MySQL Counting Records Cumulative per Day - mysql

Question

I have a table that look like this:
S_ID | DATE
1 2016-01-01
1 2016-01-02
1 2016-01-02
1 2016-01-05
1 2016-01-05
2 2017-01-02
2 2017-01-04
2 2017-01-04
2 2017-01-04
2 2017-01-04
2 2017-01-05
I am trying, in a single query to have the result below (adding cumulatively the records in time)
DATE | S_ID 1 | S_ID 2
2016-01-01 | 1 | 0
2016-01-02 | 3 | 1
2016-01-03 | 3 | 1
2016-01-04 | 3 | 5
2016-01-05 | 5 | 6
Any suggestions?

Answer 1

Try this using group on date and conditional sum along with user variables to get cumulative sum.
Select date,
#s1 := #s1 + s_id_1 s_id_1,
#s2 := #s2 + s_id_2 s_id_2
From (select
date,
sum(s_id = 1) s_id_1,
sum(s_id = 2) s_id_2
from your_table
group by date
Order by date) t cross join (select #s1 := 0, #s2 :=0 ) t2;
It uses the fact that the true is 1 and false is 0 in mysql

Answer 2

You should better use
select
t.s_ID,
t.`date`,
(SELECT SUM(1) FROM table x WHERE x.`date` <= t.`date` AND x.S_ID = t.S_ID) AS cumulative_sum
from table t
group by s_ID,`date`;
This will not provide your desired result, but a form of the result which works independent of the number of users, while the information stays the same:
s_ID | DATE | cumulative_sum
1 2016-01-01 1
1 2016-01-02 3
1 2016-01-05 5
2 2016-01-02 1
2 2016-01-04 5
2 2016-01-05 6
(If there's no entry for an id/date pair, the count didn't change that day)

Answer 3

You can arrived this with a PREPARED Statement. The first Query will generate a query with all s_id in your table. You only must change YOYOURTABLE to your table name :
see the sample
CONCAT('SELECT `date`,'
,GROUP_CONCAT(f1)
,' FROM YOURTABLE GROUP BY DATE')
INTO #myquery
FROM (
SELECT DISTINCT CONCAT('sum(s_id = ',s_id,') AS sid_',s_id) AS f1
FROM yourtable
) tab1;
-- ONLY for Test to verify the Query
SELECT #myquery;
PREPARE test FROM #myquery;
EXECUTE test;
DEALLOCATE PREPARE test;
sample
mysql> SELECT * FROM yourtable;
+------+------------+
| s_id | date |
+------+------------+
| 1 | 2016-01-01 |
| 1 | 2017-02-02 |
| 2 | 2017-01-05 |
| 4 | 2016-03-04 |
| 7 | 2016-12-12 |
+------+------------+
5 rows in set (0,01 sec)
mysql> SELECT
-> CONCAT('SELECT `date`,'
-> ,GROUP_CONCAT(f1)
-> ,' FROM YOURTABLE GROUP BY DATE')
-> INTO #myquery
-> FROM (
-> SELECT DISTINCT CONCAT('sum(s_id = ',s_id,') AS sid_',s_id) AS f1
-> FROM yourtable
-> ) tab1;
Query OK, 1 row affected (0,01 sec)
mysql> SELECT #myquery;
+----------------------------------------------------------------------------------------------------------------------------------------+
| #myquery |
+----------------------------------------------------------------------------------------------------------------------------------------+
| SELECT `date`,sum(s_id = 1) AS sid_1,sum(s_id = 2) AS sid_2,sum(s_id = 4) AS sid_4,sum(s_id = 7) AS sid_7 FROM YOURTABLE GROUP BY DATE |
+----------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0,00 sec)
mysql> PREPARE test FROM #myquery;
Query OK, 0 rows affected (0,00 sec)
mysql> EXECUTE test;
+------------+-------+-------+-------+-------+
| date | sid_1 | sid_2 | sid_4 | sid_7 |
+------------+-------+-------+-------+-------+
| 2016-01-01 | 1 | 0 | 0 | 0 |
| 2016-03-04 | 0 | 0 | 1 | 0 |
| 2016-12-12 | 0 | 0 | 0 | 1 |
| 2017-01-05 | 0 | 1 | 0 | 0 |
| 2017-02-02 | 1 | 0 | 0 | 0 |
+------------+-------+-------+-------+-------+
5 rows in set (0,00 sec)
mysql> DEALLOCATE PREPARE test;
Query OK, 0 rows affected (0,00 sec)
mysql>