Adding Missing Dates From SQL Query - sql

I am trying to add missing dates to my query so that my results look like this:
10/22/2018 15
10/21/2018 0
10/20/2018 14
Rather than this:
10/22/2018 15
10/20/2018 14
I want the past 300 days listed, even if the output value is 0.
Here is my query:
SELECT TOP (300)
CAST(createddate as DATE),
count(DISTINCT ID)
FROM table
GROUP BY CAST(createddate as DATE)
ORDER BY CAST(createddate as DATE) DESC

You can use a recursive CTE to generate the data:
WITH dates as (
SELECT MAX(CAST(createddate as date)) as dte, 1 as lev
FROM table
UNION ALL
SELECT DATEADD(day, -1, dte), lev + 1
FROM dates
WHERE lev < 300
)
SELECT COUNT(DISTINCT t.ID)
FROM dates d LEFT JOIN
table t
ON d.dte = CAST(t.createddate as DATE)
GROUP BY d.dte
ORDER BY d.dte DESC
OPTION (MAXRECURSION 0);

Related

SQL: How to create a daily view based on different time intervals using SQL logic?

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

MS-SQL how to add missing month in a table values

I have a table with the following entries,
ID
date
Frequency
1
'2012-04-30'
5
1
'2012-06-30'
4
1
'2012-07-31'
25
2
'2012-04-30'
7
2
'2012-05-31'
4
2
'2012-06-30'
1
2
'2012-07-31'
6
I need to add missing month and the date which gets added should be the last date of that month with frequency value as 0.
The expected output is
ID
date
Frequency
1
'2012-04-30'
5
1
'2012-05-31'
0
1
'2012-06-30'
4
1
'2012-07-31'
25
2
'2012-04-30'
7
2
'2012-05-31'
4
2
'2012-06-30'
1
2
'2012-07-31'
6
I need to add missing month and the date which gets added should be the last date of that
I would suggest recursive CTEs:
with cte as (
select id, date, frequency,
lead(date) over (partition by id order by date) as next_date
from t
union all
select id, eomonth(date, 1), 0, next_date
from cte
where eomonth(date, 1) < dateadd(day, -1, next_date)
)
select id, date, frequency
from cte
order by id, date;
The anchor part of the CTE calculates the end date for a given row. The recursive part then just keeps adding months to fill in the missing rows (and none if there are none). The use of eomonth(date, 1) is just a handy way of getting the last day of the next month.
Here is a db<>fiddle.
If you have all dates in the table, you can also use cross join to generate the rows and then left join to bring in the existing data:
select i.id, d.date, coalesce(t.frequency, 0) as frequency
from (select distinct id from t) i cross join
(select distinct date from t) d left join
t
on i.id = t.id and d.date = t.date
order by i.id, d.date;
If you have a large amount of data, you can compare performance. This may be a case where a recursive CTE is faster than alternative methods.

Get an interval of dates from a range of dates

I have two dates 21/10/2019 and 26/6/2031, and from these two I need a list of dates with three months interval starting from the first date like this:
22/10/2019 | 21/01/2020
22/01/2020 | 21/04/2020
22/04/2020 | 21/07/2020
22/07/2020 | 21/10/2020
...
22/01/2031 | 21/04/2031
22/04/2031 | 26/06/2031
I've tried using ROW_NUMBER() and DENSE_RANK() and LAG() to group a complete list of dates between the two dates, but I can't seem to figure it out. I think I might need to partition this somehow, but I can't get it right.
If you don't understand, please let me know. I'm pretty new at this :)
You can use a recursive query:
with cte (dt, end_dt) as (
select #start_dt, #end_dt
union all
select dateadd(month, 3, dt), end_dt from cte where dt < end_dt
)
select dt,
case when dateadd(month, 3, dt) < end_dt
then dateadd(day, -1, dateadd(month, 3, dt))
else end_dt
end as end_dt
from cte
order by dt;
If you need to generate more than 100 quarters, you need to add option (maxrecursion 0) at the very end of the query.
Demo on DB Fiddle
This could also be done using a 'tally' or numbers based approach. The upper limit of the tally_cte is 12^5=248,832 (more than recursion can produce and it could be increased as much as needed).
declare
#start_dt datetime='2019-10-21',
#end_dt datetime='2031-06-26'
;with
n(n) as (select * from (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12)) v(n)),
tally_cte(n) as (
select row_number() over (order by (select null))
from n n1 cross join n n2 cross join n n3 cross join n n4 cross join n n5)
select t.*, cast(dt.dt as date) start_dt, cast(dateadd(MONTH, 3, dt.dt) as date) end_dt
from tally_cte t
cross apply (select dateadd(month, (n-1)*3, #start_dt) dt) dt
where n<datediff(month, #start_dt, #end_dt)/3;
N start_dt end_dt
1 2019-10-21 2020-01-21
2 2020-01-21 2020-04-21
3 2020-04-21 2020-07-21
...
45 2030-10-21 2031-01-21

SQL query needed - Counting 365 days backwards

I have searched the forum many times but couldn't find a solution for my situation. I am working with an Oracle database.
I have a table with all Order Numbers and Customer Numbers by Day. It looks like this:
Day | Customer Nbr | Order Nbr
2018-01-05 | 25687459 | 256
2018-01-09 | 36478592 | 398
2018-03-07 | 25687459 | 1547
and so on....
Now I need a SQL Query which gives me a table by day and Customer Nbr and counts the number of unique Order Numbers within the last 365 days starting from column 1.
For the example above the resulting table should look like:
Day | Customer Nbr | Order Cnt
2019-01-01 | 25687459 | 2
2019-01-02 | 25687459 | 2
...
2019-03-01 | 25687459 | 1
One method is to generate values for all days of interest for each customer and then use a correlated subquery:
with dates as (
select date '2019-01-01' + rownum as dte from dual
connect by date '2019-01-01' + rownum < sysdate
)
select d.dte, t.customer_nbr,
(select count(*)
from t t2
where t2.customer_nbr = t.customer_nbr and
t2.day <= t.dte and
t2.date > t.dte - 365
) as order_cnt
from dates d cross join
(select distinct customer_nbr from t) ;
Edit:
I've just seen you clarify the question, which I've interpreted to mean:
For every day in the last year, show how many orders there were for each customer between that date, and 1 year previously. Working on an answer now...
Updated Answer:
For each customer, we count the number of records between the order day, and 365 days before it...
WITH yourTable AS
(
SELECT SYSDATE - 1 Day, 'Alex' CustomerNbr FROM DUAL
UNION ALL
SELECT SYSDATE - 2, 'Alex' FROM DUAL
UNION ALL
SELECT SYSDATE - 366, 'Alex'FROM DUAL
UNION ALL
SELECT SYSDATE - 400, 'Alex'FROM DUAL
UNION ALL
SELECT SYSDATE - 500, 'Alex'FROM DUAL
UNION ALL
SELECT SYSDATE - 1, 'Joe'FROM DUAL
UNION ALL
SELECT SYSDATE - 300, 'Chris'FROM DUAL
UNION ALL
SELECT SYSDATE - 1, 'Chris'FROM DUAL
)
SELECT Day, CustomerNbr, OrdersLast365Days
FROM yourTable t
OUTER APPLY
(
SELECT COUNT(1) OrdersLast365Days
FROM yourTable t2
WHERE t.CustomerNbr = t2.CustomerNbr
AND TRUNC(t2.Day) >= TRUNC(t.Day) - 364
AND TRUNC(t2.Day) <= TRUNC(t.Day)
)
ORDER BY t.Day DESC, t.CustomerNbr;
If you want to report on just the days you have orders for, then a simple WHERE clause should be enough:
SELECT Day, CustomerNbr, COUNT(1) OrderCount
FROM <yourTable>
WHERE TRUNC(DAY) >= TRUNC(SYSDATE -364)
GROUP BY Day, CustomerNbr
ORDER BY Day Desc;
If you want to report on every day, you'll need to generate them first. This can be done by a recursive CTE, which you then join to your table:
WITH last365Days AS
(
SELECT TRUNC (SYSDATE - ROWNUM + 1) dt
FROM DUAL CONNECT BY ROWNUM < 365
)
SELECT d.Day, COALESCE(t.CustomerNbr, 'None') CustomerNbr, SUM(CASE WHEN t.CustomerNbr IS NULL THEN 0 ELSE 1 END) OrderCount
FROM last365Days d
LEFT OUTER JOIN <yourTable> t
ON d.Day = TRUNC(t.Day)
GROUP BY d.Day, t.CustomerNbr
ORDER BY d.Day Desc;
I would probably have done it with and analytic function. In your windowing clause, you can specify a number of rows before, or a range. In this case I will use a range.
This will give you, For Each customer for each day the number of orders during one rolling year before the date displayed
WITH DATES AS (
SELECT * FROM
(SELECT TRUNC(SYSDATE)-(LEVEL-1) AS DAY FROM DUAL CONNECT BY TRUNC(SYSDATE)-(LEVEL-1) >= ( SELECT MIN(TRUNC(DAY)) FROM MY_TABLE ))
CROSS JOIN
(SELECT DISTINCT CUST_ID FROM MY_TABLE))
SELECT DISTINCT
DATES.DAY,
DATES.CUST_ID,
COUNT(ORDER_ID) OVER (PARTITION BY DATES.CUST_ID ORDER BY DATES.DAY RANGE BETWEEN INTERVAL '1' YEAR PRECEDING AND INTERVAL '1' SECOND PRECEDING)
FROM
DATES
LEFT JOIN
MY_TABLE
ON DATES.DAY=TRUNC(MY_TABLE.DAY) AND DATES.CUST_ID=MY_TABLE.CUST_ID
ORDER BY DATES.CUST_ID,DATES.DAY;

SQL Join two tables by unrelated date

I’m looking to join two tables that do not have a common data point, but common value (date). I want a table that lists the date and total number of hired/terminated employees on that day. Example is below:
Table 1
Hire Date Employee Number Employee Name
--------------------------------------------
5/5/2018 10078 Joe
5/5/2018 10077 Adam
5/5/2018 10078 Steve
5/8/2018 10079 Jane
5/8/2018 10080 Mary
Table 2
Termination Date Employee Number Employee Name
----------------------------------------------------
5/5/2018 10010 Tony
5/6/2018 10025 Jonathan
5/6/2018 10035 Mark
5/8/2018 10052 Chris
5/9/2018 10037 Sam
Desired result:
Date Total Hired Total Terminated
--------------------------------------
5/5/2018 3 1
5/6/2018 0 2
5/7/2018 0 0
5/8/2018 2 1
5/9/2018 0 1
Getting the total count is easy, just unsure as the best approach from the standpoint of "adding" a date column
If you need all dates within some window then you need to join the data to a calendar. You can then left join and sum flags for data points.
DECLARE #StartDate DATETIME = (SELECT MIN(ActionDate) FROM(SELECT ActionDate = MIN(HireDate) FROM Table1 UNION SELECT ActionDate = MIN(TerminationDate) FROM Table2)AS X)
DECLARE #EndDate DATETIME = (SELECT MAX(ActionDate) FROM(SELECT ActionDate = MAX(HireDate) FROM Table1 UNION SELECT ActionDate = MAX(TerminationDate) FROM Table2)AS X)
;WITH AllDates AS
(
SELECT CalendarDate=#StartDate
UNION ALL
SELECT DATEADD(DAY, 1, CalendarDate)
FROM AllDates
WHERE DATEADD(DAY, 1, CalendarDate) <= #EndDate
)
SELECT
CalendarDate,
TotalHired = SUM(CASE WHEN H.HireDate IS NULL THEN NULL ELSE 1 END),
TotalTerminated = SUM(CASE WHEN T.TerminationDate IS NULL THEN NULL ELSE 1 END)
FROM
AllDates D
LEFT OUTER JOIN Table1 H ON H.HireDate = D.CalendarDate
LEFT OUTER JOIN Table2 T ON T.TerminationDate = D.CalendarDate
/* If you only want dates with data points then uncomment out the where clause
WHERE
NOT (H.HireDate IS NULL AND T.TerminationDate IS NULL)
*/
GROUP BY
CalendarDate
I would do this with a union all and aggregations:
select dte, sum(is_hired) as num_hired, sum(is_termed) as num_termed
from (select hiredate as dte, 1 as is_hired, 0 as is_termed from table1
union all
select terminationdate, 0 as is_hired, 1 as is_termed from table2
) ht
group by dte
order by dte;
This does not include the "missing" dates. If you want those, a calendar or recursive CTE works. For instance:
with ht as (
select dte, sum(is_hired) as num_hired, sum(is_termed) as num_termed
from (select hiredate as dte, 1 as is_hired, 0 as is_termed from table1
union all
select terminationdate, 0 as is_hired, 1 as is_termed from table2
) ht
group by dte
),
d as (
select min(dte) as dte, max(dte) as max_dte)
from ht
union all
select dateadd(day, 1, dte), max_dte
from d
where dte < max_dte
)
select d.dte, coalesce(ht.num_hired, 0) as num_hired, coalesce(ht.num_termed) as num_termed
from d left join
ht
on d.dte = ht.dte
order by dte;
Try this one
SELECT ISNULL(a.THE_DATE, b.THE_DATE) as Date,
ISNULL(a.Total_Hire,0) as Total_Hire,
ISNULL (b.Total_Terminate,0) as Total_terminate
FROM (SELECT Hire_date as the_date, COUNT(1) as Total_Hire
FROM TABLE_HIRE GROUP BY HIRE_DATE) a
FULL OUTER JOIN (SELECT Termination_Date as the_date, COUNT(1) as Total_Terminate
FROM TABLE_TERMINATE GROUP BY HIRE_DATE) a
ON a.the_date = b.the_date