First, let's clarify what you mean by "weekly."  Do you mean that no two inspections can be more than 7 days apart?  Or, simply that there must be at least one inspection during every calendar week?  Because, these require two very different solutions.

The first one is actually the easier one.  Create a formula for each record that looks like:

If DateDiff("d",Previous({MyReport.DateCk}),{MyReport.DateCk}) > 7 Then
    "Late Inspection"
Else
    "Within Parameters"

Obviously, you should probably change the actual output.  Making this something like a condition to change the Date to red when it's late might be good.


The second one is trickier.  That's because there is no built-in function in Crystal to find a missing week.  There are a few options.

One is to do something like the above, but test it based on the week rather than the number of days between.  It might look like:

If DatePart("ww",{MyReport.DateCk}) - DatePart("ww",Previous({MyReport.Date)) > 1 Then
    "Late Inspection"
Else
    "Within Parameters"

However, all this does is flag the late inspections.  It doesn't necessarily tell you the missing dates.  (Also note that it fails on the first week of the year, but it's fairly trivial to add an extra bit to account for that.)

Another option is to add a table to your database, which includes the date of each Sunday for the year.  Or, probably, a more generic table of sequential numbers, which you use to calculate the date for each Sunday of the year.  Then, create a join between this table and your inspection table, and make a note of any Sunday that doesn't have a corresponding inspection in the next 7 days.