This is an example of some data
id CurrentMailPromoDate
1 1/1/2013
2 3/1/2013
3 6/9/2013
4 6/10/2013
5 9/18/2013
6 12/27/2013
7 12/27/2013
What I need is to extract only those id's where the date diff between the current and previous record >= 100
In other words the result set would be :
id CurrentMailPromoDate
1 1/1/2013 **initial record**
3 6/9/2013
5 9/18/2013
6 12/27/2013
7 12/27/2013
The challenge is getting the previous date. In SQL Server 2012, you can use lag(), but this is not supported in SQL Server 2008. Instead, I use a correlated subquery to get the previous date.
The following returns the set you are looking for:
with t as (
select 1 as id, CAST('2013-01-01' as DATE) as CurrentMailPromoDate union all
select 2, '2013-03-01' union all
select 3, '2013-06-09' union all
select 4, '2013-06-10' union all
select 5, '2013-09-18' union all
select 6, '2013-12-27' union all
select 7, '2013-12-27')
select t.id, t.CurrentMailPromoDate
from (select t.*,
(select top 1 CurrentMailPromoDate
from t t2
where t2.CurrentMailPromoDate < t.CurrentMailPromoDate
order by CurrentMailPromoDate desc
) as prevDate
from t
) t
where DATEDIFF(dd, prevDate, CurrentMailPromoDate) >= 100 or prevDate is null;
EDIT:
The subquery is returning the previous date for each record. For a given record, it is looking at all the dates less than the date on the record. These are sorted in descending order and the first one is picked. This is prevDate.
The where clause just filters out anything that is less than 100 days from the previous date.
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 like below
ID_NUMBER
SALEDATA
SALEAMOUNT
1
2020-09-07
47,000
2
2020-03-25
51,470
3
2021-06-12
32,000
4
2018-10-12
37,560
I want to select the rows with the 2 most recent dates only. So my desired output would be like below
ID_NUMBER
SALEDATA
SALEAMOUNT
1
2020-09-07
47,000
3
2021-06-12
32,000
Can someone please guide me on where would i start with this in SQL? I tried using MAX() but it is only giving me the most recent.
Thank you!
In Standard SQL, you would use:
select t.*
from t
order by saledata desc
offset 0 row fetch first 2 row only;
Not all databases support fetch first. It might be spelled limit or select top or something else, depending on your database.
Another option, with the rank analytic function. Sample data till line #7, query begins at line #9. See comments within code.
SQL> with test (id_number, saledata, saleamount) as
2 -- sample data
3 (select 1, date '2020-09-07', 47000 from dual union all
4 select 2, date '2020-03-25', 51470 from dual union all
5 select 3, date '2021-06-12', 32000 from dual union all
6 select 4, date '2018-10-12', 37560 from dual
7 )
8 -- sort them by date in descending order, fetch the first two rows
9 select id_number, saledata, saleamount
10 from (select t.*,
11 rank() over (order by saledata desc) rn
12 from test t
13 )
14 where rn <= 2
15 order by saledata;
ID_NUMBER SALEDATA SALEAMOUNT
---------- ---------- ----------
1 2020-09-07 47000
3 2021-06-12 32000
SQL>
select top 2 from your data set order by the saledata column descending
I am trying to run a query that will retrieve the most recent 30 days that have data (not the last 30 days)
There are can be several rows for the same date (so can't use the limit 30)
My data has the following formatting:
date count
2017-05-05 111
2017-05-05 78
2017-04-28 54
2017-01-11 124
Is there a way for me to add a WHERE clause to get the most recent 30 days with data?
Not sure if I correctly understand, though...
(this is for most recent 2 day):
with t(date, count) as(
select '2017-05-05', 111 union all
select '2017-05-05', 78 union all
select '2017-04-28', 54 union all
select '2017-01-11', 124
)
select date from t group by date order by date desc limit 2
If you want all rows, which has the same date as the last 30, distinct dates in your table, you can use the dense_rank() window function:
select (t).*
from (select t, dense_rank() over (order by date desc)
from t) s
where dense_rank <= 30
or IN, with a sub-select:
select *
from t
where date in (select distinct date
from t
order by date desc
limit 30)
http://rextester.com/ESDLIM64772
select * from tablename
where datecolumn in (select TOP 30 max(datecolumn) from tablename group by datecolumn)
I have an SQL Table in which I keep project information coming from primavera.
Suppose that i have columns for Start Date,End Date,Duration, and Total Qty as shown below .
How can i distribute Total Qty over Months using these information. What kind of additional columns, sql queries i need in order to get correct monthly distribution?
Thanks in Advance.
Columns in order:
itemname,quantity,startdate,duration,enddate
item1 -- 108 -- 2013-03-25 -- 720 -- 2013-07-26
item2 -- 640 -- 2013-03-25 -- 720 -- 2013-07-26
.
.
I think the key is to break the records apart by month. Here is an example of how to do it:
with months as (
select 1 as mon union all select 2 union all select 3 union all
select 4 as mon union all select 5 union all select 6 union all
select 7 as mon union all select 8 union all select 9 union all
select 10 as mon union all select 11 union all select 12
)
select item, m.mon, quantity / nummonths
from (select t.*, (month(enddate) - month(startdate) + 1) as nummonths
from t
) t join
months m
on month(t.startDate) <= m.mon and
months(t.endDate) >= m.mon;
This works because all the months are within the same year -- as in your example. You are quite vague on how the split should be calculated. So, I assumed that every month from the start to the end gets an equal amount.
I have a table with many IDs and many dates associated with each ID, and even a few IDs with no date. For each ID and date combination, I want to select the ID, date, and the next largest date also associated with that same ID, or null as next date if none exists.
Sample Table:
ID Date
1 5/1/10
1 6/1/10
1 7/1/10
2 6/15/10
3 8/15/10
3 8/15/10
4 4/1/10
4 4/15/10
4
Desired Output:
ID Date Next_Date
1 5/1/10 6/1/10
1 6/1/10 7/1/10
1 7/1/10
2 6/15/10
3 8/15/10
3 8/15/10
4 4/1/10 4/15/10
4 4/15/10
SELECT
mytable.id,
mytable.date,
(
SELECT
MIN(mytablemin.date)
FROM mytable AS mytablemin
WHERE mytablemin.date > mytable.date
AND mytable.id = mytablemin.id
) AS NextDate
FROM mytable
This has been tested on SQL Server 2008 R2 (but it should work on other DBMSs) and produces the following output:
id date NextDate
----------- ----------------------- -----------------------
1 2010-05-01 00:00:00.000 2010-06-01 00:00:00.000
1 2010-06-01 00:00:00.000 2010-06-15 00:00:00.000
1 2010-07-01 00:00:00.000 2010-08-15 00:00:00.000
2 2010-06-15 00:00:00.000 2010-07-01 00:00:00.000
3 2010-08-15 00:00:00.000 NULL
3 2010-08-15 00:00:00.000 NULL
4 2010-04-01 00:00:00.000 2010-04-15 00:00:00.000
4 2010-04-15 00:00:00.000 2010-05-01 00:00:00.000
4 NULL NULL
Update 1:
For those that are interested, I've compared the performance of the two variants in SQL Server 2008 R2 (one uses MIN aggregate and the other uses TOP 1 with an ORDER BY):
Without an index on the date column, the MIN version had a cost of 0.0187916 and the TOP/ORDER BY version had a cost of 0.115073 so the MIN version was "better".
With an index on the date column, they performed identically.
Note that this was testing with just these 9 records so the results could be (very) spurious...
Update 2:
The results hold for 10,000 uniformly distributed random records. The TOP/ORDER BY query takes so long to run at 100,000 records I had to cancel it and give up.
If your db is oracle, you can use lead() and lag() functions.
SELECT id, date,
LEAD(date, 1, 0) OVER (PARTITION BY ID ORDER BY Date DESC NULLS LAST) NEXT_DATE,
FROM Your_table
ORDER BY ID;
SELECT
id,
date,
( SELECT date
FROM table t1
WHERE t1.date > t2.date
ORDER BY t1.date LIMIT 1 )
FROM table t2
I think self JOIN would be faster than subselect.
WITH dates AS (
SELECT 1 AS ID, '2010-05-01' AS Date
UNION ALL SELECT 1, '2010-06-01'
UNION ALL SELECT 1, '2010-07-01'
UNION ALL SELECT 2, '2010-06-15'
UNION ALL SELECT 3, '2010-08-15'
UNION ALL SELECT 3, '2010-08-15'
UNION ALL SELECT 4, '2010-04-01'
UNION ALL SELECT 4, '2010-04-15'
UNION ALL SELECT 4, ''
)
SELECT
dates.ID,
dates.Date,
nextDates.Date AS Next_Date
FROM
dates
LEFT JOIN
dates nextDates
ON nextDates.ID = dates.ID
AND nextDates.Date > dates.Date
LEFT JOIN
dates noLower
ON noLower.ID = nextDates.ID
AND noLower.Date < nextDates.Date
AND noLower.Date > dates.Date
WHERE
dates.Date > 0
AND noLower.ID IS NULL
https://www.db-fiddle.com/f/4sWRLt2hxjik5HqiJ21ez8/1