Joined: 25 Feb 2013
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Topic: Counting Unique Records For All Groups Posted: 21 Jan 2014 at 4:32pm
Hello All,
I am trying to find the best way to sum the count of all records that appear across multiple groups.
To explain what I am trying to achieve we may have two groups in a report that both contain the same record, i.e.
Order 1
Part A
Part B
Part C
Order 2
Part A
Part B
Part D
I am attempting to generate a total at the report footer, that will contain a summary to the effect of:
Part A: 2
Part B: 2
Part C: 1
Part D: 1
When looking at our order history, I am attempting to see how many times Part A occurs with Part B on the same order. Because of this I have grouped the information by the order number record and used a selection criteria, this prevents me simply grouping around the part number field and using a count.
I could use a formula and manually enter a sum if part = A, however I'm trying to evaluate against approximately 10,000 records (different part numbers), so that would be prohibitive.
Manually I have exported the report without a total to Excel, and used Excel's Data Validation to show only unique records, and then use a COUNTIF statement to generate this information, however it is a manual process and would like to achieve a result inside Crystal Reports.
Is this possible, or am I barking up the wrong tree? I don't mind being advised to research certain topics if someone could point me in the right direction.
Joined: 30 Oct 2012
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Posts: 104
Posted: 22 Jan 2014 at 5:13am
Have you tried creating a separate report that groups the data by part type and summing the values there? You could place that report as a subreport in your report footer.
Joined: 19 Dec 2008
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Posts: 9053
Posted: 22 Jan 2014 at 5:24am
another option would be to drop a cross tab in your report header or footer,
group the CT rows on the part field and do a count of order # as the summary
Perhaps a distinct count or a sum is needed if there are part numbers amounts in each order or multiple of the same part in one order that need to be excluded or included.
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