I've got a huge table:
+------+---------+-----+-----+-------+------+---------+
|ticker|data_date|price|count|oper_id|ext_nr|oper_summ|
+------+---------+-----+-----+-------+------+---------+
|SBER |2015-08..|70.00|15 |0 |251528|1050.00 |
|AFLT |2015-08..|30.00|5 |0 |251525|150.00 |
|SBER |2015-08..|69.00|10 |1 |251521|690.00 |
|SBER |2015-08..|71.00|15 |1 |251513|1065.00 |
|SBER |2015-08..|72.00|15 |0 |251512|1080.00 |
data_date format: 2015-01-05 09:59:59
UNIQUE KEY `idx_ticker_ext_nr` (`ticker`,`ext_nr`)
I need to SELECT LAST N rows WHERE SUM(oper_summ) will be LESS THEN 10000
I've found similar topic: limiting the rows to where the sum a column equals a certain value in MySQL
SELECT
O.ext_nr,
O.price, O.count,
O.oper_summ,
(SELECT
sum(oper_summ) FROM Table1
WHERE ext_nr <= O.ext_nr) 'RunningTotal'
FROM Table1 O
HAVING RunningTotal <= 10000
but unable to make it work in my coditions...
Found a solution:
SET #msum := 0;
select t1.* from
(select m.*,
(#msum := #msum + m.oper_summ) as cumul_oper_summ from jos_fintools_data m order by m.data_date DESC )
t1 where t1.cumul_oper_summ <= 10000;
credits goes to toomanyredirects: limiting the rows to where the sum a column equals a certain value in MySQL
Use variables:
SELECT o.*
FROM (SELECT O.ext_nr, O.price, O.count, O.oper_summ,
(#os := #os + oper_summ) as RunningTotal
FROM Table1 O CROSS JOIN
(SELECT #os := 0) params
ORDER BY data_date desc
) o
HAVING RunningTotal <= 10000;
Note: you need to order by something in the subquery. I'm not sure what the right column is. My best guess is the date column.
SET #msum := 0;
select t1.* from
(select m.*,
(#msum := #msum + m.oper_summ) as cumul_oper_summ from jos_fintools_data m order by m.data_date DESC )
t1 where t1.cumul_oper_summ <= 10000;
Related
I would like to append the results of 2 queries into one result set.
SELECT n.member_no, n.surname, n.first_name
FROM `names` AS n
WHERE member_no = '1003';
SELECT s.registration
FROM `system` AS s
WHERE s.RECNUM = 1;
This must return one record with data from the names table plus data from the system (one record) table
Member_no | surname | first_name | registration
--------------------------------------------------
1003 | Brown | Peter | My registration
You can use CrossJoin:
SELECT n.member_no, n.surname, n.first_name, s.registration
FROM names AS n
CROSS JOIN system s
WHERE n.member_no = '1003' and s.RECNUM = 1;
we can correlate your registration and name tables based on row_number(). You may want to try below query.
SELECT rn.member_no, rn.surname, n.first_name, s.registration
FROM
(SELECT member_no, surname, first_name, row_number() over (order by member_no) rn
FROM `names`) n
LEFT JOIN
(SELECT row_number() over (order by RECNUM) rn, registration
FROM `system`) s on s.rn = n.rn
WHERE n.member_no = '1003'
Try this one.
SELECT DISTINCT n.member_no, n.surname, n.first_name,s.registration
FROM `names` AS n, `system` AS s
WHERE s.RECNUM = 1 AND member_no = '1003';
I have a table products, and a table product-languages.
products:
prd_id|prd_name
product-languages:
prd_id|language_id|prd_name
1 |1 |Product_German_name
1 |2 |Product_English_name
1 |4 |Product_French_name
I want to join product-languages on prd_id and then sort it by my priority list of language_id (for example: 2,1,4) - first I want the result of lang_id=2, if this is not available I want to have lang_id=1 as the first result.
Is this possible in SQL? I think my personal order list is the problem, because I would have to check if that lang_id is even available...
Subquery maybe?
Thanks
You can do this in standard SQL via case with something like:
select
*
from
products p,
product_languages l
where
p.prd_id = l.prd_id
order by
(case language_id
when 2 then 1
when 1 then 2
when 4 then 3
end) asc
This is a bit of a pain in MySQL, but you can do it. One method uses variables:
select t.*
from (select t.*,
(#rn := if(#prd_id = prd_id, #rn + 1,
if(#prd_id := prd_id, 1, 1)
)
) as rn
from producct_languages t join
(select #prd_id := -1, #rn := 0) params
order by prd_id, field(language_id, 2, 1, 4)
) t
where rn = 1;
I have a table ("lms_attendance") of users' check-in and out times that looks like this:
id user time io (enum)
1 9 1370931202 out
2 9 1370931664 out
3 6 1370932128 out
4 12 1370932128 out
5 12 1370933037 in
I'm trying to create a view of this table that would output only the most recent record per user id, while giving me the "in" or "out" value, so something like:
id user time io
2 9 1370931664 out
3 6 1370932128 out
5 12 1370933037 in
I'm pretty close so far, but I realized that views won't accept subquerys, which is making it a lot harder. The closest query I got was :
select
`lms_attendance`.`id` AS `id`,
`lms_attendance`.`user` AS `user`,
max(`lms_attendance`.`time`) AS `time`,
`lms_attendance`.`io` AS `io`
from `lms_attendance`
group by
`lms_attendance`.`user`,
`lms_attendance`.`io`
But what I get is :
id user time io
3 6 1370932128 out
1 9 1370931664 out
5 12 1370933037 in
4 12 1370932128 out
Which is close, but not perfect. I know that last group by shouldn't be there, but without it, it returns the most recent time, but not with it's relative IO value.
Any ideas?
Thanks!
Query:
SQLFIDDLEExample
SELECT t1.*
FROM lms_attendance t1
WHERE t1.time = (SELECT MAX(t2.time)
FROM lms_attendance t2
WHERE t2.user = t1.user)
Result:
| ID | USER | TIME | IO |
--------------------------------
| 2 | 9 | 1370931664 | out |
| 3 | 6 | 1370932128 | out |
| 5 | 12 | 1370933037 | in |
Note that if a user has multiple records with the same "maximum" time, the query above will return more than one record. If you only want 1 record per user, use the query below:
SQLFIDDLEExample
SELECT t1.*
FROM lms_attendance t1
WHERE t1.id = (SELECT t2.id
FROM lms_attendance t2
WHERE t2.user = t1.user
ORDER BY t2.id DESC
LIMIT 1)
No need to trying reinvent the wheel, as this is common greatest-n-per-group problem. Very nice solution is presented.
I prefer the most simplistic solution (see SQLFiddle, updated Justin's) without subqueries (thus easy to use in views):
SELECT t1.*
FROM lms_attendance AS t1
LEFT OUTER JOIN lms_attendance AS t2
ON t1.user = t2.user
AND (t1.time < t2.time
OR (t1.time = t2.time AND t1.Id < t2.Id))
WHERE t2.user IS NULL
This also works in a case where there are two different records with the same greatest value within the same group - thanks to the trick with (t1.time = t2.time AND t1.Id < t2.Id). All I am doing here is to assure that in case when two records of the same user have same time only one is chosen. Doesn't actually matter if the criteria is Id or something else - basically any criteria that is guaranteed to be unique would make the job here.
Based in #TMS answer, I like it because there's no need for subqueries but I think ommiting the 'OR' part will be sufficient and much simpler to understand and read.
SELECT t1.*
FROM lms_attendance AS t1
LEFT JOIN lms_attendance AS t2
ON t1.user = t2.user
AND t1.time < t2.time
WHERE t2.user IS NULL
if you are not interested in rows with null times you can filter them in the WHERE clause:
SELECT t1.*
FROM lms_attendance AS t1
LEFT JOIN lms_attendance AS t2
ON t1.user = t2.user
AND t1.time < t2.time
WHERE t2.user IS NULL and t1.time IS NOT NULL
Already solved, but just for the record, another approach would be to create two views...
CREATE TABLE lms_attendance
(id int, user int, time int, io varchar(3));
CREATE VIEW latest_all AS
SELECT la.user, max(la.time) time
FROM lms_attendance la
GROUP BY la.user;
CREATE VIEW latest_io AS
SELECT la.*
FROM lms_attendance la
JOIN latest_all lall
ON lall.user = la.user
AND lall.time = la.time;
INSERT INTO lms_attendance
VALUES
(1, 9, 1370931202, 'out'),
(2, 9, 1370931664, 'out'),
(3, 6, 1370932128, 'out'),
(4, 12, 1370932128, 'out'),
(5, 12, 1370933037, 'in');
SELECT * FROM latest_io;
Click here to see it in action at SQL Fiddle
If your on MySQL 8.0 or higher you can use Window functions:
Query:
DBFiddleExample
SELECT DISTINCT
FIRST_VALUE(ID) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS ID,
FIRST_VALUE(USER) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS USER,
FIRST_VALUE(TIME) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS TIME,
FIRST_VALUE(IO) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS IO
FROM lms_attendance;
Result:
| ID | USER | TIME | IO |
--------------------------------
| 2 | 9 | 1370931664 | out |
| 3 | 6 | 1370932128 | out |
| 5 | 12 | 1370933037 | in |
The advantage I see over using the solution proposed by Justin is that it enables you to select the row with the most recent data per user (or per id, or per whatever) even from subqueries without the need for an intermediate view or table.
And in case your running a HANA it is also ~7 times faster :D
Ok, this might be either a hack or error-prone, but somehow this is working as well-
SELECT id, MAX(user) as user, MAX(time) as time, MAX(io) as io FROM lms_attendance GROUP BY id;
select b.* from
(select
`lms_attendance`.`user` AS `user`,
max(`lms_attendance`.`time`) AS `time`
from `lms_attendance`
group by
`lms_attendance`.`user`) a
join
(select *
from `lms_attendance` ) b
on a.user = b.user
and a.time = b.time
I have tried one solution which works for me
SELECT user, MAX(TIME) as time
FROM lms_attendance
GROUP by user
HAVING MAX(time)
I have a very large table and all of the other suggestions here were taking a very long time to execute. I came up with this hacky method that was much faster. The downside is, if the max(date) row has a duplicate date for that user, it will return both of them.
SELECT * FROM mb_web.devices_log WHERE CONCAT(dtime, '-', user_id) in (
SELECT concat(max(dtime), '-', user_id) FROM mb_web.devices_log GROUP BY user_id
)
select result from (
select vorsteuerid as result, count(*) as anzahl from kreditorenrechnung where kundeid = 7148
group by vorsteuerid
) a order by anzahl desc limit 0,1
I have done same thing like below
SELECT t1.*
FROM lms_attendance t1
WHERE t1.id in (SELECT max(t2.id) as id
FROM lms_attendance t2
group BY t2.user)
This will also reduce memory utilization.
Thanks.
Possibly you can do group by user and then order by time desc. Something like as below
SELECT * FROM lms_attendance group by user order by time desc;
Try this query:
select id,user, max(time), io
FROM lms_attendance group by user;
This worked for me:
SELECT user, time FROM
(
SELECT user, time FROM lms_attendance --where clause
) AS T
WHERE (SELECT COUNT(0) FROM table WHERE user = T.user AND time > T.time) = 0
ORDER BY user ASC, time DESC
I want to do the following. I have a table in the database, I am working on a table called asistencia and this table has 3 columns
id_asistencia as a int AUTOINCREMENT
nro_matricula as an int which I took it from another table called
alumnos
fecha as a date
This is a sketch of the database
id_asistencia | nro_matricula | fecha
1 | 0001| 2015-01-10
2 | 0002| 2015-01-10
3 | 0002| 2015-02-10 (another date )
The thing is I have to do a percentage
select all id_1 records in my nro_matricula column and see how many times its repeated in my rows and do a percentage respect all the dates in my database
EG : id_1 came to class day(whatever day) and he/she did not came to class the next day so id_1 has 50% assistance
Expected result
nro_matricula | percentage
0001| 50
0002| 100
The question is how can I make this query. If can be done in PHP its even better but i feel that this can be done in SQL
PS : The Database wasn't created by me
And excuse my English is not the better and i expect it to be understandable for you to help me
You can use sql statement like this:
select (
sum (if nro_matricula = '001' ,1,0 )
/ count(*)
from asistencia
--where nro_matricula = '001'
Maybe just simply:
select al.nro_matricula,
100 * count(distinct al.fecha) / (select count(distinct al1.fecha) from alumnos al1) as percentage
from alumnos al
group by al.nro_matricula
I did found the answer to my question. Thank you all for helping me out
SELECT
asistencia.nro_matricula as matricula,
COUNT( DISTINCT asistencia.fecha)* 100 /
COUNT( DISTINCT asistencia.nro_matricula) / (SELECT COUNT(DISTINCT asistencia.fecha)
FROM asistencia
ORDER BY COUNT(*) DESC
LIMIT 1 )
as porcentaje_asistencia
FROM asistencia
JOIN alumno
WHERE asistencia.nro_matricula = alumno.nro_matricula AND alumno.id_curso = 'basica6a'
Tried this in Oracle. Should work in MySQL too.
SELECT aa.NRO_MATRICULA , days_present/total_count* 100 FROM
(SELECT DISTINCT NRO_MATRICULA,
COUNT(*) as days_present FROM ASISTENCIA GROUP BY NRO_MATRICULA ) AA
,
(SELECT COUNT(*) as total_count FROM (SELECT DISTINCT(FECHA) FROM ASISTENCIA GROUP BY FECHA)) BB
Ouptut
nro_matricula percentage
0001 50
0002 100
The query (SELECT COUNT(*) FROM (SELECT DISTINCT(FECHA) FROM ASISTENCIA AA GROUP BY FECHA)) will give count of distinct date (2 in your case). Then we are getting distinct nro_matricula group by nro_matricula to get its count which will give the days it was present. Then divide both values from above steps to get percentage.
I need help to write MySQL query.
I have table full of logs where one of the column is unix timestamp.
I want to group (GROUP BY) those records so that events that were made in close range time (i.e. 5 sec) between each of them are in one group.
For example:
Table:
timestamp
----------
1429016966
1429016964
1429016963
1429016960
1429016958
1429016957
1429016950
1429016949
1429016943
1429016941
1429016940
1429016938
Become to groups like that:
GROUP_CONCAT(timestamp) | COUNT(*)
-----------------------------------------------------------------------------
1429016966,1429016964,1429016963,1429016960,1429016958,1429016957 | 6
1429016950,1429016949 | 2
1429016943,1429016941,1429016940,1429016938 | 4
Of course I can work with the data array afterwards in php, but I think that mysql would do it faster.
I started by using a variable to get the position of each row, where 1 is the highest time column and ending with the lowest, like this:
SET #a := 0;
SELECT timeCol, #a := #a + 1 AS position
FROM myTable
ORDER BY timeCol DESC;
For simplicity, we will call this positionsTable so that the rest of the query will be more readable. Once I created that table, I used a 'time_group' variable that checked if a previous row was within the last 5 seconds. If it was, we keep the same time_group. It sounds ugly, and looks kind of ugly, but it's like this:
SELECT m.timeCol, m.position,
CASE WHEN (SELECT p.timeCol FROM positionsTable p WHERE p.position = m.position - 1) <= m.timeCol + 5
THEN #time_group
ELSE #time_group := #time_group + 1 END AS timeGroup
FROM positionsTable m;
And then ultimately, using that as a subquery, you can group them:
SELECT GROUP_CONCAT(timeCol), COUNT(*)
FROM(
SELECT m.timeCol, m.position,
CASE WHEN (SELECT p.timeCol FROM positionsTable p WHERE p.position = m.position - 1) <= m.timeCol + 5
THEN #time_group
ELSE #time_group := #time_group + 1 END AS timeGroup
FROM positionsTable m) tmp
GROUP BY timeGroup;
Here is an SQL Fiddle example.
http://sqlfiddle.com/#!9/37d88/20
SELECT GROUP_CONCAT(t1.t) as `time`,
COUNT(*)
FROM (SELECT *
FROM table1
ORDER BY t) as t1
GROUP BY CASE WHEN (#start+5)>=t THEN #start
ELSE #start:=t END