From the data below I need to select the record nearest to a specified date for each Linked ID using SQL Server 2005:
ID Date Linked ID
...........................
1 2010-09-02 25
2 2010-09-01 25
3 2010-09-08 39
4 2010-09-09 39
5 2010-09-10 39
6 2010-09-10 34
7 2010-09-29 34
8 2010-10-01 37
9 2010-10-02 36
10 2010-10-03 36
So selecting them using 01/10/2010 should return:
1 2010-09-02 25
5 2010-09-10 39
7 2010-09-29 34
8 2010-10-01 37
9 2010-10-02 36
I know this must be possible, but can't seem to get my head round it (must be too near the end of the day :P) If anyone can help or give me a gentle shove in the right direction it would be greatly appreciated!
EDIT: Also I have come across this sql to get the closest date:
abs(DATEDIFF(minute, Date_Column, '2010/10/01'))
but couldn't figure out how to incorporate into the query properly...
Thanks
you can try this.
DECLARE #Date DATE = '10/01/2010';
WITH cte AS
(
SELECT ID, LinkedID, ABS(DATEDIFF(DD, #date, DATE)) diff,
ROW_NUMBER() OVER (PARTITION BY LinkedID ORDER BY ABS(DATEDIFF(DD, #date, DATE))) AS SEQUENCE
FROM MyTable
)
SELECT *
FROM cte
WHERE SEQUENCE = 1
ORDER BY ID
;
You didn't indicate how you want to handle the case where multiple rows in a LinkedID group represent the closest to the target date. This solution will only include one row And, in this case you can't guarantee which row of the multiple valid values is included.
You can change ROW_NUMBER() with RANK() in the query if you want to include all rows that represent the closest value.
You want to look at the absolute value of the DATEDIFF function (http://msdn.microsoft.com/en-us/library/ms189794.aspx) by days.
The query can look something like this (not tested)
with absDates as
(
select *, abs(DATEDIFF(day, Date_Column, '2010/10/01')) as days
from table
), mdays as
(
select min(days) as mdays, linkedid
from absDates
group by linkedid
)
select *
from absdates
inner join mdays on absdays.linkedid = mdays.linkedid and absdays.days = mdays.mdays
You can also try to do it with a subquery in the select statement:
select [LinkedId],
(select top 1 [Date] from [Table] where [LinkedId]=x.[LinkedId] order by abs(DATEDIFF(DAY,[Date],#date)))
from [Table] X
group by [LinkedId]
Related
Following is the table:
start_date
recorded_date
id
2021-11-10
2021-11-01
1a
2021-11-08
2021-11-02
1a
2021-11-11
2021-11-03
1a
2021-11-10
2021-11-04
1a
2021-11-10
2021-11-05
1a
I need a query to find the total day changes in aggregate for a given id. In this case, it changed from 10th Nov to 8th Nov so 2 days, then again from 8th to 11th Nov so 3 days and again from 11th to 10th for a day, and finally from 10th to 10th, that is 0 days.
In total there is a change of 2+3+1+0 = 6 days for the id - '1a'.
Basically for each change there is a recorded_date, so we arrange that in ascending order and then calculate the aggregate change of days grouped by id. The final result should be like:
id
Agg_Change
1a
6
Is there a way to do this using SQL. I am using vertica database.
Thanks.
you can use window function lead to get the difference between rows and then group by id
select id, sum(daydiff) Agg_Change
from (
select id, abs(datediff(day, start_Date, lead(start_date,1,start_date) over (partition by id order by recorded_date))) as daydiff
from tablename
) t group by id
It's indeed the use of LAG() to get the previous date in an OLAP query, and an outer query getting the absolute date difference, and the sum of it, grouping by id:
WITH
-- your input - don't use in real query ...
indata(start_date,recorded_date,id) AS (
SELECT DATE '2021-11-10',DATE '2021-11-01','1a'
UNION ALL SELECT DATE '2021-11-08',DATE '2021-11-02','1a'
UNION ALL SELECT DATE '2021-11-11',DATE '2021-11-03','1a'
UNION ALL SELECT DATE '2021-11-10',DATE '2021-11-04','1a'
UNION ALL SELECT DATE '2021-11-10',DATE '2021-11-05','1a'
)
-- real query starts here, replace following comma with "WITH" ...
,
w_lag AS (
SELECT
id
, start_date
, LAG(start_date) OVER w AS prevdt
FROM indata
WINDOW w AS (PARTITION BY id ORDER BY recorded_date)
)
SELECT
id
, SUM(ABS(DATEDIFF(DAY,start_date,prevdt))) AS dtdiff
FROM w_lag
GROUP BY id
-- out id | dtdiff
-- out ----+--------
-- out 1a | 6
I was thinking lag function will provide me the answer, but it kept giving me wrong answer because I had the wrong logic in one place. I have the answer I need:
with cte as(
select id, start_date, recorded_date,
row_number() over(partition by id order by recorded_date asc) as idrank,
lag(start_date,1) over(partition by id order by recorded_date asc) as prev
from table_temp
)
select id, sum(abs(date(start_date) - date(prev))) as Agg_Change
from cte
group by 1
If someone has a better solution please let me know.
My data is something like this
Count years
1 2020-08-11
1 2020-07-11
1 2019-09-01
1 2019-08-16
1 2019-05-04
1 2018-06-11
I'm writing a query where I have to find the count of year for eg <= 04 May 2019
,I need to find the count of all the dates lesser than that date i.e '2019-05-04'
count will come as 1 and then add 1 to the count.
I've written the query like this:
with sum_count as(
select count(*) as 'Counts', years from [practice].[dbo].[People]
where years<='2019-05-04'
group by years)
select sum(Counts) + 1 as Sum
from sum_count
Could you please help to do the same for all the dates like for 2020-08-11
the count shall come as 5 and sum a 6
You could achieve this quite simply by a select statement without the need for a cte:
Declare #td datetime = '20190504'
SELECT COUNT([years])+1 FROM [practice].[dbo].[People] Where [years] <= #td
If this something you would be repeating a lot you can have it as a stored procedure
CREATE PROC proc_name (#dt datetime)
as SELECT COUNT([years])+1 FROM [practice].[dbo].[People] Where [years] <= #td
and you would call it as such
exec proc_name ('20200801')
If I understand correctly, you simply want a window function. The following enumerates each row within each year:
select p.*,
row_number() over (partition by year(years) order by years) as seqnum
from [practice].[dbo].[People] p;
No stored procedure or auxiliary function is necessary.
Writing procedure can be one way, but if your hell bent on using a query you can use this.
select count(1)+1 as sum,a.year1
from
(
select distinct year as year1
from [practice].[dbo].[People]
) a
inner join
[practice].[dbo].[People] b on b.year<=a.year1
group by a.year1
Cheers!!!
I have s table that lists absences(holidays) of all employees, and what we would like to find out is who is away today, and the date that they will return.
Unfortunately, absences aren't given IDs, so you can't just retrieve the max date from an absence ID if one of those dates is today.
However, absences are given an incrementing ID per day as they are inputt, so I need a query that will find the employeeID if there is an entry with today's date, then increment the AbsenceID column to find the max date on that absence.
Table Example (assuming today's date is 11/11/2014, UK format):
AbsenceID EmployeeID AbsenceDate
100 10 11/11/2014
101 10 12/11/2014
102 10 13/11/2014
103 10 14/11/2014
104 10 15/11/2014
107 21 11/11/2014
108 21 12/11/2014
120 05 11/11/2014
130 15 20/11/2014
140 10 01/03/2015
141 10 02/03/2015
142 10 03/03/2015
143 10 04/03/2015
So, from the above, we'd want the return dates to be:
EmployeeID ReturnDate
10 15/11/2014
21 12/11/2014
05 11/11/2014
Edit: note that the 140-143 range couldn't be included in the results as they appears in the future, and none of the date range of the absence are today.
Presumably I need an iterative sub-function running on each entry with today's date where the employeeID matches.
So based on what I believe you're asking, you want to return a list of the people that are off today and when they are expected back based on the holidays that you have recorded in the system, which should only work only on consecutive days.
SQL Fiddle Demo
Schema Setup:
CREATE TABLE EmployeeAbsence
([AbsenceID] int, [EmployeeID] int, [AbsenceDate] DATETIME)
;
INSERT INTO EmployeeAbsence
([AbsenceID], [EmployeeID], [AbsenceDate])
VALUES
(100, 10, '2014-11-11'),
(101, 10, '2014-11-12'),
(102, 10, '2014-11-13'),
(103, 10, '2014-11-14'),
(104, 10, '2014-11-15'),
(107, 21, '2014-11-11'),
(108, 21, '2014-11-12'),
(120, 05, '2014-11-11'),
(130, 15, '2014-11-20')
;
Recursive CTE to generate the output:
;WITH cte AS (
SELECT EmployeeID, AbsenceDate
FROM dbo.EmployeeAbsence
WHERE AbsenceDate = CAST(GETDATE() AS DATE)
UNION ALL
SELECT e.EmployeeID, e.AbsenceDate
FROM cte
INNER JOIN dbo.EmployeeAbsence e ON e.EmployeeID = cte.EmployeeID
AND e.AbsenceDate = DATEADD(d,1,cte.AbsenceDate)
)
SELECT cte.EmployeeID, MAX(cte.AbsenceDate)
FROM cte
GROUP BY cte.EmployeeID
Results:
| EMPLOYEEID | Return Date |
|------------|---------------------------------|
| 5 | November, 11 2014 00:00:00+0000 |
| 10 | November, 15 2014 00:00:00+0000 |
| 21 | November, 12 2014 00:00:00+0000 |
Explanation:
The first SELECT in the CTE gets employees that are off today with this filter:
WHERE AbsenceDate = CAST(GETDATE() AS DATE)
This result set is then UNIONED back to the EmployeeAbsence table with a join that matches EmployeeID as well as the AbsenceDate + 1 day to find the consecutive days recursively using:
-- add a day to the cte.AbsenceDate from the first SELECT
e.AbsenceDate = DATEADD(d,1,cte.AbsenceDate)
The final SELECT simply groups the cte results by employee with the MAX AbsenceDate that has been calculated per employee.
SELECT cte.EmployeeID, MAX(cte.AbsenceDate)
FROM cte
GROUP BY cte.EmployeeID
Excluding Weekends:
I've done a quick test based on your comment and the below modification to the INNER JOIN within the CTE should exclude weekends when adding the extra days if it detects that adding a day will result in a Saturday:
INNER JOIN dbo.EmployeeAbsence e ON e.EmployeeID = cte.EmployeeID
AND e.AbsenceDate = CASE WHEN datepart(dw,DATEADD(d,1,cte.AbsenceDate)) = 7
THEN DATEADD(d,3,cte.AbsenceDate)
ELSE DATEADD(d,1,cte.AbsenceDate) END
So when you add a day: datepart(dw,DATEADD(d,1,cte.AbsenceDate)) = 7, if it results in Saturday (7), then you add 3 days instead of 1 to get Monday: DATEADD(d,3,cte.AbsenceDate).
You'd need to do a few things to get this data into a usable format. You need to be able to work out where a group begins and ends. This is difficult with this example because there is no straight forward grouping column.
So that we can calculate when a group starts and ends, you need to create a CTE containing all the columns and also use LAG() to get the AbsenceID and EmployeeID from the previous row for each row. In this CTE you should also use ROW_NUMBER() at the same time so that we have a way to re-order the rows into the same order again.
Something like:
WITH
[AbsenceStage] AS (
SELECT [AbsenceID], [EmployeeID], [AbsenceDate]
,[RN] = ROW_NUMBER() OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
,[AbsenceID_Prev] = LAG([AbsenceID]) OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
,[EmployeeID_Prev] = LAG([EmployeeID]) OVER (ORDER BY [EmployeeID] ASC, [AbsenceDate] ASC, [AbsenceID] ASC)
FROM [HR_Absence]
)
Now that we have this we can compare each row to the previous to see if the current row is in a different "group" to the previous row.
The condition would be something like:
[EmployeeID_Prev] IS NULL -- We have a new group if the previous row is null
OR [EmployeeID_Prev] <> [EmployeeID] -- Or if the previous row is for a different employee
OR [AbsenceID_Prev] <> ([AbsenceID]-1) -- Or if the AbsenceID is not sequential
You can then use this to join the CTE to it's self to find the first row in each group with something like:
....
FROM [AbsenceStage] AS [Row]
INNER JOIN [AbsenceStage] AS [First]
ON ([First].[RN] = (
-- Get the first row before ([RN] Less that or equal to) this one where it is the start of a grouping
SELECT MAX([RN]) FROM [AbsenceStage]
WHERE [RN] <= [Row].[RN] AND (
[EmployeeID_Prev] IS NULL
OR [EmployeeID_Prev] <> [EmployeeID]
OR [AbsenceID_Prev] <> ([AbsenceID]-1)
)
))
...
You can then GROUP BY the [First].[RN] which will now act like a group id and allow you to get the start and end date of each absence group.
SELECT
[Row].[EmployeeID]
,MIN([Row].[AbsenceDate]) AS [Absence_Begin]
,MAX([Row].[AbsenceDate]) AS [Absence_End]
...
-- FROM and INNER JOIN from above
...
GROUP BY [First].[RN], [Row].[EmployeeID];
You could then put all that into a view giving you the EmployeeID with the Start and End date of each absence. You can then easily pull out the Employee's currently off with a:
WHERE CAST(CURRENT_TIMESTAMP AS date) BETWEEN [Absence_Begin] AND [Absence_End]
SQL Fiddle
Like another answer here, I'm going to create the leave intervals, but via a different method. First the code:
declare #today date = getdate(); --use whatever date here
with g as (
select *, dateadd(day, -1 * row_number() over (partition by employeeid order by absencedate), AbsenceDate) as group_number
from employeeabsence
) , leave_intervals as (
select employeeid, min(absencedate) as [start], max(absencedate) as [end]
from g
group by EmployeeID, group_number
)
select employeeid, [start], [end]
from leave_intervals
where #today between [start] and [end]
By way of explanation, we first put a date value into a variable. I chose today, but this code will work for any date passed in. Next, we create a common table expression (CTE) that will add on a grouping column to your table. This is the meat of the solution, so it bears some treatment. Within a given interval, the AbsenceDate increases at a rate of one day per row. row_number() also increases at a rate of one per row. So, if we subtract a row_number() number of days from the AbsenceDate, we'll get another (arbitrary) date. The key here is to realize that that arbitrary date will be the same for every row in the interval, so we can use it to group by. From there, it's just a matter of doing just that; get the min and max per interval. Lastly, we find what intervals contain #today.
Assuming I have two records, both with a date and a count:
--Date-- --Count--
2011-09-20 00:00:00 5
2011-09-16 00:00:00 8
How would you select this for filling the time gaps, always taking the last previous record?
So the output would be:
--Date-- --Count--
2011-09-20 00:00:00 5
2011-09-19 00:00:00 8
2011-09-18 00:00:00 8
2011-09-17 00:00:00 8
2011-09-16 00:00:00 8
I couldn't figure out a neat solution for this, yet.
I guess this could be done with DATEDIFF, and a for-loop, but I hope this can be done easier.
You have 2 issues you're trying to resolve. The first issue is how to fill the gaps. The second issue is populating the Count field for those missing records.
Issue 1: This can be resolved by either using a Dates Lookup table or by creating a recursive common table expression. I would recommend creating a Dates Lookup table for this if that is an option. If you cannot create such a table, then you're going to need something like this.
WITH CTE AS (
SELECT MAX(dt) maxdate, MIN(dt) mindate
FROM yourtable
),
RecursiveCTE AS (
SELECT mindate dtfield
FROM CTE
UNION ALL
SELECT DATEADD(day, 1, dtfield)
FROM RecursiveCTE R
JOIN CTE T
ON R.dtfield < T.maxdate
)
That should create you a list of dates starting with the MIN date in your table and ending in the MAX.
Issue 2: Here is where a correlated subquery would come in handy (as much as I generally stay away from them) to get the last cnt from your original table:
SELECT r.dtfield,
(SELECT TOP 1 cnt
FROM yourtable
WHERE dt <= r.dtfield
ORDER BY dt DESC) cnt
FROM RecursiveCTE r
SQL Fiddle Demo
My solution goes like this.
Step 1: Have a Date table which has all the dates. - you can use many methods ex: Get a list of dates between two dates
Step 2: Do a Left outer from the date table to your result set. - which would result you with the below resultset: Call this table as "TEST_DATE_COUnt"
--Date-- --Count--
2011-09-20 00:00:00 5
2011-09-19 00:00:00 0
2011-09-18 00:00:00 0
2011-09-17 00:00:00 0
2011-09-16 00:00:00 8
Step 3: Do a Recursive query like below:
SELECT t1.date_x, t1.count_x,
(case when count_x=0 then (SELECT max(COUNT_X)
FROM TEST_DATE_COUNT r
WHERE r.DATE_X <= t1.DATE_X)
else COUNT_X
end)
cnt
FROM TEST_DATE_COUNT t1
Please let me know if this works. I tested and it worked.
If I had a large table (100000 + entries) which had service records or perhaps admission records. How would I find all the instances of re-occurrence within a set number of days.
The table setup could be something like this likely with more columns.
Record ID Customer ID Start Date Time Finish Date Time
1 123456 24/04/2010 16:49 25/04/2010 13:37
3 654321 02/05/2010 12:45 03/05/2010 18:48
4 764352 24/03/2010 21:36 29/03/2010 14:24
9 123456 28/04/2010 13:49 31/04/2010 09:45
10 836472 19/03/2010 19:05 20/03/2010 14:48
11 123456 05/05/2010 11:26 06/05/2010 16:23
What I am trying to do is work out a way to select the records where there is a re-occurrence of the field [Customer ID] within a certain time period (< X days). (Where the time period is Start Date Time of the 2nd occurrence - Finish Date Time of the first occurrence.
This is what I would like it to look like once it was run for say x=7
Record ID Customer ID Start Date Time Finish Date Time Re-occurence
9 123456 28/04/2010 13:49 31/04/2010 09:45 1
11 123456 05/05/2010 11:26 06/05/2010 16:23 2
I can solve this problem with a smaller set of records in Excel but have struggled to come up with a SQL solution in MS Access. I do have some SQL queries that I have tried but I am not sure I am on the right track.
Any advice would be appreciated.
I think this is a clear expression of what you want. It's not extremely high performance but I'm not sure that you can avoid either correlated sub-query or a cartesian JOIN of the table to itself to solve this problem. It is standard SQL and should work in most any engine, although the details of the date math may differ:
SELECT * FROM YourTable YT1 WHERE EXISTS
(SELECT * FROM YourTable YT2 WHERE
YT2.CustomerID = YT1.CustomerID AND YT2.StartTime <= YT2.FinishTime + 7)
In order to accomplish this you would need to make a self join as you are comparing the entire table to itself. Assuming similar names it would look something like this:
select r1.customer_id, min(start_time), max(end_time), count(1) as reoccurences
from records r1,
records r2
where r1.record_id > r2.record_id -- this ensures you don't double count the records
and r1.customer_id = r2.customer_id
and r1.finish_time - r2.start_time <= 7
group by r1.customer_id
You wouldn't be able to easily get both the record_id and the number of occurences, but you could go back and find it by correlating the start time to the record number with that customer_id and start_time.
This will do it:
declare #t table(Record_ID int, Customer_ID int, StartDateTime datetime, FinishDateTime datetime)
insert #t values(1 ,123456,'2010-04-24 16:49','2010-04-25 13:37')
insert #t values(3 ,654321,'2010-05-02 12:45','2010-05-03 18:48')
insert #t values(4 ,764352,'2010-03-24 21:36','2010-03-29 14:24')
insert #t values(9 ,123456,'2010-04-28 13:49','2010-04-30 09:45')
insert #t values(10,836472,'2010-03-19 19:05','2010-03-20 14:48')
insert #t values(11,123456,'2010-05-05 11:26','2010-05-06 16:23')
declare #days int
set #days = 7
;with a as (
select record_id, customer_id, startdatetime, finishdatetime,
rn = row_number() over (partition by customer_id order by startdatetime asc)
from #t),
b as (
select record_id, customer_id, startdatetime, finishdatetime, rn, 0 recurrence
from a
where rn = 1
union all
select a.record_id, a.customer_id, a.startdatetime, a.finishdatetime,
a.rn, case when a.startdatetime - #days < b.finishdatetime then recurrence + 1 else 0 end
from b join a
on b.rn = a.rn - 1 and b.customer_id = a.customer_id
)
select record_id, customer_id, startdatetime, recurrence from b
where recurrence > 0
Result:
https://data.stackexchange.com/stackoverflow/q/112808/
I just realize it should be done in access. I am so sorry, this was written for sql server 2005. I don't know how to rewrite it for access.