Community Tip - Did you get an answer that solved your problem? Please mark it as an Accepted Solution so others with the same problem can find the answer easily. X
On pages 22 and 23, Section 14.1 of this file are calculations for 7 different cases. Each case has a boolean expression to satisfy. I need to create some kind of loop that will start with the expression for case 1 and if that is satisfied return case = case 1. If case 1 doesnt satisfy the expression i need to look at case 2 and if that works return case = case 2 an dif not continue to case 3. I need the loop to follow this logic to find the first expression that is satisfied and return that case or go until it reaches case 7. Any help would be appreciated. Thanks.
Solved! Go to Solution.
David DePolo wrote:
On pages 22 and 23, Section 14.1 of this file are calculations for 7 different cases. Each case has a boolean expression to satisfy. I need to create some kind of loop that will start with the expression for case 1 and if that is satisfied return case = case 1. If case 1 doesnt satisfy the expression i need to look at case 2 and if that works return case = case 2 an dif not continue to case 3. I need the loop to follow this logic to find the first expression that is satisfied and return that case or go until it reaches case 7. Any help would be appreciated. Thanks.
Many ways of skinning this particular cat.
On the assumption that you wish to retain the look and style of the section, see attached. There are a few other alternative approaches given as well.
Stuart
Sounds like a good use of the "while" loop in a program. Do the case calculations in the loop until your criterion is satisfied. Update the case number in the loop, and then the final case number will be sent to the output variables.
David DePolo wrote:
On pages 22 and 23, Section 14.1 of this file are calculations for 7 different cases. Each case has a boolean expression to satisfy. I need to create some kind of loop that will start with the expression for case 1 and if that is satisfied return case = case 1. If case 1 doesnt satisfy the expression i need to look at case 2 and if that works return case = case 2 an dif not continue to case 3. I need the loop to follow this logic to find the first expression that is satisfied and return that case or go until it reaches case 7. Any help would be appreciated. Thanks.
Many ways of skinning this particular cat.
On the assumption that you wish to retain the look and style of the section, see attached. There are a few other alternative approaches given as well.
Stuart
Stuart thank you very much for the examples. I chose to use the last vector solution which really simplified the calculations. Thanks again for such a quick detailed response.
Dave