SELECT DISTINCT HAVING Count unique conditions

Question

I\'ve searched for an answer on this but can\'t find quite how to get this distinct recordset based on a condition. I have a table with the following sample data:

Answer 1

Use a COUNT(DISTINCT Location) and join against a subquery on Type and Color The GROUP BY and HAVING clauses as you have attempted to use them will do the job.

/* Be sure to use DISTINCT in the outer query to de-dup */
SELECT DISTINCT
   MyTable.Type,
   MyTable.Color,
   Location
FROM 
  MyTable
  INNER JOIN (
    /* Joined subquery returns type,color pairs having COUNT(DISTINCT Location) > 1 */
    SELECT
      Type,
      Color,
      /* Don't actually need to select this value - it could just be in the HAVING */
      COUNT(DISTINCT Location) AS UniqueLocations
    FROM
      MyTable
    GROUP BY Type, Color
    /* Note: Some RDBMS won't allow the alias here and you 
       would have to use the expanded form
       HAVING COUNT(DISTINCT Location) > 1
     */
    HAVING UniqueLocations > 1
  /* JOIN back against the main table on Type, Color */
  ) subq ON MyTable.Type = subq.Type AND MyTable.Color = subq.Color

Here is a demonstration

Answer 2

You could write your first query as this:

Select Type, Color, Count(Distinct Location) As UniqueLocations
From Table
Group By Type, Color
Having Count(Distinct Location) > 1

(if you're using MySQL you could use the alias UniqueLocations in your having clause, but on many other systems the aliases are not yet available as the having clause is evaluated before the select clause, in this case you have to repeat the count on both clauses).

And for the second one, there are many different ways to write that, this could be one:

Select Distinct Type, Color, Location
From Table
Where
  Exists (
    Select
      *
    From
      Table Table_1
    Where
      Table_1.Type = Table.Type
      and Table_1.Color = Table.Color
    Group By
      Type, Color
    Having
      Count(Distinct Location) > 1
  )