counting records, this might be in the wrong place

[jordan gibson]

Hi

I think this is the area I should be posting this in but if not I will appologise in advance.

I am new to Filemaker and trying to get my feet, I have gone through some tutorials online and started to build my first "solution"

What I want to achive is a layout that shows how many records a user has created on that day.

I have created six layouts each one is unrelated and has a a few fields of data we want to capture. I am also auto entering the username(account) the created and modified date.

So basicaly what im trying to achieve is a front page that count records where the :username and the ::created_date match get (AccountName) and todays date.

I thought this would be simple and im sure it is if I was looking in the right place.

Any help would be great!!!

Jord

[LaRetta]

Hi Jord, welcome to FM Forums!

When you say, "I have created six layouts each one is unrelated and has a a few fields of data we want to capture" it is a bit confusing and concerning. Layouts are not tables. Are all these records in ONE table or do you have six different tables (for six different users)? If so, I would stop and re-group right there ...

You should have one table with a field for AccountName and a field for CreationDate. And this table should also have a unique auto-enter serial number (important). Then you would create a second occurrence of this table in your graph (just click the ++ and select the same table again and name it 'self join.'

You will need to create a calculation called cToday in this main table with: Get ( CurrentDate ) and set the calculation result to type DATE.

You would then relate these tables as:

MainTable::AccountName = SelfJoin::AccountName

AND

MainTable::cToday = SelfJoin::CreationDate

Then create a summary field called sCountIDs with Count ( thatUniqueSerial )

On your main layout, place your SelfJoin::sCountIDs field.

[jordan gibson]

Hi.

I had to read your reply several times and to be hoenest im still lost.

Yes i created 6 tables and a layout for each, the thinking behind that was that I didnt think i would need to report any information between each table.

in sudo script my thinking would be something along the lines of

count records where username = get (AccountName) and created_date = //

Is this possible with a script or should i be looking at starting over with joined tables.

[LaRetta]

called cToday in this main table with: Get ( CurrentDate ) and set the calculation result to type DATE.

I apologize that I should have been clear that this calculation must be (in Storage Options) checked as 'Do not store calculation results."

[LaRetta]

What is different about the six tables? If the only difference is the Account Name then it should be ONE table. Seriously. It doesn't matter if you currently don't see that you will ever need to report information between each table.

What if the Owner wants ONE report showing the total entries per Account Name? If you have the records in one table then you can do that easily. And that is my point about displaying the counts as well ...

Can you tell us what type of data these tables contain and provide us with a bit of context and purpose?

[jordan gibson]

Cheers for your patience

Each table contains a few fields (mostly defined as drop downs etc)

Currently im auto entering the username and creation date on each record as well.

What we wanted to do was show the user how many records they had entered that day on the bottom of the layout (ideally on each of the six layouts, one for each table.)

Ive managed to do a subsummary report that shows total jobs per user but due to not linking them I have one report per job/table.

edit*

Sorry I should have been clearer. Each table and associated layout is a job of sorts. Its not table per user , its say thirty users each filling in differant jobs each day.

Edited June 12, 2009 by Guest

[comment]

What is different about the six tables?

[jordan gibson]

They have differant fields on them? appart from username and date_created which each table has