Query: getting the last record for each member - sql

Question

Given a table ("Table") as follows (sorry about the CSV style since I don't know how to make it look like a table with the Stack Overflow editor):
id,member,data,start,end
1,001,abc,12/1/2012,12/31/2999
2,001,def,1/1/2009,11/30/2012
3,002,ghi,1/1/2009,12/31/2999
4,003,jkl,1/1/2012,10/31/2012
5,003,mno,8/1/2011,12/31/2011
If using Ruby Sequel, how should I write my query so I will get the following dataset in return.
id,member,data,start,end
1,001,abc,12/1/2012,12/31/2999
3,002,ghi,1/1/2009,12/31/2999
4,003,jkl,1/1/2012,10/31/2012
I get the most current (largest end date value) record for EACH (distinct) member from the original table.
I can get the answer if I convert the table to an Array, but I am looking for a solution in SQL or Ruby Sequel query, if possible. Thank you.
Extra credit: The title of this post is lame...but I can't come up with a good one. Please offer a better title if you have one. Thank you.

Answer 1

The Sequel version of this is a bit scary. The best I can figure out is to use a subselect and, because you need to join the table and the subselect on two columns, a "join block" as described in Querying in Sequel. Here's a modified version of Knut's program above:
require 'csv'
require 'sequel'
# Create Test data
DB = Sequel.sqlite()
DB.create_table(:mytable){
field :id
String :member
String :data
String :start # Treat as string to keep it simple
String :end # Ditto
}
CSV.parse(<<xx
1,"001","abc","2012-12-01","2999-12-31"
2,"001","def","2009-01-01","2012-11-30"
3,"002","ghi","2009-01-01","2999-12-31"
4,"003","jkl","2012-01-01","2012-10-31"
5,"003","mno","2011-08-01","2011-12-31"
xx
).each{|x|
DB[:mytable].insert(*x)
}
# That was all setup, here's the query
ds = DB[:mytable]
result = ds.join(ds.select_group(:member).select_append{max(:end).as(:end)}, :member=>:member) do |j, lj, js|
Sequel.expr(Sequel.qualify(j, :end) => Sequel.qualify(lj, :end))
end
puts result.all
This gives you:
{:id=>1, :member=>"001", :data=>"abc", :start=>"2012-12-01", :end=>"2999-12-31"}
{:id=>3, :member=>"002", :data=>"ghi", :start=>"2009-01-01", :end=>"2999-12-31"}
{:id=>4, :member=>"003", :data=>"jkl", :start=>"2012-01-01", :end=>"2012-10-31"}
In this case it's probably easier to replace the last four lines with straight SQL. Something like:
puts DB[
"SELECT a.* from mytable as a
join (SELECT member, max(end) AS end FROM mytable GROUP BY member) as b
on a.member = b.member and a.end=b.end"].all
Which gives you the same result.

Answer 2

What's the criteria for your result?
If it is the keys 1,3 and 4 you may use DB[:mytable].filter( :id => [1,3,4]) (complete example below)
For more information about filtering with sequel, please refer the sequel documentation, especially Dataset Filtering.
require 'csv'
require 'sequel'
#Create Test data
DB = Sequel.sqlite()
DB.create_table(:mytable){
field :id
field :member
field :data
field :start #should be date, not implemented in example
field :end #should be date, not implemented in example
}
CSV.parse(<<xx
id,member,data,start,end
1,001,abc,12/1/2012,12/31/2999
2,001,def,1/1/2009,11/30/2012
3,002,ghi,1/1/2009,12/31/2999
4,003,jkl,1/1/2012,10/31/2012
5,003,mno,8/1/2011,12/31/2011
xx
).each{|x|
DB[:mytable].insert(*x)
}
#Create Test data - end -
puts DB[:mytable].filter( :id => [1,3,4]).all

Answer 3

In my opinion, you're approaching the problem from the wrong side. ORMs (and Sequel as well) represent a nice, DSL-ish layer above the database, but, underneath, it's all SQL down there. So, I would try to formulate the question and the answer in a way to get SQL query which would return what you need, and then see how it would translate to Sequel's language.
You need to group by member and get the latest record for each member, right?
I'd go with the following idea (roughly):
SELECT t1.*
FROM table t1
LEFT JOIN table t2 ON t1.member = t2.member AND t2.end > t1.end
WHERE t2.id IS NULL
Now you should see how to perform left joins in Sequel, and you'll need to alias tables as well. Shouldn't be that hard.