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Generating A Random Password


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Hi guys

On creation of a new record I need FM to generate a unique random password with the format of four numerics (0-9) and then six alpha characters (a-z). Can anyone suggest how I might go about doing this?

Seems like something so simple but cannot think where to start.

Thanks.

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Your calc is ...

Middle("0123456789"; Int(Random*10; 0)+1;1) &

Middle("0123456789"; Int(Random*10; 0)+1;1) &

Middle("0123456789"; Int(Random*10; 0)+1;1) &

Middle("0123456789"; Int(Random*10; 0)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25; 0)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25; 0)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25; 0)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25; 0)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25; 0)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25; 0)+1;1)

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Hiya jfmcel

One thing I noticed on your fantastic calculation was that when the first random numeric was a zero it was being omitted in the field display. So I changed the first line of your script to dle("123456789"; Int(Random*9; 0)+1;1) &

This way the first generated number will never be a zero. See below.

Middle("123456789"; Round(Random*9; 0)+1;1) & Middle("0123456789"; Round(Random*10; 0)+1;1) & Middle("0123456789"; Round(Random*10; 0)+1;1) & Middle("0123456789"; Round(Random*10; 0)+1;1) & Middle("qwertyuiopasdfghjkzxcvbnm"; Round(Random*25; 0)+1;1) & Middle("qwertyuiopasdfghjkzxcvbnm"; Round(Random*25; 0)+1;1) & Middle("qwertyuiopasdfghjkzxcvbnm"; Round(Random*25; 0)+1;1) & Middle("qwertyuiopasdfghjkzxcvbnm"; Round(Random*25; 0)+1;1) & Middle("qwertyuiopasdfghjkzxcvbnm"; Round(Random*25; 0)+1;1) & Middle("qwertyuiopasdfghjkzxcvbnm"; Round(Random*25; 0)+1;1)

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Oops - I modified my calc to fix a problem introduced by using Round() and not Int(), but I screwed it up in the process. The correct answer is...

Middle("123456789"; Int(Random*9)+1;1) &

Middle("0123456789"; Int(Random*10)+1;1) &

Middle("0123456789"; Int(Random*10)+1;1) &

Middle("0123456789"; Int(Random*10)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25)+1;1) &

Middle("qwertyuiopasdfghjkzxcvbnm"; Int(Random*25)+1;1)

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By definition, you cannot have both random and unique.

Once having arrived at a random number by the methods described, you need to check for uniqueness, probably with a self-join lookup. Then re-do the calc if it's not unique. I think you could do this in a calculation field but it's scotch time, not mind time so the solution is left as an exercise ......

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The chances of the password not being unique are so extremely small that it should not ever be a concern. [For this method of generating a random password, the chances of any two passwords being the same is 1/(9*10^3*25^6) or about one in 2 trillion.]

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The chance of getting a duplicate is not relevant. I recently had reason to develop some thousands of random six-digit numbers and was amazed at the frequency of duplicates. (Try throwing a simple dice and see how often you get duplicates.) Depending on how important uniqueness is, checking for duplicates could be essential.

We have a lottery here which has a jackpot based on the winning number being drawn on a second draw. The odds are, I think, 1 in 1 trillion. It has been won and there have not been anything like 1 trillion lotteries.

One of the classic problems in program development is trying to cover all those so-called 'impossible'/'never happens'/'extremely unlikely' situations. They happen on day 1 of live running.

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We are talking about passwords here. So why should duplicate passwords ever be a concern. [FM 7 Account Names would be a different story. The solution there is to let FM return an error when you try to create the account.]

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I must have been asleep when I wrote that last post.

Your calculation of the 'odds' is not correct. The apparent 'odds' are much lower than than that - 1 in zillions of trillion. However, your calculation does not generate genuine random numbers.

The chances of getting a duplicate depend on the number of records being generated.

To use some simple numbers -

a 10-sided dice

need 5 random numbers

Throw the dice 5 times and you will almost certainlky get duplicates. I can't remember the specific formula but the chance of getting two identical numbers in 5 tries is close to 1. (Ever seen the party trick of finding out how many people ina room have the same birthday? In a room of forty, you have almost certainy.)

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This topic is 7259 days old. Please don't post here. Open a new topic instead.

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