MySQL version 8.0
I want to calculate time difference between two datetime column.
And get rows where duration >= 12:00:00.
which I would normally do:
select id
, start_time
, end_time
, timediff(end_time, start_time) as duration
from table;
which I would get something like this:
id start_time end_time duration
0 1 2020-06-01 01:00:00 2020-06-01 14:00:00 13:00:00
1 2 2020-06-01 01:00:00 2020-06-01 18:00:00 17:00:00
2 3 2020-06-01 19:00:00 2020-06-02 10:00:00 15:00:00
3 4 2020-06-02 04:00:00 2020-06-02 16:00:00 12:00:00
For duration column I don't want times between 00:00:00 ~ 04:00:00 to be added towards the duration. So for the first row duration = 10:00:00 since 01:00:00~14:00:00 = 10:00:00, ignoring times between 00:00:00 ~ 04:00:00
same for second row we substract 3 hours from duration.
so my desired output would be:
id start_time end_time duration
0 1 2020-06-01 01:00:00 2020-06-01 14:00:00 10:00:00
1 2 2020-06-01 01:00:00 2020-06-01 18:00:00 14:00:00
2 3 2020-06-01 19:00:00 2020-06-02 10:00:00 11:00:00
3 4 2020-06-02 04:00:00 2020-06-02 16:00:00 12:00:00
There are lots of rows where times include minutes and seconds too.
Thanks in advance!
I've grabbed all rows where duration >= 12:00:00.
Then separated data into 4 regions depending on their start_time.
a_region = 00~04
b_region = 04~12
c_region = 12~16
d_region = 16~24
For a_region I've subtracted 04:00:00 - start_time which is time we should compensate to duration in a_region.
compensation = 04:00:00 - start_time
compensated_time = duration - compensation.
For b_region it needs no compensation if it has passed 00~04 it means it already passed duration = 12:00:00.
For c_region,
compensation = 16:00:00 - start_time
compensated_time = duration - compensation
For d_region since we've grabbed duration >= 12:00:00
it will pass all of 00~04 therefore
compensated_time = duration - 04:00:00.
I solved it using Python but above is the logic I've used.
One option uses greatest():
select id
, start_time
, end_time
, timediff(
greatest(,
end_time,
date_format(end_time, '%Y-%m-%d 04:00:00')
),
greatest(
start_time,
date_format(start_time, '%Y-%m-%d 04:00:00')
)
) as duration
from table;
Related
MariaDB version 10.4.10.
I have a stock scraper script that fetches stock data every hour and inserts it into a MySQL database. The database structure is similar to this:
stocks( time_fetched DATETIME, fetch_id INT, name VARCHAR, price INT )
And some example data:
**time_fetched fetch_id name price**
2020-03-10 09:00:00 1 stock1 10
2020-03-10 09:00:00 1 stock2 15
2020-03-10 10:00:00 2 stock1 12
2020-03-10 10:00:00 2 stock2 20
2020-03-10 11:00:00 3 stock1 8
2020-03-10 11:00:00 3 stock2 18
I want a way to get price change for each stock between, say, stocks fetched at 09:00 and 10:00, or between 09:00 and 11:00. Something like this (pseudo code):
SELECT *, DIFF( current_price, price_one_hour_ago) AS change_1h, DIFF( current_price, price_two_hours_ago) AS change_2h
Is it possible to do this directly in MySQL? I am using a PHP script to display the data, so I might have to resort to do it in PHP instead.
You can use a CTE to generate prices delayed by 1 or 2 hours, and then compute the changes using those values (potentially using COALESCE to make NULL values into 0):
WITH prices AS (
SELECT time_fetched, name, price,
LAG(price, 1) OVER(PARTITION BY name ORDER BY time_fetched) AS price_1h,
LAG(price, 2) OVER(PARTITION BY name ORDER BY time_fetched) AS price_2h
FROM stocks
)
SELECT time_fetched, name, price,
COALESCE(price - price_1h, 0) AS change_1h,
COALESCE(price - price_2h, 0) AS change_2h
FROM prices
Output:
time_fetched name price change_1h change_2h
2020-03-10 09:00:00 stock1 10 0 0
2020-03-10 10:00:00 stock1 12 2 0
2020-03-10 11:00:00 stock1 8 -4 -2
2020-03-10 09:00:00 stock2 15 0 0
2020-03-10 10:00:00 stock2 20 5 0
2020-03-10 11:00:00 stock2 18 -2 3
Demo on dbfiddle also showing query results without COALESCE.
My table have fields that represent starting and ending working period as datetime.
I need to find related entries that match a total of 14hours min over a sliding period of 24 hours.
I think window function will (maybe) save me, but MariadDB (i use) doesn't implement yet Range time intervals in window function.
here is some example data:
id starting_hour ending_hour
-- ------------------- -------------------
1 2018-09-02 06:00:00 2018-09-02 08:30:00
2 2018-09-03 08:30:00 2018-09-03 10:00:00
4 2018-09-03 11:00:00 2018-09-03 15:00:00
5 2018-09-02 15:30:00 2018-09-02 16:00:00
6 2018-09-02 16:15:00 2018-09-02 17:00:00
7 2018-09-20 00:00:00 2018-09-20 08:00:00
8 2018-09-19 10:00:00 2018-09-19 12:00:00
9 2018-09-19 12:00:00 2018-09-19 16:00:00
10 2018-10-08 12:00:00 2018-10-08 14:00:00
11 2018-10-29 09:00:00 2018-10-29 10:00:00
So how to find rows where in a 24 hours window their sum a more or equal to 14 hours.
thanks
Edit:
SELECT
id,
starting_hour,
ending_hour,
TIMEDIFF (ending_hour, starting_hour) AS duree,
(
SELECT SUM(TIMEDIFF(LEAST(ending_hour, DATE_ADD(a.starting_hour, INTERVAL 24 HOUR)), starting_hour)) / 10000
FROM `table` b
WHERE b.starting_hour BETWEEN a.starting_hour AND DATE_ADD(a.starting_hour, INTERVAL 24 HOUR)
) AS duration
FROM
`table` a
HAVING duration >= 14
ORDER BY starting_hour ASC
;
This returns Id 8 but i want the whole period. (eg: Id 8, Id 9 and Id 7)
EDIT2:
The expected results are ranges of working time where they are in a window of 24 hours and where their sum are more or equal to 14 hours.
EDIT 3:
In fact under MySQL 8 this seems to work.
SELECT * FROM (
SELECT
*,
SEC_TO_TIME(SUM(TIME_TO_SEC(TIMEDIFF(hs.`ending_hour`, hs.`starting_hour`))) OVER (ORDER BY hs.starting_hour RANGE BETWEEN INTERVAL '12' HOUR PRECEDING AND INTERVAL '12' HOUR following)) AS tot
FROM
table hs
WHERE hs.`starting_hour` > DATE_SUB(NOW(), INTERVAL 50 DAY) AND hs.`ending_hour` <= NOW()
ORDER BY hs.`starting_hour` ASC
) t1
HAVING tot >= '14:00:00'
;
Is there a way to do it under MariaDB 10.2 without window function ? Or without window range function ?
I need Time Wise Sales Report in mysql,I get the Report like
Time Amount
19:00:00 2
20:00:00 5
21:00:00 0
22:00:00 0
23:00:00 14
but I need sum of Amount like
Time Amount
19:00:00 2
20:00:00 5
21:00:00 5
22:00:00 5
23:00:00 14
This is my Query
SELECT concat(MID(date_of_call, 12, 2),':00:00') TimeIntervel,COUNT(*) TotalEntries,date_format(date_of_call,'%d/%m/%Y') FROM ICC_MM_setup_suply
WHERE DATE_FORMAT(date_of_call,'%d/%m/%Y')='31/01/2017'
GROUP BY concat(MID(date_of_call, 12, 2),':00:00')
You can use User variables:
set #lastval := 0;
select
time,
#lastval := if(Amount = 0, #lastval, Amount) Amount
from t
order by time;
Suppose I have 5 records for a sales table.
ID Name datetime_col
1 ABC 2016-09-15 02:07:56
2 HSJ 2016-09-31 11:45:45
3 JSD 2016-11-26 07:09:56
4 JUH 2016-12-31 12:00:00
5 IGY 2017-01-13 14:00:07
I want to find how many records are there in sales table for each hour between 2016-09-15 AND 2017-01-13
Then result should be like
Hour sales_at_this_hour
2016-09-15 01:00:00 0
2016-09-15 02:00:00 1
2016-09-15 03:00:00 0
...
...
2017-01-13 01:00:00 0
2017-01-13 02:00:00 0
2017-01-13 03:00:00 0
....
2017-01-13 14:00:00 1
Then find the average of sales_at_this_hour using MySQL
EDIT: sorry not fully understand the question at first.
Use DATE_FORMAT
select
DATE_FORMAT(datetime_col, '%Y-%m-%d %h:00:00') as date,
count(id) as count
from table_name
group by date;
Get result with hours that has sales_at_this_hour > 1 (not exactly what you ask for)
datetime_col count
2016-02-04 05:00:00 5
2016-02-04 07:00:00 1
2016-02-04 08:00:00 5
2016-02-04 10:00:00 10
2016-02-04 11:00:00 1
Provide start_date and end_date, and then use DATEDIFF to calculate total time interval for the average calculation.
set #start_date = '2016-01-01', #end_date = '2017-01-01';
select
DATE_FORMAT(group_by_date.datetime, '%h:00:00') as hour,
AVG(group_by_date.count) / DATEDIFF(#end_date, #start_date) as average
from (
select
DATE_FORMAT(created_dtm, '%Y-%m-%d %h:00:00') as datetime,
count(id) as count
from table_name
where created_dtm > #start_date
and created_dtm < #end_date
group by datetime
) group_by_date
group by hour;
For each hour,
average sale count per day = total sale count / total days
hour average
01:00:00 0.03841209
02:00:00 0.01653005
03:00:00 0.0306716
04:00:00 0.01147541
05:00:00 0.01179831
Need help for MySQL query for retrieving record of class schedule that is conflict in the given time.
Ex.
SchedID StartTime EndTime
1 09:00:00 13:00:00
2 08:30:00 10:00:00
3 11:00:00 15:00:00
4 07:30:00 08:30:00
5 11:30:00 13:00:00
I would like to retrieve the list that is conflict in this given time
Start Time = 09:00:00
End Time = 11:00:00
The record will yield following result:
SchedID StartTime EndTime
1 09:00:00 13:00:00
2 08:30:00 10:00:00
3 11:00:00 15:00:00
Thank you.
Try this
SELECT DISTINCT a.schedid, a.starttime, a.endtime
FROM
tbl_name a, tbl_name b
WHERE ((b.starttime > a.starttime AND b.starttime < a.endtime)
OR (b.endtime > a.starttime AND b.endtime < a.endtime))
ORDER BY schedid
i just found my answer here Determine Whether Two Date Ranges Overlap
Heres my query:
SELECT * FROM tableName WHERE (start_time < givenEndTime)
AND (end_time > givenStartTime);
This works perfectly fine for me.