1 Reply Latest reply on Dec 31, 2012 3:06 PM by philmodjunk

    Is this a complicated Find or do I need a Script?



      Is this a complicated Find or do I need a Script?


           Trying to make a new database for scheduling workflow at a commercial print shop. There are multiple options in our Bindery department, each step of which is scheduled by date. I'd like to create a Find or a Report or something that lets me print out one sheet of paper for a particular day with a column for each of the 10 tasks and a list of all jobs scheduled for that particular task in each column. So under column 1 (cutting) you'd have three jobs listed for 1/2/13 - Job 1, Job 2 and Job 4. Under column 2 (numbering) you'd have 2 jobs listed for 1/2/13 - Job 3 and Job 5. And so on. Only the tasks scheduled for that day would be listed. [rough sample attached]

           Right now I use the Quick Search for a certain date and have created a layout that shows all the columns by task across the top, all the jobs by name on the left, with a grid between them with dates (or blank spaces if nothing is scheduled for that task). The only problem with this is that if Job 1 has Cutting scheduled for 1/2/13, and Numbering for 1/3/13, both dates appear on the sheet in their appropriate columns. So I have to go through with a highlighter to note the ones that are actually for 1/2/13. Obviously this is less than ideal.

           I'm making a new database for the new year, so now's the time to throw out the old format that doesn't really work and come up with something that does. I bought the Missing Manual for Filemaker Pro 12, and it's taught me a lot about databases, but I'm not sure how to go about setting up a brand new database so that I can print out our To Do List every day.

           Can somebody point me in the right direction?


        • 1. Re: Is this a complicated Find or do I need a Script?

               This looks like more of a data structure and relationships issue than a find.

               I'd consider using a different portal for each column shown here so that either a relationship or a portal filter or both in combination can result in a list of Jobs for that task and day. This also assumes that your data is structured something like this:


               Jobs::__pkJobID = Job_Task::_fkJobID
               Tasks::_pkTaskID = Job_Task::_fkTaskID

               Job_Task would be the table that lists the assigned/scheduled tasks for a given job (one record for each such task) and a date field in Job_Task can be part of what is used in the relationship or portal filter to control which jobs are listed for a given day.

               Here's one possiblility:

               DailyTasks::Today = Job_Task::TaskDate

               This relationship could be used in all of the portals, but each portal can include a filter for a specific task.

               So if "Cutting" has a taskID number of 1, The portal filter expression for Cutting tasks might be:

               Job_Taks::_fkTaskID = 1

               Today can be either a date field that auto-enters today's date when you create a new record or an unstored calcualtion: Get ( CurrentDate ).