Find 20 random records

[satinsilhouette]

I need to find 20 random records out of thousands. In excel this is easy. How do I do it in Crystal 11?

Thanks!

Thanks so much!
satinsilhouette

 

[lbass]

Create a formula:

rnd()

Add this to your detail section and to report->sort records. Then use a detail suppression formula like this:

recordnumber > 20

-LB

 

[satinsilhouette]

Thank you, I thought that was round.

Thanks so much!
satinsilhouette

 

[satinsilhouette]

Okay - that gives me 20 records, but not 20 unique records. How do I get 20 unique records?

Thanks so much!
satinsilhouette

 

[Dharma1955]

Add this to your detail section:

//@Random_Value
Whileprintingrecords
Numbervar random
If previous {result of RND()}<> { result of RND()} then random_:=random+1 else random:=random

Then use a detail suppression formula like this:

Previous({result of RND()})={result of RND()} or {@Random_Value}>20

 

[dgillz]

That IS 20 unique records. It is not random however, it just shows you the first 20 records. Please post samples of your data and desired results.

Software Sales, Training, Implementation and Support for Macola, eSynergy, and Crystal Reports

http://www.gainfocus.biz

"What version of URGENT!!! are you using?

 

[Dharma1955]

If you accept that RND() gives a random number the approach I outlined gives 20 unique random numbers.

 

[Dharma1955]

I should have said random values.

 

[satinsilhouette]

Thanks folks! Appreciate the help!

Thanks so much!
satinsilhouette

 

[dgillz]

Dharma,

My bad,I misread your post. That would indeed give you 20 random records.

Software Sales, Training, Implementation and Support for Macola, eSynergy, and Crystal Reports

http://www.gainfocus.biz

"What version of URGENT!!! are you using?