Reputation: 59
I have added one more id column for the order purpose, if the table is like this
+-----+-------+
| id | cs_id |
+-----+-------+
| 1 | a |
| 2 | b |
| 3 | a |
| 4 | a |
| 5 | a |
| 6 | b |
| 7 | b |
| 8 | b |
| 9 | b |
+-----+-------+
i want the continuous occurrence of cs_id order by id column
+-----+-------+---------------------------------
| id | cs_id | continuous_occurrence_cs_id
+-----+-------+---------------------------------|
| 1 | a | 1
| 2 | b | 1
| 3 | a | 1
| 4 | a | 2
| 5 | a | 3
| 6 | b | 1
| 7 | b | 2
| 8 | b | 3
| 9 | b | 4
+-----+-------+---------------------------------+
Upvotes: 0
Views: 788
Reputation: 36087
First of all, in SQL by the definition data has no any order unless the ORDER BY
is used.
See: Wikipedia - Order By
ORDER BY is the only way to sort the rows in the result set.
Without this clause, the relational database system may return the rows in any order.
You must provide an additional column to your table that determines the order, and can be used in ORDER BY
clause, for exampleRN
column in the below example:
RN CS_ID
---------- ----------
1 a
2 b
3 a
4 a
5 a
6 b
7 b
8 b
9 b
For the above data you can use Common Table Expression (recursive query) to get required result, for example the below query works on Oracle database:
WITH my_query( RN, cs_id , cont ) AS (
SELECT t.rn, t.cs_id, 1
FROM My_table t
WHERE rn = 1
UNION ALL
SELECT t.rn, t.cs_id,
case when t.cs_id = m.cs_id
then m.cont + 1
else 1
end
FROM My_table t
JOIN my_query m
ON t.rn = m.rn + 1
)
select * from my_query
order by rn;
RN CS_ID CONT
---------- ---------- ----------
1 a 1
2 b 1
3 a 1
4 a 2
5 a 3
6 b 1
7 b 2
8 b 3
9 b 4
Upvotes: 3