Question

I need help with an SQL-Query. I have one table with many entries and I want to query all the entries which have the same values for the last 3 columns.

My table looks as follows:

|Refrigator|98C08A|2011-08-06 00:00:30|126|126
|Refrigator|B7BE29|2011-08-06 00:00:30|73|70
|Refrigator|599393|2011-08-06 00:00:30|126|126
|Refrigator|B7BE29|2011-08-06 00:00:29|73|70
|Refrigator|599393|2011-08-06 00:00:29|126|126
|Refrigator|599393|2011-08-06 00:00:29|126|126
|Refrigator|98C08A|2011-08-06 00:00:29|126|126
|Refrigator|98C08A|2011-08-06 00:00:29|126|126
|Refrigator|599393|2011-08-06 00:00:28|126|126

So I want to get all rows, which have the exact same values for the last 3 columns, so the result should look like:

|Refrigator|98C08A|2011-08-06 00:00:30|126|126
|Refrigator|599393|2011-08-06 00:00:30|126|126
|Refrigator|599393|2011-08-06 00:00:29|126|126
|Refrigator|599393|2011-08-06 00:00:29|126|126 (if possible without this duplicate)
|Refrigator|98C08A|2011-08-06 00:00:29|126|126
|Refrigator|98C08A|2011-08-06 00:00:29|126|126 (if possible without this duplicate)

Does anyone have an idea how to manage this?
What I tried so far was:

SELECT * 
FROM smtab 
WHERE Datetime IN (
      SELECT Datetime 
      FROM smtab 
      GROUP BY Datetime 
      HAVING count(Datetime) >1) 
AND Power1 IN (
      SELECT Power1 
      FROM smtab 
      GROUP BY Power1 
      HAVING count(Power1) >1) 
AND Power8 IN (
      SELECT Power8 
      FROM smtab 
      GROUP BY Power8 
      HAVING count(Power8) >1) 
ORDER BY Datetime DESC;

but I didn’t work!!!

Hope someone can help me! thx in advance…

Answer 1

SELECT DISTINCT *
FROM   smtab NATURAL JOIN (
  SELECT   Datetime, Power1, Power8
  FROM     smtab
  GROUP BY Datetime, Power1, Power8
  HAVING   COUNT(*) > 1
) AS t