_RANDOM :)

Getting a COUNT of unique document numbers loaded

What do i need to use in the Mapping in order to get the following results :-
I am loading from a Purchase Order Item table (with Purch.Ord. number as the key). As this is an item detail table, there will be multiple records for 1 purchase order no. the result i need is a COUNT of the unique list of PO numbers loaded from this table. how can i achieve this ?
even if i load from the PO header table (which will give me unique PO numbers for each record), how do i get the COUNT of records loaded from the table and use that result to put into my data warehouse target ?
thanks for any replies in advance !

thanks for the answer Igor. you have partially answered my question (as i am a newbie to OWB, i have to test out your suggestion to see if i can achieve the results you mentioned).
maybe an example will make my question clearer.
Assume 1 have records for 2 POs in table PURCHORD:-
PONUM POITEMNO PRDCODE QTY
10000 001 PRD_A 10
10000 002 PRD_B 2
10001 001 PRD_A 5
10001 002 PRD_B 1
10001 003 PRD_C 10
the result i expect if i extract records from this table i => 2 unique Purchase Orders (my measure in the Cube is count of POs)! it is basically a COUNT on the "DISTINCT (GROUP by PONUM)". i just can't figure out how to implement this in OWB.
thanks again

Unique Random numbers

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