I have a table as shown here:
USER
ROI
DATE
1
5
2021-11-24
1
4
2021-11-26
1
6
2021-11-29
I want to get the ROI for the dates in between the other dates, expected result will be as below
From 2021-11-24 to 2021-11-30
USER
ROI
DATE
1
5
2021-11-24
1
5
2021-11-25
1
4
2021-11-26
1
4
2021-11-27
1
4
2021-11-28
1
6
2021-11-29
1
6
2021-11-30
You may use a calendar table approach here. Create a table containing all dates and then join with it. Sans an actual table, you may use an inline CTE:
WITH dates AS (
SELECT '2021-11-24' AS dt UNION ALL
SELECT '2021-11-25' UNION ALL
SELECT '2021-11-26' UNION ALL
SELECT '2021-11-27' UNION ALL
SELECT '2021-11-28' UNION ALL
SELECT '2021-11-29' UNION ALL
SELECT '2021-11-30'
),
cte AS (
SELECT USER, ROI, DATE, LEAD(DATE) OVER (ORDER BY DATE) AS NEXT_DATE
FROM yourTable
)
SELECT t.USER, t.ROI, d.dt
FROM dates d
INNER JOIN cte t
ON d.dt >= t.DATE AND (d.dt < t.NEXT_DATE OR t.NEXT_DATE IS NULL)
ORDER BY d.dt;
Related
Here is an example:
Id|price|Date
1|2|2022-05-21
1|3|2022-06-15
1|2.5|2022-06-19
Needs to look like this:
Id|Date|price
1|2022-05-21|2
1|2022-05-22|2
1|2022-05-23|2
...
1|2022-06-15|3
1|2022-06-16|3
1|2022-06-17|3
1|2022-06-18|3
1|2022-06-19|2.5
1|2022-06-20|2.5
...
Until today
1|2022-08-30|2.5
I tried using the lag(price) over (partition by id order by date)
But i can't get it right.
I'm not familiar with Azure, but it looks like you need to use a calendar table, or generate missing dates using a recursive CTE.
To get started with a recursive CTE, you can generate line numbers for each id (assuming multiple id values) in the source data ordered by date. These rows with row number equal to 1 (with the minimum date value for the corresponding id) will be used as the starting point for the recursion. Then you can use the DATEADD function to generate the row for the next day. To use the price values from the original data, you can use a subquery to get the price for this new date, and if there is no such value (no row for this date), use the previous price value from CTE (use the COALESCE function for this).
For SQL Server query can look like this
WITH cte AS (
SELECT
id,
date,
price
FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) AS rn
FROM tbl
) t
WHERE rn = 1
UNION ALL
SELECT
cte.id,
DATEADD(d, 1, cte.date),
COALESCE(
(SELECT tbl.price
FROM tbl
WHERE tbl.id = cte.id AND tbl.date = DATEADD(d, 1, cte.date)),
cte.price
)
FROM cte
WHERE DATEADD(d, 1, cte.date) <= GETDATE()
)
SELECT * FROM cte
ORDER BY id, date
OPTION (MAXRECURSION 0)
Note that I added OPTION (MAXRECURSION 0) to make the recursion run through all the steps, since the default value is 100, this is not enough to complete the recursion.
db<>fiddle here
The same approach for MySQL (you need MySQL of version 8.0 to use CTE)
WITH RECURSIVE cte AS (
SELECT
id,
date,
price
FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) AS rn
FROM tbl
) t
WHERE rn = 1
UNION ALL
SELECT
cte.id,
DATE_ADD(cte.date, interval 1 day),
COALESCE(
(SELECT tbl.price
FROM tbl
WHERE tbl.id = cte.id AND tbl.date = DATE_ADD(cte.date, interval 1 day)),
cte.price
)
FROM cte
WHERE DATE_ADD(cte.date, interval 1 day) <= NOW()
)
SELECT * FROM cte
ORDER BY id, date
db<>fiddle here
Both queries produces the same results, the only difference is the use of the engine's specific date functions.
For MySQL versions below 8.0, you can use a calendar table since you don't have CTE support and can't generate the required date range.
Assuming there is a column in the calendar table to store date values (let's call it date for simplicity) you can use the CROSS JOIN operator to generate date ranges for the id values in your table that will match existing dates. Then you can use a subquery to get the latest price value from the table which is stored for the corresponding date or before it.
So the query would be like this
SELECT
d.id,
d.date,
(SELECT
price
FROM tbl
WHERE tbl.id = d.id AND tbl.date <= d.date
ORDER BY tbl.date DESC
LIMIT 1
) price
FROM (
SELECT
t.id,
c.date
FROM calendar c
CROSS JOIN (SELECT DISTINCT id FROM tbl) t
WHERE c.date BETWEEN (
SELECT
MIN(date) min_date
FROM tbl
WHERE tbl.id = t.id
)
AND NOW()
) d
ORDER BY id, date
Using my pseudo-calendar table with date values ranging from 2022-05-20 to 2022-05-30 and source data in that range, like so
id
price
date
1
2
2022-05-21
1
3
2022-05-25
1
2.5
2022-05-28
2
10
2022-05-25
2
100
2022-05-30
the query produces following results
id
date
price
1
2022-05-21
2
1
2022-05-22
2
1
2022-05-23
2
1
2022-05-24
2
1
2022-05-25
3
1
2022-05-26
3
1
2022-05-27
3
1
2022-05-28
2.5
1
2022-05-29
2.5
1
2022-05-30
2.5
2
2022-05-25
10
2
2022-05-26
10
2
2022-05-27
10
2
2022-05-28
10
2
2022-05-29
10
2
2022-05-30
100
db<>fiddle here
I have a table that shows when a user signs up for a subscription and when their membership will expire. A user can purchase a new subscription even if their current one is in force.
userid|purchasedate|expirydate
1 |2019-01-01 |2019-02-01
2 |2019-01-02 |2019-02-02
3 |2019-01-03 |2019-02-03
3 |2019-01-04 |2019-03-03
I need a SQL query that will GROUP BY the date and return the number of active subscriptions on that date. So it would return:
date |count
2019-01-01|1
2019-01-02|2
2019-01-03|3
2019-01-04|3
Below is for BigQuery Standard SQL
#standardSQL
SELECT day, COUNT(DISTINCT userid) active_subscriptions
FROM (SELECT AS STRUCT MIN(purchasedate) min_date, MAX(expirydate) max_date FROM `project.dataset.table`),
UNNEST(GENERATE_DATE_ARRAY(min_date, max_date)) day
JOIN `project.dataset.table`
ON day BETWEEN purchasedate AND expirydate
GROUP BY day
You can test, play with above using dummy data from your question as in below example
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2019-01-01' purchasedate, DATE '2019-02-01' expirydate UNION ALL
SELECT 2, '2019-01-02', '2019-02-02' UNION ALL
SELECT 3, '2019-01-03', '2019-02-03' UNION ALL
SELECT 3, '2019-01-04', '2019-03-03'
)
SELECT day, COUNT(DISTINCT userid) active_subscriptions
FROM (SELECT AS STRUCT MIN(purchasedate) min_date, MAX(expirydate) max_date FROM `project.dataset.table`),
UNNEST(GENERATE_DATE_ARRAY(min_date, max_date)) day
JOIN `project.dataset.table`
ON day BETWEEN purchasedate AND expirydate
GROUP BY day
with below output
Row day active_subscriptions
1 2019-01-01 1
2 2019-01-02 2
3 2019-01-03 3
4 2019-01-04 3
5 2019-01-05 3
6 2019-01-06 3
... ... ...
... ... ...
31 2019-01-31 3
32 2019-02-01 3
33 2019-02-02 2
34 2019-02-03 1
35 2019-02-04 1
... ... ...
... ... ...
61 2019-03-02 1
62 2019-03-03 1
You need a list of dates and count(distinct):
select d.dte, count(distinct t.userid) as num_users
from (select distinct purchase_date as dte from t) d left join
t
on d.dte >= t.dte and
d.dte <= t.expiry_date
group by d.dte
order by d.dte;
EDIT:
BigQuery can be fickle about inequalities in the on clause. Here is another approach:
select dte, count(distinct t.userid) as num_users
from t cross join
unnest(generate_date_array(t.purchase_date, t.expiry_date, interval 1 day)) dte
group by dte
order by dte;
You can use a where clause to filter down to particular dates.
I make the table name 'test_expirydate' and use your data
and this one work
select
tb1.expirydate,
count(*) as total
from test_expirydate as tb1
left join (
select
expirydate
from test_expirydate as tb2
group by userid
) as tb2
on tb1.expirydate >= tb2.expirydate
group by tb1.expirydate
I don't sure is it work in other case or not but it fine with current data
Oh, I interpret that the left column should be the expiration date.
I need to create a table that contains records with 1) all 365 days of the year and 2) a counter representing which business day of the month the day is. Non-business days should be represented with a 0. For example:
Date | Business Day
2019-10-01 1
2019-10-02 2
2019-10-03 3
2019-10-04 4
2019-10-05 0 // Saturday
2019-10-06 0 // Sunday
2019-10-07 5
....
2019-11-01 1
2019-11-02 0 // Saturday
2019-11-03 0 // Sunday
2019-11-04 2
So far, I've been able to create a table that contains all dates of the year.
CREATE TABLE ${TMPID}_days_of_the_year
(
`theDate` STRING
);
INSERT OVERWRITE TABLE ${TMPID}_days_of_the_year
select
dt_set.theDate
from
(
-- last 0~99 months
select date_sub('2019-12-31', a.s + 10*b.s + 100*c.s) as theDate
from
(
select 0 as s union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9
) a
cross join
(
select 0 as s union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9
) b
cross join
(
select 0 as s union all select 1 union all select 2 union all select 3
) c
) dt_set
where dt_set.theDate between '2019-01-01' and '2019-12-31'
order by dt_set.theDate DESC;
And I also have a table that contains all of the weekend days and holidays (this data is loaded from a file, and the date format is YYYY-MM-DD)
CREATE TABLE ${TMPID}_company_holiday
(
`holidayDate` STRING
)
;
LOAD DATA LOCAL INPATH '${FILE}' INTO TABLE ${TMPID}_company_holiday;
My question is.... how do I join these tables together while creating the business day counter column shown as in the sample data above?
You can use row_number() for the enumeration. This is a little tricky, because it needs to be conditional, but the information you need is provided by a left join:
select dy.*,
(case when ch.holiday_date is null
then row_number() over (partition by trunc(dy.date, 'MONTH'), ch.holiday_date
order by dy.date
)
else 0
end) as business_day
from days_of_the_year dy left join
company_holiday ch
on dy.date = ch.holiday_date;
I have records of No. of calls coming to a call center. When a call comes into a call center a ticket is open.
So, let's say ticket 1 (T1) is open on 8/1/19 and it stays open till 8/5/19. So, if a person ran a query everyday then on 8/1 it will show 1 ticket open...same think on day 2 till day 5....I want to get records by day to see how many tickets were open for each day.....
In short, Frequency Distribution by Day.
Ticket Open_date Close_date
T1 8/1/2019 8/5/2019
T2 8/1/2019 8/6/2019
Result:
Result
Date # Tickets_Open
8/1/2019 2
8/2/2019 2
8/3/2019 2
8/4/2019 2
8/5/2019 2
8/6/2019 1
8/7/2019 0
8/8/2019 0
8/9/2019 0
8/10/2019 0
We can handle your requirement via the use of a calendar table, which stores all dates covering the full range in your data set.
WITH dates AS (
SELECT '2019-08-01' AS dt UNION ALL
SELECT '2019-08-02' UNION ALL
SELECT '2019-08-03' UNION ALL
SELECT '2019-08-04' UNION ALL
SELECT '2019-08-05' UNION ALL
SELECT '2019-08-06' UNION ALL
SELECT '2019-08-07' UNION ALL
SELECT '2019-08-08' UNION ALL
SELECT '2019-08-09' UNION ALL
SELECT '2019-08-10'
)
SELECT
d.dt,
COUNT(t.Open_date) AS num_tickets_open
FROM dates d
LEFT JOIN tickets t
ON d.dt BETWEEN t.Open_date AND t.Close_date
GROUP BY
d.dt;
Note that in practice if you expect to have this reporting requirement in the long term, you might want to replace the dates CTE above with a bona-fide table of dates.
This solution generates the list of dates from the tickets table using CTE recursion and calculates the count:
WITH Tickets(Ticket, Open_date, Close_date) AS
(
SELECT "T1", "8/1/2019", "8/5/2019"
UNION ALL
SELECT "T2", "8/1/2019", "8/6/2019"
),
Ticket_dates(Ticket, Dates) as
(
SELECT t1.Ticket, CONVERT(DATETIME, t1.Open_date)
FROM Tickets t1
UNION ALL
SELECT t1.Ticket, DATEADD(dd, 1, CONVERT(DATETIME, t1.Dates))
FROM Ticket_dates t1
inner join Tickets t2 on t1.Ticket = t2.Ticket
where DATEADD(dd, 1, CONVERT(DATETIME, t1.Dates)) <= CONVERT(DATETIME, t2.Close_date)
)
SELECT CONVERT(varchar, Dates, 1), count(*)
FROM Ticket_dates
GROUP by Dates
ORDER by Dates
A "general purpose" trick is to generate a series of numbers, which can be done using CTE's but there are many alternatives, and from that create the needed range of dates. Once that exists then you can left join your ticket data to this and then count by date.
CREATE TABLE mytable(
Ticket VARCHAR(8) NOT NULL PRIMARY KEY
,Open_date DATE NOT NULL
,Close_date DATE NOT NULL
);
INSERT INTO mytable(Ticket,Open_date,Close_date) VALUES ('T1','8/1/2019','8/5/2019');
INSERT INTO mytable(Ticket,Open_date,Close_date) VALUES ('T2','8/1/2019','8/6/2019');
Also note I am using a cross apply in this example to "attach" the min and max dates of your tickets to each numbered row. You would need to include your own logic on what data to select here.
;WITH
cteDigits AS (
SELECT 0 AS digit UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL
SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9
)
, cteTally AS (
SELECT
[1s].digit
+ [10s].digit * 10
+ [100s].digit * 100 /* add more like this as needed */
AS num
FROM cteDigits [1s]
CROSS JOIN cteDigits [10s]
CROSS JOIN cteDigits [100s] /* add more like this as needed */
)
select
n.num + 1 rownum
, dateadd(day,n.num,ca.min_date) as on_date
, count(t.Ticket) as tickets_open
from cteTally n
cross apply (select min(Open_date), max(Close_date) from mytable) ca (min_date, max_date)
left join mytable t on dateadd(day,n.num,ca.min_date) between t.Open_date and t.Close_date
where dateadd(day,n.num,ca.min_date) <= ca.max_date
group by
n.num + 1
, dateadd(day,n.num,ca.min_date)
order by
rownum
;
result:
+--------+---------------------+--------------+
| rownum | on_date | tickets_open |
+--------+---------------------+--------------+
| 1 | 01.08.2019 00:00:00 | 2 |
| 2 | 02.08.2019 00:00:00 | 2 |
| 3 | 03.08.2019 00:00:00 | 2 |
| 4 | 04.08.2019 00:00:00 | 2 |
| 5 | 05.08.2019 00:00:00 | 2 |
| 6 | 06.08.2019 00:00:00 | 1 |
+--------+---------------------+--------------+
Say I've got a table with two columns (date and price). If I select over a range of dates, then is there a way to count the number of price changes over time?
For instance:
Date | Price
22-Oct-11 | 3.20
23-Oct-11 | 3.40
24-Oct-11 | 3.40
25-Oct-11 | 3.50
26-Oct-11 | 3.40
27-Oct-11 | 3.20
28-Oct-11 | 3.20
In this case, I would like it to return a count of 4 price changes.
Thanks in advance.
You can use the analytic functions LEAD and LAG to access to prior and next row of a result set and then use that to see if there are changes.
SQL> ed
Wrote file afiedt.buf
1 with t as (
2 select date '2011-10-22' dt, 3.2 price from dual union all
3 select date '2011-10-23', 3.4 from dual union all
4 select date '2011-10-24', 3.4 from dual union all
5 select date '2011-10-25', 3.5 from dual union all
6 select date '2011-10-26', 3.4 from dual union all
7 select date '2011-10-27', 3.2 from dual union all
8 select date '2011-10-28', 3.2 from dual
9 )
10 select sum(is_change)
11 from (
12 select dt,
13 price,
14 lag(price) over (order by dt) prior_price,
15 (case when lag(price) over (order by dt) != price
16 then 1
17 else 0
18 end) is_change
19* from t)
SQL> /
SUM(IS_CHANGE)
--------------
4
Try this
select count(*)
from
(select date,price from table where date between X and Y
group by date,price )
Depending on the Oracle version use either analytical functions (see answer from Justin Cave) or this
SELECT
SUM (CASE WHEN PREVPRICE != PRICE THEN 1 ELSE 0 END) CNTCHANGES
FROM
(
SELECT
C.DATE,
C.PRICE,
MAX ( D.PRICE ) PREVPRICE
FROM
(
SELECT
A.Date,
A.Price,
(SELECT MAX (B.DATE) FROM MyTable B WHERE B.DATE < A.DATE) PrevDate
FROM MyTable A
WHERE A.DATE BETWEEN YourStartDate AND YourEndDate
) C
INNER JOIN MyTable D ON D.DATE = C.PREVDATE
GROUP BY C.DATE, C.PRICE
)