constrain found set

[Mike Summer]

I am a new member struggling with the following problem. I want to search on two fields and I’m using the ‘constrain found’ set to do this. However it is not working. Can you help? The script is as follows.

Enter Find Mode[]

Set Field [Customer::User;Get(Accountname)]

Perform Find []

Constrain Found Set []

Enter Find Mode []

Set Field [Customer::DateCreated; Get(CurrentDate)]

Perform Find []

The above script returns the records by CurrentDate, but not CurrentDate and User.

I would appreciate your thoughts.

[Ted S]

How about this:

Enter Find Mode[]

Set Field [Customer::User;Get(Accountname)]

Set Field [Customer::DateCreated; Get(CurrentDate)]

Perform Find []

[ssimons]

Why not try setting both fields before performing the find.

Enter Find Mode

Set Field [Customer::User; Get(Accountname)]

Set Field [Customer::DateCreated; Get(CurrentDate)]

Perform Find

Edited January 23, 2006 by Guest
Sorry, I think Ted and I were replying at the same time.

[mr_vodka]

Try it like this:

Enter Find Mode[]

Set Field [Customer::User;Get(Accountname)]

Set Field [Customer::DateCreated; Get(CurrentDate)]

Perform Find []

[comment]

Just to explain why your version didn't work, it should be:

Enter Find Mode[]

Set Field [Customer::User;Get(Accountname)]

Perform Find []

Enter Find Mode []

Set Field [Customer::DateCreated; Get(CurrentDate)]

Constrain Found Set []

IOW, Constrain Found Set [] comes INSTEAD OF Perform Find [].

[Mike Summer]

Thanks. It worked fine. I think I overcomplicated it, thinking I had to constrain a found set first and then do another find on that set.

[Mike Summer]

Thanks. It worked fine. I think I overcomplicated it, thinking I had to constrain a found set first and then do another find on that set.

[Mike Summer]

Thanks. I was looking for an explanation of why this did not work - the constrain found set[] replaces perform find []. Just one more question. It appears to make more sense to put the constrain found set [] after the first find, rather than at the end of the second find, does it not? My logic was to constrain on the first find and then do a second find on the constrained set.

[comment]

The first find does Perform Find[], which REPLACES the current found set. The second one does Constrain Found Set, which REDUCES the (new) current found set. Each one needs to have search criteria BEFORE it can do anything.

[Mike Summer]

Gotcha. Much appreciated.