Input Table :
date_1 date_2 ID
2019-01-01 2019-06-30 1
2019-05-01 2019-05-31 1
2019-06-01 2019-07-30 1
2019-01-02 2019-02-28 2
2019-03-01 2019-08-30 2
2019-01-02 2019-02-28 3
2019-02-06 2019-08-30 3
I am working on a complex HIVE problem of dates.
I need to changes dates of date_1 column and date_2 column for same ID.
I want to copy date_2's date to date_1's date in next row based on a condition. And all this I have to do for each ID, i.e. partition By ID.
Note : Data is sorted by ID asc, date_1 asc, date_2 asc.
For example :
Consider 2nd row, date_1 date is '2019-05-01' and now see its previous row for same ID 1 , here date_2 date is '2019-06-30'.
So check IF date_2 value of any row's previous row is greater than current row's value of date_1 , which is true in case of second row of ID 1.
When true then replace date_1 value of second row with date_2 value of previous row.
i.e. change 2019-05-01 to 2019-06-30, otherwise keep it as it is. Same do it for 3rd row and so on.
when considering 3rd row , then look for its previous row 2nd . And same goes for other rows.
Consider 2nd row of ID 2.
Here 2019-02-28is not greater than 2019-03-01 , so keep it as it is.
Expected Output :
date_1 date_2 ID
2019-01-01 2019-06-30 1
2019-06-30 2019-05-31 1
2019-06-01 2019-07-30 1
2019-01-02 2019-02-28 2
2019-03-01 2019-08-30 2
2019-01-02 2019-02-28 3
2019-02-28 2019-08-30 3
I think you want lag() like this
select date_add(lag(date2, 1, date1) over (partition by id order by date1), 1) as date1,
date2,
id
from t;
Related
This is very different from doing an SQL order by 2 date columns (or for proper way to sort sql columns, which is only for 1 column). There, we would do something like:
ORDER BY CASE WHEN date_1 > date_2
THEN date_2 ELSE date_1 END
FYI, I'm using YYY-MM-DD in this example for brevity, but I also need it to work for
TIMESTAMP (YYYY-MM-DD HH:MI:SS)
I have this table:
id
name
date_1
date_2
date_3
date_4
date_5
date_6
date_7
date_8
1
John
2008-08-11
2008-08-12
2009-08-11
2009-08-21
2009-09-11
2017-08-11
2017-09-12
2017-09-30
2
Bill
2008-09-12
2008-09-12
2008-10-12
2011-09-12
2008-09-13
2022-05-20
2022-05-21
2022-05-22
3
Andy
2008-10-13
2008-10-13
2008-10-14
2008-10-15
2008-11-01
2008-11-02
2008-11-03
2008-11-04
4
Hank
2008-11-14
2008-11-15
2008-11-16
2008-11-17
2008-12-31
2009-01-01
2009-01-02
2009-01-02
5
Alex
2008-12-15
2018-12-15
2018-12-15
2018-12-16
2018-12-17
2018-12-18
2018-12-25
2008-12-31
... But, the permutations of that give me a headache, just to think about them.
This Answer had more of a "general solution", but that was to SELECT, not to ORDER BY...
SELECT MAX(date_col)
FROM(
SELECT MAX(date_col1) AS date_col FROM some_table
UNION
SELECT MAX(date_col2) AS date_col FROM some_table
UNION
SELECT MAX(date_col3) AS date_col FROM some_table
...
)
Is there something more like that, such as could be created by iterating a loop in, say PHP or Node.js? I need something a scalable solution.
I only need to list each row once.
I want to order them each by whichever col has the most recent date of those I list on that row.
Something like:
SELECT * FROM some_table WHERE
(
GREATEST OF date_1
OR date_2
OR date_3
OR date_4
OR date_5
OR date_6
OR date_7
OR date_8
)
You can use the GREATEST function to achieve it.
SELECT GREATEST(date_1,date_2,date_3,date_4,date_5,date_6,date_7,date_8) max_date,t.*
FROM Tab t
ORDER BY GREATEST(date_1,date_2,date_3,date_4,date_5,date_6,date_7,date_8) Desc;
DB Fiddle: Try it here
max_date
id
name
date_1
date_2
date_3
date_4
date_5
date_6
date_7
date_8
2022-05-22
2
Bill
2008-09-12
2008-09-12
2008-10-12
2011-09-12
2008-09-13
2022-05-20
2022-05-21
2022-05-22
2018-12-25
5
Alex
2008-12-15
2018-12-15
2018-12-15
2018-12-16
2018-12-17
2018-12-18
2018-12-25
2008-12-31
2017-09-30
1
John
2008-08-11
2008-08-12
2009-08-11
2009-08-21
2009-09-11
2017-08-11
2017-09-12
2017-09-30
2009-01-02
4
Hank
2008-11-14
2008-11-15
2008-11-16
2008-11-17
2008-12-31
2009-01-01
2009-01-02
2009-01-02
2008-11-04
3
Andy
2008-10-13
2008-10-13
2008-10-14
2008-10-15
2008-11-01
2008-11-02
2008-11-03
2008-11-04
In the event of a NULL value, GREATEST could throw-off the ORDER.
Based on this Answer from a Question about GREATEST handling NULL, this would apply these tables, based on the approved Answer:
SELECT COALESCE (
GREATEST(date_1,date_2,date_3,date_4,date_5,date_6,date_7,date_8),
date_1,date_2,date_3,date_4,date_5,date_6,date_7,date_8
) max_date,t.*
FROM TAB t
ORDER BY COALESCE (
GREATEST(date_1,date_2,date_3,date_4,date_5,date_6,date_7,date_8),
date_1,date_2,date_3,date_4,date_5,date_6,date_7,date_8
) DESC;
I have the below table in bigquery:
Timestamp variant_id activity
2020-04-02 08:50 1 active
2020-04-03 07:39 1 not_active
2020-04-04 07:40 1 active
2020-04-05 10:22 2 active
2020-04-07 07:59 2 not_active
I want to query this subset of data to get the number of active variant per day.
If variant_id 1 is active at date 2020-04-04, it still active the follwing dates also 2020-04-05, 2020-04-06 until the value activity column is not_active , the goal is to count each day the number of variant_id who has the value active in the column activity, but I should take into account that each variant_id has the value of the last activity on a specific date.
for example the result of the desired query in the subset data must be:
Date activity_count
2020-04-02 1
2020-04-03 0
2020-04-04 1
2020-04-05 2
2020-04-06 2
2020-04-07 1
2020-04-08 1
2020-04-09 1
2020-04-10 1
any help please ?
Consider below approach
select date, count(distinct if(activity = 'active', variant_id, null)) activity_count
from (
select date(timestamp) date, variant_id, activity,
lead(date(timestamp)) over(partition by variant_id order by timestamp) next_date
from your_table
), unnest(generate_date_array(date, ifnull(next_date - 1, '2020-04-10'))) date
group by date
if applied to sample data in your question - output is
I've got a problem with my SQL task and didn't find any answers yet.
I've got table with this sample data:
ID
Value
Date
1
1
2020-01-01
1
2
2020-03-02
1
1
2020-03-21
1
1
2020-04-14
1
3
2020-05-01
1
1
2020-08-09
1
1
2020-09-12
1
1
2020-10-12
1
3
2020-12-04
All I want to get is:
ID
Value
Date
1
1
2020-01-01
1
2
2020-03-02
1
1
2020-03-21
1
3
2020-05-01
1
1
2020-08-09
1
3
2020-12-04
Some kind of changing value history, but only if the value was changed - when value on new record is the same, get value with min date.
I tried with grouping and row_number, but got no positive results. Any ideas how to do that?
One way to articulate your logic is to say that you want to retain a record when the previous record, as ordered by the date (within a given ID), has a different value than the current record.
WITH cte AS (
SELECT *, LAG(Value) OVER (PARTITION BY ID ORDER BY Date) LagValue
FROM yourTable
)
SELECT ID, Value, Date
FROM cte
WHERE LagValue <> Value OR LagValue IS NULL
ORDER BY Date;
Demo
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I have a CONTRACTOR_SCHEDULER table. Letters in columns "Schedule" mean working time (m - 8:00-20:00, n - 20:00-8:00(next day), d - 8:00-8:00(next day), h-day off). (Name, Begin_date) is unique.
Name
Schedule
Begin_date
End_date
John
nhmh
2019-01-01
2019-01-08
John
nnh
2019-01-09
2019-01-25
Kate
dh
2019-01-01
2019-01-07
Kate
mnhdh
2019-01-08
2019-01-14
Mike
nh
2019-01-01
2019-02-01
Mike
mh
2019-02-02
2019-12-31
I need a SQL Server stored procedure that creates new table CONTRACTOR_WORK_DAY of working days using CONTRACTOR_SCHEDULER rows (days off do not appear in table).
Example:
First row - John has schedule nhm. First letter is n so begin_date - 2019-01-01 20:00, End_date - 2019-01-02 08:00. Next letter is h, skip it as a day off. Last letter is m so begin_date - 2019-01-03 08:00, End_date - 2019-01-03 20:00. Repeat schedule until 2019-01-08 - end_date in first table.
Table for the first row of CONTRACTOR_SCHEDULER would be:
Name
Begin_date
End_date
John
2019-01-01 20:00
2019-01-02 08:00
John
2019-01-03 08:00
2019-01-03 20:00
John
2019-01-05 20:00
2019-01-06 08:00
John
2019-01-08 08:00
2019-01-08 20:00
I wrote this in python using some loop over schedule string etc. but can not figure out how to do it in T-SQL for SQL Server.
Yet another option.
Not sure if I agree with the last record of the desired results. I have 2019-01-07 while you have 2019-01-08
Example or dbFiddle
Select A.[Name]
,[Begin_Date] = convert(datetime,left(dateadd(DAY,N ,[Begin_date]),10)+' '+BegTime)
,[End_Date] = convert(datetime,left(dateadd(DAY,N+NxtDay,[Begin_date]),10)+' '+EndTime)
From YourTable A
Cross Apply ( values ( left(replicate(Schedule,10),datediff(DAY,Begin_Date,End_Date)) ) )B(S)
Cross Apply (
Select N=N-1
,Subs=substring(B.S,N,1)
From ( Select Top (len(S)+1) N=Row_Number() Over (Order By (Select Null)) From master..spt_values n1 ) B1
) C
Join ( values ('m','08:00','20:00',0)
,('n','20:00','08:00',1)
,('d','08:00','08:00',1)
) D(SchdCd,BegTime,EndTime,NxtDay) on Subs=SchdCd
Order By [Begin_Date]
Results
Name Begin_Date End_Date
John 2019-01-01 20:00:00.000 2019-01-02 08:00:00.000
John 2019-01-03 08:00:00.000 2019-01-03 20:00:00.000
John 2019-01-05 20:00:00.000 2019-01-06 08:00:00.000
John 2019-01-07 08:00:00.000 2019-01-07 20:00:00.000
WITH
numbers AS
(
-- Generate a table of values (from 0 upwards)
-- Must be at least as long as the longest schedule string
SELECT
ROW_NUMBER() OVER (ORDER BY sv.number) - 1 AS id
FROM
master..spt_values sv
),
pivotted AS
(
-- Create one row for each day in the date range
-- Extract the relevant character from the schedule for each day
SELECT
c.*,
n.id AS date_offset,
SUBSTRING(c.schedule, (n.id % LEN(c.schedule)) + 1, 1) AS schedule_char
FROM
CONTRACTOR_SHERULER c
INNER JOIN
numbers n
ON n.id <= DATEDIFF(DAY, c.begin_date, c.end_date)
)
-- Add a number of days to the begin_date
-- Then add a number of hours based on the current character from the schedule
SELECT
pivotted.*,
DATEADD(
HOUR,
CASE pivotted.schedule_char WHEN 'm' THEN 8
WHEN 'n' THEN 20
WHEN 'd' THEN 8 END,
DATEADD(DAY, date_offset, pivotted.begin_date)
)
AS begin_datetime,
DATEADD(
HOUR,
CASE pivotted.schedule_char WHEN 'm' THEN 20
WHEN 'n' THEN 32
WHEN 'd' THEN 32 END,
DATEADD(DAY, date_offset, pivotted.begin_date)
)
AS end_datetime
FROM
pivotted
WHERE
pivotted.schedule_char <> 'h'
Demo : https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=f6c52e6718329f2ae013dc93edad5d78
[SQL Novice] I have a table that looks like this:
id date
1 2019-01-01
1 2019-01-02
2 2019-03-01
2 2019-05-01
I want to only filter the id column on 2 where date is between 2019-04-01 and 2019-05-01 without impacting id equals 1.
The new table should look like this:
id date
1 2019-01-01
1 2019-01-02
2 2019-03-01
I tried this:
select * from table1 where id =2 and date between 2019-03-01 and 2019-04-01
And get this data set:
id date
2 2019-03-01
I think you want or:
where id = 1 or
(id = 2 and date between '2019-03-01' and '2019-04-01')
for your desired result need
select * from table1 where [date] >= '2019-01-01' and [date] <= '2019-03-01'