I just want to ask if grouping rows with the same value but came from different columns is possible.
I have a scenario that we should sum up the total minutes if the records are found "continuous" transactions by checking if the STARTDATETIME column matches the previous data of ENDDATETIME column if they are the same. See image link below for reference.
Thanks guys.
I modified Gordon Linoff's solution ( see my comment under the question):
SELECT
c.employee_id
,MIN(c.start_date) AS start_date
,MAX(c.end_date) AS end_date
,COUNT(*) AS numcontracts,
TIMESTAMPDIFF(minute,MIN(c.start_date),MAX(c.end_date)) AS timediff
FROM
(
SELECT
c0.*
,(#rn := #rn + COALESCE(startflag, 0)) AS cumestarts
FROM
(SELECT c1.*,
(NOT EXISTS (SELECT 1
FROM contracts c2
WHERE c1.employee_id = c2.employee_id AND
c1.start_date = c2.end_date
)
) AS startflag
FROM contracts c1
ORDER BY employee_id, start_date
) c0 CROSS JOIN (SELECT #rn := 0) params
) c
GROUP BY c.employee_id, c.cumestarts
http://rextester.com/VOGMU19779
timediff contains the minutes passed in the combined interval.
Related
Get only the biggest date:
These are check-in and check-out records of employees, some times they do twice or more entries on the system in a row. In this sample there were two check-out in a row. Assuming these rows always gonna be ordered, in the case of check-out I would like have the biggest date, and in the case of the check-in the smallest date.
In that case I would like to have this:
The smaller date was excluded:
DEMO
Try this, in this big CASE statement I increment column by one, if checkin switches from null to not null and the other way around. Then it's enough to group by this column taking max and min of checkout and checkin respectively:
select #checkinLag := null, #rn := 0;
select max(id),
functionario,
loja,
min(checkin),
max(checkout)
from (
select case when (checkinLag is null and checkin is not null) or
(checkinLag is not null and checkin is null)
then #rn := #rn + 1 else #rn end rn,
checkin,
checkout,
loja,
id,
functionario
from (
select #checkinLag checkinLag,
#checkinLag := checkin,
checkin,
checkout,
loja,
id,
functionario
from dummyTable
order by coalesce(checkin, checkout)
) a
) a group by functionario, loja, rn
I have used subqueries, to guarantee order of evaluating expressions (assigning and using of #checkinLag), as Gordon Linoff pointed.
Demo
My solution:
Select
*
from dummyTable base
where (base.checkout is null or not exists (
select
1
from dummyTable co
where co.checkout between base.checkout and DATE_ADD(base.checkout, INTERVAL 5 SECOND)
and base.id <> co.id
and base.functionario = co.functionario
and base.loja = co.loja
)) and (base.checkin is null or not exists (
select
1
from dummyTable ci
where ci.checkin between DATE_SUB(base.checkin, INTERVAL 5 SECOND) and base.checkin
and base.id <> ci.id
and base.functionario = ci.functionario
and base.loja = ci.loja
));
you can test the query here. There is no need that the rows are orderd. I choose 5 seconds as the interval where check-in/outs should be ignored.
I'm trying in MySql to count the number of users created each day and then get an accumulative figure on a row by row basis. I have followed other suggestions on here, but I cannot seem to get the accumulation to be correct.
The problem is that it keeps counting from the base number of 200 and not taking account of previous rows.
Where was I would expect it to return
My Sql is as follows;
SELECT day(created_at), count(*), (#something := #something+count(*)) as value
FROM myTable
CROSS JOIN (SELECT #something := 200) r
GROUP BY day(created_at);
To create the table and populate it you can use;
CREATE TABLE myTable (
id INT AUTO_INCREMENT,
created_at DATETIME,
PRIMARY KEY (id)
);
INSERT INTO myTable (created_at)
VALUES ('2018-04-01'),
('2018-04-01'),
('2018-04-01'),
('2018-04-01'),
('2018-04-02'),
('2018-04-02'),
('2018-04-02'),
('2018-04-03'),
('2018-04-03');
You can view this on SqlFiddle.
Use a subquery:
SELECT day, cnt, (#s := #s + cnt)
FROM (SELECT day(created_at) as day, count(*) as cnt
FROM myTable
GROUP BY day(created_at)
) d CROSS JOIN
(SELECT #s := 0) r;
GROUP BY and variables have not worked together for a long time. In more recent versions, ORDER BY also needs a subquery.
I have a SQL table, one row is the revenue in the specific day, and I want to add a new column in the table, the value is the incremental (could be positive or negative) revenue between a specific day and the previous day, and wondering how to implement by SQL?
Here is an example,
original table,
...
Day1 100
Day2 200
Day3 150
...
new table (add incremental column at the end, and for first column, could assign zero),
Day1 100 0
Day2 200 100
Day3 150 -50
I am using MySQL/MySQL Workbench.
thanks in advance,
Lin
SELECT a.day, a.revenue , a.revenue-COALESCE(b.revenue,0) as previous_day_rev
FROM DailyRevenue a
LEFT JOIN DailyRevenue b on a.day=b.day-1
the query assume that each day has one record in the table. If there could be more than 1 row for each day you need to create a view that sums up all days grouping by day.
If you're okay with re-ordering the columns slightly, something like this is pretty simple to understand:
SET #prev := 0;
SELECT day, revenue - #prev AS diff, #prev := revenue AS revenue
FROM revenue ORDER BY day ASC;
The trick is that we calculate the difference to the previous first, then set the previous to the current and display it as the current in one step.
Note, this depends on the order being correct since the calculations are done during the returning of the rows, so you need to make sure you have an ORDER BY clause that returns the days in the correct order.
Try;
select
t.date_col, t.val_col,
case when t1.val_col is null then 0
else t.val_col - t1.val_col end diff
from (
select t.* , #r := #r + 1 lev
from tbl t,
(select #r := 0) r
order by t.date_col
) t
left join (
select t.* , #r1 := #r1 + 1 lev
from tbl t,
(select #r1 := 1) r
order by t.date_col
) t1
on t.lev = t1.lev
This will calculate value diff even if there is a missing date
There is a table with the columns :
USE 'table';
insert into person values
('11','xxx','1976-05-10','p1'),
('11','xxx ','1976-06-11','p1'),
('11','xxx ','1976-07-21','p2'),
('11','xxx ','1976-08-31','p2'),
Can anyone suggest me a query to get the start and the end date of the person with respect to the place he changed chronologically.
The query I wrote
SELECT PId,Name,min(Start_Date) as sdt, max(Start_Date) as edt, place
from **
group by Place;
only gives me the first two rows of my answer. Can anyone suggest the query??
This isn't pretty, and performance might be horrible, but at least it works:
select min(sdt), edt, place
from (
select A.Start_Date sdt, max(B.Start_Date) edt, A.place
from person A
inner join person B on A.place = B.place
and A.Start_Date <= B.Start_Date
left join person C on A.place != C.place
and A.Start_Date < C.Start_Date
and C.Start_Date < B.Start_Date
where C.place is null
group by A.Start_Date, A.place
) X
group by edt, place
The idea is that A and B represent all pairs of rows. C will be any row in between these two which has a different place. So after the C.place is null restriction, we know that A and B belong to the same range, i.e. a group of rows for one place with no other place in between them in chronological order. From all these pairs, we want to identify those with maximal range, those which encompass all others. We do so using two nested group by queries. The inner one will choose the maximal end date for every possible start date, whereas the outer one will choose the minimal start date for every possible end date. The result are maximal ranges of chronologically subsequent rows describing the same place.
This can be achived by:
SELECT Id, PId,
MIN(Start_Date) AS sdt,
MAX(Start_Date) as edt,
IF(`place` <> #var_place_prev, (#var_rank:= #var_rank + 1), #var_rank) AS rank,
(#var_place_prev := `place`) AS `place`
FROM person, (SELECT #var_rank := 0, #var_place_prev := "") dummy
GROUP BY rank, Place;
Example: SQLFiddle
If you want records to be ordered by ID then:
SELECT Id, PId,
MIN(Start_Date) AS sdt,
MAX(Start_Date) as edt,
`place`
FROM(
SELECT Id, PId,
Start_Date
IF(`place` <> #var_place_prev,(#var_rank:= #var_rank + 1),#var_rank) AS rank,
(#var_place_prev := `place`) AS `place`
FROM person, (SELECT #var_rank := 0, #var_place_prev := "") dummy
ORDER BY ID ASC
) a
GROUP BY rank, Place;
I have a MySQL table with the structure:
beverages_log(id, users_id, beverages_id, timestamp)
I'm trying to compute the maximum streak of consecutive days during which a user (with id 1) logs a beverage (with id 1) at least 5 times each day. I'm pretty sure that this can be done using views as follows:
CREATE or REPLACE VIEW daycounts AS
SELECT count(*) AS n, DATE(timestamp) AS d FROM beverages_log
WHERE users_id = '1' AND beverages_id = 1 GROUP BY d;
CREATE or REPLACE VIEW t AS SELECT * FROM daycounts WHERE n >= 5;
SELECT MAX(streak) AS current FROM ( SELECT DATEDIFF(MIN(c.d), a.d)+1 AS streak
FROM t AS a LEFT JOIN t AS b ON a.d = ADDDATE(b.d,1)
LEFT JOIN t AS c ON a.d <= c.d
LEFT JOIN t AS d ON c.d = ADDDATE(d.d,-1)
WHERE b.d IS NULL AND c.d IS NOT NULL AND d.d IS NULL GROUP BY a.d) allstreaks;
However, repeatedly creating views for different users every time I run this check seems pretty inefficient. Is there a way in MySQL to perform this computation in a single query, without creating views or repeatedly calling the same subqueries a bunch of times?
This solution seems to perform quite well as long as there is a composite index on users_id and beverages_id -
SELECT *
FROM (
SELECT t.*, IF(#prev + INTERVAL 1 DAY = t.d, #c := #c + 1, #c := 1) AS streak, #prev := t.d
FROM (
SELECT DATE(timestamp) AS d, COUNT(*) AS n
FROM beverages_log
WHERE users_id = 1
AND beverages_id = 1
GROUP BY DATE(timestamp)
HAVING COUNT(*) >= 5
) AS t
INNER JOIN (SELECT #prev := NULL, #c := 1) AS vars
) AS t
ORDER BY streak DESC LIMIT 1;
Why not include user_id in they daycounts view and group by user_id and date.
Also include user_id in view t.
Then when you are queering against t add the user_id to the where clause.
Then you don't have to recreate your views for every single user you just need to remember to include in your where clause.
That's a little tricky. I'd start with a view to summarize events by day:
CREATE VIEW BView AS
SELECT UserID, BevID, CAST(EventDateTime AS DATE) AS EventDate, COUNT(*) AS NumEvents
FROM beverages_log
GROUP BY UserID, BevID, CAST(EventDateTime AS DATE)
I'd then use a Dates table (just a table with one row per day; very handy to have) to examine all possible date ranges and throw out any with a gap. This will probably be slow as hell, but it's a start:
SELECT
UserID, BevID, MAX(StreakLength) AS StreakLength
FROM
(
SELECT
B1.UserID, B1.BevID, B1.EventDate AS StreakStart, DATEDIFF(DD, StartDate.Date, EndDate.Date) AS StreakLength
FROM
BView AS B1
INNER JOIN Dates AS StartDate ON B1.EventDate = StartDate.Date
INNER JOIN Dates AS EndDate ON EndDate.Date > StartDate.Date
WHERE
B1.NumEvents >= 5
-- Exclude this potential streak if there's a day with no activity
AND NOT EXISTS (SELECT * FROM Dates AS MissedDay WHERE MissedDay.Date > StartDate.Date AND MissedDay.Date <= EndDate.Date AND NOT EXISTS (SELECT * FROM BView AS B2 WHERE B1.UserID = B2.UserID AND B1.BevID = B2.BevID AND MissedDay.Date = B2.EventDate))
-- Exclude this potential streak if there's a day with less than five events
AND NOT EXISTS (SELECT * FROM BView AS B2 WHERE B1.UserID = B2.UserID AND B1.BevID = B2.BevID AND B2.EventDate > StartDate.Date AND B2.EventDate <= EndDate.Date AND B2.NumEvents < 5)
) AS X
GROUP BY
UserID, BevID