Question

I have the following table in SQL Server 2008:

CREATE TABLE tbl (ID INT, dtIn DATETIME2, dtOut DATETIME2, Type INT)

INSERT tbl VALUES
(1, '05:00', '6:00', 1),
(2, '05:00', '7:00', 1),
(3, '05:01', '8:00', 1),
(4, '05:00', '8:00', 1),
(5, '05:00', '6:00', 2),
(6, '05:00', '7:00', 2)

that selects IDs of all records of the same type, with the same dtIn date, ordered by stOut in ascending order:

SELECT DISTINCT tbl.id FROM tbl   
  LEFT JOIN tbl AS t1
  ON tbl.type = t1.type AND
     tbl.dtIn = t1.dtIn
  ORDER BY tbl.dtOut ASC

But it gives me an error:

ORDER BY items must appear in the select list if SELECT DISTINCT is
specified

I tried putting that ORDER BY in different places and it all doesn’t seem to work. What am I doing wrong here?

Answer 1

When you narrow it down to individual id’s, you create the possibility that each id might have more than one dtOut associated with it. When that happens, how will Sql Server know which order to use?

You could try:

SELECT t1.id
FROM tbl t1
LEFT JOIN  tbl t2 on t1.type = t2.type AND t1.dtIn = t2.dtIn
GROUP BY t1.id, t2.dtOut
ORDER BY t2.dtOut

However, as I mentioned above this can open the possibility of having the same id listed more than once, if it matches to more than one record on the right-side table.