SQL Query for events that happend in a specific order - sql

Question

I have the following table:
+--------+-------+------+--+
| Object | Event | Time | |
+--------+-------+------+--+
| Obj1 | A | 1 | |
| Obj1 | B | 3 | |
| Obj2 | A | 7 | |
| Obj2 | B | 4 | |
+--------+-------+------+--+
My goal is to get all objects that both had the event A & B with the condition that A happened first (in time). So far I only came up with the query to find me all objects that had A & B without including the time:
SELECT DISTINCT Object
FROM
(SELECT *
FROM
(SELECT *
FROM table
INNER JOIN
(SELECT Object Obj
FROM table
WHERE event LIKE '%A%' AS temp_table) ON table.Object = temp_table.Obj) AS temp_final
WHERE event LIKE '%B%') AS temp2;
So the end result would be that I get a table that includes only:
Obj1
Since this is the only Object that fulfills all criteria.
The time column is a Date stamp in real life, but for simplicity I used integers.
Thanks you for the help

Answer 1

If you are only tracking two events that happened one after the other, than you can solve this with a single JOIN.
This will work regardless of the number of events Obj1 has, as how you mentioned, you are only interested in A and B existing and being one after the other, respectively.
select distinct t1.object
from TABLE t1
inner join TABLE t2 on t1.object = t2.object
and t2.time > t1.time
and t1.event = 'A'
and t2.event = 'B'
Here is a sample of the result of the code:
declare #tbl table (obj varchar(10), event varchar(1), time int)
insert #tbl values ('Obj1', 'A', 1), ('Obj1', 'B', 3), ('Obj2', 'A', 7), ('Obj2', 'B', 4)
select distinct t1.obj
from #tbl t1
inner join #tbl t2 on t1.obj = t2.obj
and t2.time > t1.time
and t1.event = 'A'
and t2.event = 'B'

Answer 2

Here is a compact solution which should run across most RDBMS. This solution does not assume that there are only two events, and should run for any number of events.
SELECT t1.Object
FROM yourTable t1
INNER JOIN
(
SELECT Object, MIN(Time) AS Time
FROM yourTable
GROUP BY Object
) t2
ON t1.Object = t2.Object AND
((t1.Event = 'A' AND t1.Time = t2.Time) OR
t1.Event <> 'A')
GROUP BY t1.Object
HAVING COUNT(*) = 2 -- change this count to match # of events
Demo on MySQL:
SQLFiddle

Answer 3

Try this:
SELECT DISTINCT object
FROM yourtable t
WHERE EXISTS
(SELECT FROM yourtable t3
WHERE t3.object = t.object
AND t3.event = 'A'
AND EXISTS
(SELECT 'B'
FROM yourtbale t4
WHERE t4.object = t3.object
AND t4.event = 'B'
AND t4.time > t3.time)
)

Answer 4

If you are using sql-server:
SELECT
A.[Object]
, A.[Time]
, B.[Time]
FROM
(SELECT
Distinct [Object]
FROM
[table] AS A
WHERE
A.[Event] = 'A'
) AS A
CROSS APPLY
(SELECT
TOP 1 *
FROM
[table] AS B
WHERE
[Event] = 'B'
AND
B.[Object] = A.[Object]
AND
A.[Time] < B.[Time]) AS B

Answer 5

For SQL Server:
;with A as
(select Object, MIN(Time) as Time from table where Event='A' group by Object)
, B as
(select Object, MIN(Time) aS Time from table where Event='B' group by Object)
Select A.Object from A inner join B on B.Object=A.Object where A.Time < B.Time