This is a follow-up to:
Dynamic minimum value for specfic range (mysql)
I do have the query to fetch the third column (lowest of the last 3 days) "Low_3_days" via SELECT command:
-----------------------------------------
| Date | Unit_ | Lowest_in_last_|
| | price | 3_days |
|----------------------------------------
| 2015-01-01 | 15 | 15 |
| 2015-01-02 | 17 | 15 |
| 2015-01-03 | 21 | 15 |
| 2015-01-04 | 18 | 17 |
| 2015-01-05 | 12 | 12 |
| 2015-01-06 | 14 | 12 |
| 2015-01-07 | 16 | 12 |
|----------------------------------------
select S.Date,Unit_price,
(select S.Date, Unit_price,
(SELECT min(s2.Unit_Price)
FROM table s2
WHERE s2.DATE BETWEEN s.DATE - interval 3 day and
s.DATE - interval 1 day
) as min_price_3_days
FROM table S;
My new challenge is - what is the best way to use UPDATE-SET-WHERE so I could add the ("Lowest_in_last_3_days") values to a new column in a table (instead of having temporary results displayed to me via SELECT).
By following the UPDATE-SET-WHERE syntax, the query would be:
UPDATE table
SET min_price_3_days =
(select S.Date, Unit_price,
(SELECT min(s2.Unit_Price)
FROM table s2
WHERE s2.DATE BETWEEN s.DATE - interval 3 day and
s.DATE - interval 1 day
) as min_price_3_days
but I have difficulties constructing the correct query.
What would be the correct approach to this? I do recognize this one is a tough one to solve.
Your UPDATE should look like:
update table set low_3_days=
(SELECT min(Unit_Price)
FROM (select unit_price, date as date2 from table) as s2
WHERE s2.date2 BETWEEN date - interval 3 day and date - interval 1 day
);
You can check it in SQLFiddle
In Fiddle I used different names for table and column. I prefer not to use SQL keywords as names
Related
I have a table full of monthly contracts. There is a monthly price, a start date, and an end date for each. I am trying to graph each month's total revenue and am wondering if it's possible to do this in one query (vs. a query for each month).
I know how to group by month and year in mysql, but this requires a more complex solution that "understands" whether to include in the sum for a given month/year based on the start and end date of the contract.
Shorthand example
| contract_id | price | start_date | end_date |
| 1 | 299 | 1546318800 (1/1/19) | 1554004800 (3/31/19) |
| 2 | 799 | 1551416400 (3/1/19) | 1559275200 (5/31/19) |
With this example, there's an overlap in March. Both contracts are running in March, so the sum returned for that month should be 1098.
I'd like to be able to produce a report that includes every month between two dates, so in this case I'd send 1/1/19 - 12/31/19, the full year of 2019 and would hope to see 0 results as well.
| month | year | price_sum |
| 1 | 2019 | 299 |
| 2 | 2019 | 299 |
| 3 | 2019 | 1098 |
| 4 | 2019 | 799 |
| 5 | 2019 | 799 |
| 6 | 2019 | 0 |
| 7 | 2019 | 0 |
| 8 | 2019 | 0 |
| 9 | 2019 | 0 |
| 10 | 2019 | 0 |
| 11 | 2019 | 0 |
| 12 | 2019 | 0 |
Here is a full working script for your problem, which uses a calendar table approach to represent every month in 2019. Specifically, we represent each month using the first of that month. Then, a given price from your table is applicable to that month if there is overlap with the start and end range.
WITH yourTable AS (
SELECT 1 AS contract_id, 299 AS price, '2019-01-01' AS start_date, '2019-03-31' AS end_date UNION ALL
SELECT 2, 799, '2019-03-01', '2019-05-31'
),
dates AS (
SELECT '2019-01-01' AS dt UNION ALL
SELECT '2019-02-01' UNION ALL
SELECT '2019-03-01' UNION ALL
SELECT '2019-04-01' UNION ALL
SELECT '2019-05-01' UNION ALL
SELECT '2019-06-01' UNION ALL
SELECT '2019-07-01' UNION ALL
SELECT '2019-08-01' UNION ALL
SELECT '2019-09-01' UNION ALL
SELECT '2019-10-01' UNION ALL
SELECT '2019-11-01' UNION ALL
SELECT '2019-12-01'
)
SELECT
d.dt,
SUM(t.price) AS price_sum
FROM dates d
LEFT JOIN yourTable t
ON d.dt < t.end_date
AND DATE_ADD(d.dt, INTERVAL 1 MONTH) > t.start_date
GROUP BY
d.dt;
Demo
Notes:
If your dates are actually stored as UNIX timestamps, then just call FROM_UNIXTIME(your_date) to convert them to dates, and use the same approach I gave above.
I had to use the overlapping date range formula here, because the criteria for overlap in a given month is that the range of that month intersects the range given by a start and end date. Have a look at this SO question for more information on that.
My code is for MySQL 8+, though in practice you may wish to create a bona fide calendar table (the CTE version of which I called dates above), which contains the range of months/years which you want to cover your data set.
I understand that you will be given a range of dates for which you will need to report. My solution requires you to initialize a temporary table, such as date_table with the first day of each month for which you want to report on:
create temporary table date_table (
d date,
primary key(d)
);
set #start_date = '2019-01-01';
set #end_date = '2019-12-01';
set #months = -1;
insert into date_table(d)
select DATE_FORMAT(date_range,'%Y-%c-%d') AS result_date from (
select (date_add(#start_date, INTERVAL (#months := #months +1 ) month)) as date_range
from mysql.help_topic a limit 0,1000) a
where a.date_range between #start_date and last_day(#end_date);
Then this should do it:
select month(dt.d) as month, year(dt.d) as year, ifnull(sum(c.price), 0) as price_sum
from date_table dt left join contract c on
dt.d >= date(from_unixtime(c.start_date)) and dt.d < date(from_unixtime(c.end_date))
group by dt.d
order by dt.d
;
Resulting in:
+-------+------+-----------+
| month | year | price_sum |
+-------+------+-----------+
| 1 | 2019 | 299 |
| 2 | 2019 | 299 |
| 3 | 2019 | 1098 |
| 4 | 2019 | 799 |
| 5 | 2019 | 799 |
| 6 | 2019 | 0 |
| 7 | 2019 | 0 |
| 8 | 2019 | 0 |
| 9 | 2019 | 0 |
| 10 | 2019 | 0 |
| 11 | 2019 | 0 |
| 12 | 2019 | 0 |
+-------+------+-----------+
See demo
I am not sure about the semantics of the column end_date. Right now I am comparing the first a follows: start_date <= first_of_month < end_date. Perhaps the test should be start_date <= first_of_month <= end_date, in which case:
dt.d >= date(from_unixtime(c.start_date)) and dt.d < date(from_unixtime(c.end_date))
becomes:
dt.d between date(from_unixtime(c.start_date)) and date(from_unixtime(c.end_date))
With end_date being the last day of the month, it would not matter either way.
I have a table with 'ON' and 'OFF' values in column activity and another column datetime.
id(AUTOINCREMENT) id_device activity datetime
1 a ON 2017-05-26 22:00:00
2 b ON 2017-05-26 05:00:00
3 a OFF 2017-05-27 04:00:00
4 b OFF 2017-05-26 08:00:00
5 a ON 2017-05-28 12:00:00
6 a OFF 2017-05-28 15:00:00
I need to get total ON time by day
day id_device total_minutes_on
2017-05-26 a 120
2017-05-26 b 180
2017-05-27 a 240
2017-05-27 b 0
2017-05-28 a 180
2017-05-28 b 0
i have searched and tried answers for another posts, i tried TimeDifference and i get correct total time.
I don't find the way to get total time grouped by date
i appreciate your help
I'm not posting this as a definite answer rather it's an experiment for me and hopefully you'll find is useful in your case. Also I would like to mention that the MySQL database version I'm working with is quite old so the method I'm using is also very manual to say the least.
First of all lets extract your expected output:
The date value in day need to be repeated twice fro each of id_device a and b.
Minutes are calculated based on the activity; if activity is 'ON' until tomorrow, it needs to be calculated until the day end at 24:00:00 while the next day will calculate minutes until the activity is OFF.
What I come up with is this:
Creating condition (1):
SELECT * FROM
(SELECT DATE(datetime) dtt FROM mytable GROUP BY DATE(datetime)) a,
(SELECT id_device FROM mytable GROUP BY id_device) b
ORDER BY dtt,id_device;
The query above will return the following result:
+------------+-----------+
| dtt | id_device |
+------------+-----------+
| 2017-05-26 | a |
| 2017-05-26 | b |
| 2017-05-27 | a |
| 2017-05-27 | b |
| 2017-05-28 | a |
| 2017-05-28 | b |
+------------+-----------+
*Above will only work with all the dates you have in the table. If you want all date regardless if there's activity or not, I suggest you create a calendar table (refer: Generating a series of dates).
So this become the base query. Then I've added an outer query to left join the query above with the original data table:
SELECT v.*,
GROUP_CONCAT(w.activity ORDER BY w.datetime SEPARATOR ' ') activity,
GROUP_CONCAT(TIME_TO_SEC(TIME(w.datetime)) ORDER BY w.datetime SEPARATOR ' ') tr
FROM
-- this was the first query
(SELECT * FROM
(SELECT DATE(datetime) dtt FROM mytable GROUP BY DATE(datetime)) a,
(SELECT id_device FROM mytable GROUP BY id_device) b
ORDER BY a.dtt,b.id_device) v
--
LEFT JOIN
mytable w
ON v.dtt=DATE(w.datetime) AND v.id_device=w.id_device
GROUP BY DATE(v.dtt),v.id_device
What's new in the query is the addition of GROUP_CONCAT operation on both activity and time value extracted from datetime column which is converted into seconds value. You notice that in both of the GROUP_CONCAT there's a similar ORDER BY condition which is important in order to get the exact corresponding value.
The query above will return the following result:
+------------+-----------+----------+-------------+
| dtt | id_device | activity | tr |
+------------+-----------+----------+-------------+
| 2017-05-26 | a | ON | 79200 |
| 2017-05-26 | b | ON OFF | 18000 28800 |
| 2017-05-27 | a | OFF | 14400 |
| 2017-05-27 | b | (NULL) | (NULL) |
| 2017-05-28 | a | ON OFF | 43200 54000 |
| 2017-05-28 | b | (NULL) | (NULL) |
+------------+-----------+----------+-------------+
From here, I've added another query outside to calculate how many minutes and attempt to get the expected result:
SELECT dtt,id_device,
CASE
WHEN SUBSTRING_INDEX(activity,' ',1)='ON' AND SUBSTRING_INDEX(activity,' ',-1)='OFF'
THEN (SUBSTRING_INDEX(tr,' ',-1)-SUBSTRING_INDEX(tr,' ',1))/60
WHEN activity='ON' THEN 1440-(tr/60)
WHEN activity='OFF' THEN tr/60
WHEN activity IS NULL AND tr IS NULL THEN 0
END AS 'total_minutes_on'
FROM
-- from the last query
(SELECT v.*,
GROUP_CONCAT(w.activity ORDER BY w.datetime SEPARATOR ' ') activity,
GROUP_CONCAT(TIME_TO_SEC(TIME(w.datetime)) ORDER BY w.datetime SEPARATOR ' ') tr
FROM
-- this was the first query
(SELECT * FROM
(SELECT DATE(datetime) dtt FROM mytable GROUP BY DATE(datetime)) a,
(SELECT id_device FROM mytable GROUP BY id_device) b
ORDER BY a.dtt,b.id_device) v
--
LEFT JOIN
mytable w
ON v.dtt=DATE(w.datetime) AND v.id_device=w.id_device
GROUP BY DATE(v.dtt),v.id_device
--
) z
The last part I do is if the activity value have both ON and OFF on the same day then (OFF-ON)/60secs=total minutes. If activity value is only ON then minutes value for '24:00:00' > 24 hr*60 min= 1440-(ON/60secs)= total minutes, and if activity only OFF, I just convert seconds to minutes because the day starts at 00:00:00 anyhow.
+------------+-----------+------------------+
| dtt | id_device | total_minutes_on |
+------------+-----------+------------------+
| 2017-05-26 | a | 120 |
| 2017-05-26 | b | 180 |
| 2017-05-27 | a | 240 |
| 2017-05-27 | b | 0 |
| 2017-05-28 | a | 180 |
| 2017-05-28 | b | 0 |
+------------+-----------+------------------+
Hopefully this will give you some ideas. ;)
I know it's possible to get yesterday records, most common way using SUBDATE(CURDATE(), 1) or maybe simply use CURDATE() - 1 and use LIMIT and ORDER to retrieve the last record of yesterday.
But here, I need to get the last record of yesterday in the first row and the rest will be all records of today. I need to run this within single query.
For example, I have following records in one of my table:
--------------------------------------------------
| value | created_at |
--------------------------------------------------
| 70 | 1/1/2017 |
| 300 | 1/1/2017 |
| 100 | 1/1/2017 |
| 235 | 1/2/2017 |
| 45 | 1/2/2017 |
--------------------------------------------------
The created_at column is a timestamp, if today is 1/2/2017 (2th January 2017) then the result of the query should be:
--------------------------------------------------
| value | created_at |
--------------------------------------------------
| 100 | 1/1/2017 |
| 235 | 1/2/2017 |
| 45 | 1/2/2017 |
--------------------------------------------------
So far, I only able to retrieve the records of today with following query:
SELECT * FROM my_table WHERE created_at >= CURDATE();
What query I need to accomplish this?
Hoping you have id as primary key
select * from
(select
*
from
tbl
where date(created_at) =date(DATE_ADD(now(), INTERVAL -1 DAY))
order by id desc limit 0,1
)tmp
UNION
select * from tbl where date(created_at)=date(now())
I am looking for a solution to SELECT (or otherwise derive) the values for Column C (minimum price for last 3 days only, not for the whole column).
----------------------------------------
Date | Unit_ | Low_3_days |
| price | |
----------------------------------------
2015-01-01 | 15 | should be: 15 |
2015-01-02 | 17 | should be: 15 |
2015-01-03 | 21 | should be: 15 |
2015-01-04 | 18 | should be: 17 |
2015-01-05 | 12 | should be: 12 |
2015-01-06 | 14 | should be: 12 |
2015-01-07 | 16 | should be: 12 |
----------------------------------------
My thought revolves around the following, but yielding an error:
select S.Date,Unit_price,
(SELECT min(LOW_3_days)
FROM table
where S.DATE BETWEEN S.DATE-1
and S.DATE-3)
AS min_price_3_days
FROM table AS S
What is the correct query to get this to work? Database used MySQL.
You are pretty close. When working with correlated subqueries, always use table aliases to be absolutely clear about where the columns are coming from:
select S.Date, Unit_price,
(SELECT min(s2.Unit_Price)
FROM table s2
WHERE s2.DATE BETWEEN s.DATE - interval 3 day and
s.DATE - interval 1 day
) as min_price_3_days
FROM table S;
here's what im trying to achive:
i have 2 sql tables:
transactions and payplans
bellow is the structures of 2 tables:
transactions
uid | plan | date | payid | status
------------------------------------
12 | 3 | 1388534400 | 334 | 1
699 | 4 | 1388214400 | 335 | 1
payplans:
plan | plan_price
-------------------
3 | 9.99
4 | 19.99
with this query:
SELECT SUM(plan_price)
FROM transations AS t
INNER JOIN payplans AS p
ON t.plan = p.plan
WHERE t.status = '1'
i was able to calculate total sum of all "plan_price" rows,
but i would like to have the price sum for every month starting jan 2013
for example:
jan-13 | 9.99
feb-13 | 29.99
etc.
For MySQL
SELECT date_format(FROM_UNIXTIME(t.date), '%b-%y') as mnth,
SUM(plan_price)
FROM transations AS t
INNER JOIN payplans AS p
ON t.plan = p.plan
WHERE t.status = '1'
GROUP BY mnth;
SQLFiddle
You converting unix_timestamp to date using FROM_UNIXTIME
formatting it into 'MON-YY' format with DATE_FORMAT
then grouping by month.