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.