I have a table like this:
ID cbk due_16_30 due_31_60
1 2018-06-19 5 200
2 2018-06-19 100 -5
1 2018-06-19 -2 2
2 2018-06-18 20 Null
2 2018-06-18 50 22
1 2018-06-18 30 150
3 2018-06-18 20 70
I want to select for each specific ID a max due_16_30 and a max due_31_60 from the latest date, where date is between some start date and end date. How can I do that in PostgreSQL?
P.S. This is how the 2nd part is solved: https://stackoverflow.com/a/51493567/8495108
One method uses distinct on:
select distinct on (id) id, max(due_16_30), max(due_31_60)
from t
where date >= ? and date < ?
group by id, date
order by id, date desc;
You can do :
select t.id, t.cbk, max(t.due_16_30), max(t.due_31_60)
from table t
where cbk = (select t1.cbk
from table t1
where t1.cbk >= start_dt and t1.cbk <= end_dt
order by t1.cbk desc
limit 1
)
group by t.id, cbk
order by t.id desc
limit 1;
You need to consider myID parameter for both subquery and outer query as :
select ID,
cbk,
max(due_16_30) as max_due_16_30,
max(due_31_60) as max_due_31_60
from tab
where cbk in
(
select max(cbk)
from tab
where cbk between start_date and end_date
and ID = myID
)
and ID = myID
group by ID, cbk;
We may try for id = 3 in the demo, since there's no id = 3 for the latest date 2018-06-19 for whole table :
DB-Fiddle Demo
Solved like this (simplified):
SELECT v.id, max(v.due_31_60), max(v.due_61_90), v.cbk
FROM my_table as v
JOIN (select id, max(cbk) as max_date from my_table
WHERE (cbk >= start_date and cbk <= end_date )
GROUP BY id) as q
ON q.id = v.id and v.cbk = q.cbk
GROUP BY v.id, v.cbk
Related
I have a table which contains a lot of duplicated rows like this:
id_emp id date ch_in ch_out
1 34103 2019-09-01
1 34193 2019-09-01 17:00
1 34194 2019-09-02 07:03:21 16:59:26
1 34104 2019-09-02 07:03:21 16:59:26
1 33361 2019-09-02 NULL NULL
I want just one row for each date and others must delete with condition like I want the output must be:
id_emp id date ch_in ch_out
1 34193 2019-09-01 17:00
1 34104 2019-09-02 07:03:21 16:59:26
I tried to use distinct but nothing working:
select distinct id_emp, id, date_1, ch_in,ch_out
from ch_inout
where id_emp=1 order by date_1 asc
And I tried too using this query to delete:
select *
from (
select *, rn=row_number() over (partition by date_1 order by id)
from ch_inout
) x
where rn > 1;
But nothing is working the result is empty.
You can use aggregation:
select id_emp, max(id) as id, date, min(ch_in), max(ch_out)
from ch_inout
group by id_emp, date;
This returns the maximum id for each group of rows. That is not exactly what is returned in the question, but you don't specify the logic.
EDIT:
If you want to delete all but the largest id for each id_emp/date combination, you can use:
delete c from ch_inout c
where id < (select max(c2.id)
from ch_inout c2
where c2.id_emp = c.id_emp and c2.date = c.date
);
You can use ROW_NUMBER() to identify the records you want to delete. Assuming that you want to keep the record with the lowest id on each date:
SELECT *
FROM (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY date ORDER BY id) rn
FROM ch_inout t
) x
WHERE rn > 1
You can easily turn this into a DELETE statement:
WITH cte AS (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY date ORDER BY id) rn
FROM ch_inout t
)
DELETE FROM cte WHERE rn > 1
Example:
id Pricemoney time/date
1 100 01/20/2017
1 10 01/21/2017
1 1000 01/21/20147
2 10 01/23/2017
2 100 01/24/2017
3 1000 01/19/2017
3 100 01/22/2017
3 10 01/24/2017
I want to run a SQL query where I can display all the Id and it's pricemoney BUT NOT include the first record (based on time/date) per unique
Just to clarify what I do not want to be displayed
userid Pricemoney issuedate
1 100 01/20/2017 -- not included
1 10 01/21/2017
1 1000 01/21/20147
2 10 01/23/2017 --- not inlcuded
2 100 01/24/2017
3 1000 01/19/2017 -- not included
3 100 01/22/2017
3 10 01/24/2017
Expected result:
id Pricemoney time/date
1 10 01/21/2017
1 1000 01/21/20147
2 100 01/24/2017
3 100 01/22/2017
3 10 01/24/2017
You can use row_number():
select t.*
from (select t.*,
row_number() over (partition by id order by time_date asc) as seqnum
from <tablename> t
) t
where seqnum > 1;
If you want to keep single rows, you can do:
select t.*
from (select t.*,
row_number() over (partition by id order by time_date asc) as seqnum,
count(*) over (partition by id) as cnt
from <tablename> t
) t
where seqnum > 1 and cnt > 1;
You may use EXISTS
select t1.*
from data t1
where exists (
select 1
from data t2
where t1.id = t2.id and t2.time_date < t1.time_date
)
you can try this :
select data1.id,data1.Date,data1.Pricemoney from data1
left join (
select id ,min(Date) date from data1
group by id
) as t
on data1.date= t.date and t.id = data1.id
where t.id is null
group by data1.id,data1.Date,data1.Pricemoney
above query not duplicated records also ignore, if want
not duplicated records then use having count(id) > 1 in left query e,g.
select data1.id,data1.Date,data1.Pricemoney from data1
left join (
select id ,min(Date) date from data1
group by id
having COUNT(id) > 1
) as t
on data1.date= t.date and t.id = data1.id
where t.id is null
group by data1.id,data1.Date,data1.Pricemoney
When I perform "SELECT * FROM table" I got results like below:
ID Date Time Type
----------------------------------
60 03/03/2013 8:55:00 AM 1
60 03/03/2013 2:10:00 PM 2
110 17/03/2013 9:15:00 AM 1
67 24/03/2013 9:00:00 AM 1
67 24/03/2013 3:05:00 PM 2
as you see each ID has a transaction Type 1 and 2 in the same Date
except ID 110 HAS only Type 1
So how could I just get result like this:
ID Date Time Type
----------------------------------
110 17/03/2013 9:15:00 AM 1
as only one record are returned from the first result
Change the partition definition (partition by id,date) according to your needs
select *
from (select t.*
,count(*) over (partition by id,date) as cnt
from mytable t
) t
where t.cnt = 1
;
You can use this:
select * from my_table t
where exists (
select 1 from my_table
where id = t.id
group by id
having count(*) = 1
)
If you want only type 1, then compare the minimum and maximum values. I prefer window functions:
select t.*
from (select t.*, min(type) over (partition by id) as mintype,
max(type) over (partition by id) as maxtype
from t
) t
where mintype = maxtype and mintype = 1;
If you want only records of the same type (and not specifically type = 1), then remove that condition.
If you want only records on the same day, then include the date in the partition by.
Under some circumstances, not exists can be faster:
select t.*
from t
where not exists (select 1 from t t2 where t2.id = t.id and t2.type <> 1);
I need to select data base upon three conditions
Find the latest date (StorageDate Column) from the table for each record
See if there is more then one entry for date (StorageDate Column) found in first step for same ID (ID Column)
and then see if DuplicateID is = 2
So if table has following data:
ID |StorageDate | DuplicateTypeID
1 |2014-10-22 | 1
1 |2014-10-22 | 2
1 |2014-10-18 | 1
2 |2014-10-12 | 1
3 |2014-10-11 | 1
4 |2014-09-02 | 1
4 |2014-09-02 | 2
Then I should get following results
ID
1
4
I have written following query but it is really slow, I was wondering if anyone has better way to write it.
SELECT DISTINCT(TD.RecordID)
FROM dbo.MyTable TD
JOIN (
SELECT T1.RecordID, T2.MaxDate,COUNT(*) AS RecordCount
FROM MyTable T1 WITH (nolock)
JOIN (
SELECT RecordID, MAX(StorageDate) AS MaxDate
FROM MyTable WITH (nolock)
GROUP BY RecordID)T2
ON T1.RecordID = T2.RecordID AND T1.StorageDate = T2.MaxDate
GROUP BY T1.RecordID, T2.MaxDate
HAVING COUNT(*) > 1
)PT ON TD.RecordID = PT.RecordID AND TD.StorageDate = PT.MaxDate
WHERE TD.DuplicateTypeID = 2
Try this and see how the performance goes:
;WITH
tmp AS
(
SELECT *,
RANK() OVER (PARTITION BY ID ORDER BY StorageDate DESC) AS StorageDateRank,
COUNT(ID) OVER (PARTITION BY ID, StorageDate) AS StorageDateCount
FROM MyTable
)
SELECT DISTINCT ID
FROM tmp
WHERE StorageDateRank = 1 -- latest date for each ID
AND StorageDateCount > 1 -- more than 1 entry for date
AND DuplicateTypeID = 2 -- DuplicateTypeID = 2
You can use analytic function rank , can you try this query ?
Select recordId from
(
select *, rank() over ( partition by recordId order by [StorageDate] desc) as rn
from mytable
) T
where rn =1
group by recordId
having count(*) >1
and sum( case when duplicatetypeid =2 then 1 else 0 end) >=1
I have a table Attendance in my database.
Date | Present
------------------------
20/11/2013 | Y
21/11/2013 | Y
22/11/2013 | N
23/11/2013 | Y
24/11/2013 | Y
25/11/2013 | Y
26/11/2013 | Y
27/11/2013 | N
28/11/2013 | Y
I want to count the most consecutive occurrence of a value Y or N.
For example in the above table Y occurs 2, 4 & 1 times. So I want 4 as my result.
How to achieve this in SQL Server?
Any help will be appreciated.
Try this:-
The difference between the consecutive date will remain constant
Select max(Sequence)
from
(
select present ,count(*) as Sequence,
min(date) as MinDt, max(date) as MaxDt
from (
select t.Present,t.Date,
dateadd(day,
-(row_number() over (partition by present order by date))
,date
) as grp
from Table1 t
) t
group by present, grp
)a
where Present ='Y'
SQL FIDDLE
You can do this with a recursive CTE:
;WITH cte AS (SELECT Date,Present,ROW_NUMBER() OVER(ORDER BY Date) RN
FROM Table1)
,cte2 AS (SELECT Date,Present,RN,ct = 1
FROM cte
WHERE RN = 1
UNION ALL
SELECT a.Date,a.Present,a.RN,ct = CASE WHEN a.Present = b.Present THEN ct + 1 ELSE 1 END
FROM cte a
JOIN cte2 b
ON a.RN = b.RN+1)
SELECT TOP 1 *
FROM cte2
ORDER BY CT DESC
Demo: SQL Fiddle
Note, the date's in the demo got altered due to the format you posted the dates in your question.