I know this might belong in Relationships or Define Fields, but it also belongs here, I think.

I have a multi-file database full of portals and relationships. One is between Agencies and Addresses. I decided to put addresses into a separate file because some of our contacts have two or three locations, most have just one, and a few have 25 or so.

I'd like to have an Agency Summary layout to print. I'd like it show all appropriate addresses, but I don't want to make 25 portal rows just in case and

have them all be blank 90% of the time.

I've considered building a different layout for each number of addresses, but it's an awful lot of work, and I'm not sure how to identify the necessary number of rows.

My questions are:

1) is there a Status field or other simple way to find out how many sub-records a given record is related to? That is, it would spit out 1, or 3, or 25 and I could use it in scripts or calcs?

2) Is there a clever work-around someone knows to make portals more flexible about this? Or a way to drop them altogether and use look-up fields or something?

Thanks!

1. You can use the Count(theRelationship::anyField) function to find how many related records there are for a parent record.

2. You should be printing from the related file if possible, rather than trying to print portals. However, you can set the portals to shrink, and it will reduce the number of portal rows to the required number when printing.