Re: can I use fetch spec to filter an array?
Re: can I use fetch spec to filter an array?
- Subject: Re: can I use fetch spec to filter an array?
- From: Jonathan Rochkind <email@hidden>
- Date: Tue, 30 Dec 2003 14:49:07 -0600
Not an EOFetchSpecification, but an EOQualifier. See
EOQualifier.filteredArrayWithQualifier.
I don't understand exactly what you want to do, so I'm not sure
exactly how to do it, but if you want to filter an array in memory,
EOQualifier.filteredArrayWithQualifier is what you are looking for.
At 9:34 AM +1300 12/31/03, Denis Stanton wrote:
I thought I knew this, but now I can't find any mention in the API
Can I use a fetch specification to filter an array in memory, as
opposed to an entity in a database?
I have written an application for a travel company. They need a
report showing the total value of bookings received from each agent
over a data range. Actually there are about 11 totals per agent and
say 300 agents.
For my first attempt I used a fetch spec inside a loop. For each
agent I defined a fetch spec with agent name, start date and end
date and then fetched all the matching records from the bookings
file into an array named bookings. Then I used
bookings.valueForKey("@sum.nettprice") to accumulate the total of
booking.netprice(), and so on through all 11 sub totals.
Too slow.
I wondered if there was any need to re-read the database. Since
there is a one-to-many relationship between agent and booking, the
bookings for an agent are already in agent.bookings(), except that
they have not been selected by date range. So can I apply a filter
to the agent.bookings() array in memory instead of reading the
Booking entity from the database using:
fetchspec = new EOFetchSpecification("Booking", qualifier, null);
MSMutableArray bookings = new
NSMutableArray(editingContext.objectsWithFetchSpecification(fetchspec)
I have also tried filtering the agent.bookings() array by means of a
for loop to select and copy each of the the bookings within range to
a new array, but that is also slow and seems crude.
Is there a proper way to do this?
Denis
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