different conditions on different groups - sql

I have a table like this. I would like to get the first row grouped by id where acc1 is not null and if all the rows in the acc1 are null then i would want to get all the rows.
id acc1 acc2
1 null 1
1 1 1
1 1 2
1 2 2
2 null 1
2 null 2
2 null 3
2 null 4
I would like to get an output like this:
id acc1 acc2
1 1 1
2 null 1
2 null 2
2 null 3
2 null 4

Assuming acc1 is unique when not null (for each id):
select t.*
from (select t.*,
rank() over (partition by id
order by (case when acc1 is null then 2 else 1 end), acct1
) as seqnum
from t
) t
where seqnum = 1;
If it is not unique, this just requires a little more work:
select t.*
from (select t.*,
row_number() over (partition by id
order by acct1, acct2
) as seqnum,
count(acct1) over (partition by id) as cnt
from t
) t
where seqnum = 1 or cnt = 0;
This assumes that "first" is based on acct1, acct2. SQL tables are inherently unordered, so you need a column that specifies the ordering.

SELECT *
FROM mytable
QUALIFY Max(acc1) Over (PARTITION BY id) IS NULL -- only NULLs
OR Row_Number() Over (PARTITION BY id -- or the first non-null value
ORDER BY acc1 NULLS LAST) = 1

Related

How to Generate Row number Partition by two column match in sql

Tbl1
---------------------------------------------------------
Id Date Qty ReOrder
---------------------------------------------------------
1 1-1-18 1 3
2 2-1-18 0 3
3 3-1-18 2 3
4 4-1-18 3< >3
5 5-1-18 2 3
6 6-1-18 0 3
7 7-1-18 1 3
8 8-1-18 0 3
---------------------------------------------------------
I want the result like below
---------------------------------------------------------
Id Date Qty ReOrder
---------------------------------------------------------
1 1-1-18 1 3
5 5-1-18 2 3
---------------------------------------------------------
if ReOrder not same with Qty then date will be same upto after reorder=Qty
You can use cumulative approach with row_number() function :
select top (1) with ties *
from (select *, max(case when qty = reorder then 'v' end) over (order by id desc) grp
from table
) t
order by row_number() over(partition by grp order by id);
Unfortunately this will require SQL Server, But you can also do:
select *
from (select *, row_number() over(partition by grp order by id) seq
from (select *, max(case when qty = reorder then 'v' end) over (order by id desc) grp
from table
) t
) t
where seq = 1;

SQL: How do I display all records per unique id, but not the first record ever recorded in SQL

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

SQL update all records except the last one with a value

I need to make a query where only the last line of each user that has a car gets a license plate number.
ID UserId LicensePlate HasCar
1 1 ABC123 1
2 1 ABC123 1
3 2 NULL 0
4 3 UVW789 1
5 3 UVW789 1
Should become:
ID UserId LicensePlate HasCar
1 1 NULL 1
2 1 ABC123 1
3 2 NULL 0
4 3 NULL 1
5 3 UVW789 1
So I basically need to find all users with a licenseplate and change all but the last one and make the LicensePlate NULL
Assuming the ID column is an identity column so it can provide the ordering, something like this should do the trick:
;WITH CTE AS
(
SELECT Id,
UserId,
LicensePlate,
ROW_NUMBER() OVER(PARTITION BY UserId ORDER BY Id DESC) rn
FROM Table
WHERE HasCar = 1
)
UPDATE CTE
SET LicensePlate = NULL
WHERE rn > 1
You can try this
UPDATE l
SET l.LicensePlate = null
FROM Car l
INNER JOIN (SELECT UserId, Max(Id) AS max_id
FROM Car
GROUP BY UserId) m ON m.UserId = l.UserId
AND m.max_id <> l.id
You can do it with a join on the table itself like that :
UPDATE car c
INNER JOIN car c2 ON c.userId = c2.userId AND c.id < c2.id AND c.HasCar = 1 AND c2.HasCar = 1
SET c.LicensePlate = NULL
The condition c.id < c2.id will avoid to select the last line
By using LAG Function also you can achieve it.
;WITH License(ID,UserId,LicensePlate,HasCar)
as
(
SELECT 1,1,'ABC123',1 UNION ALL
SELECT 2,1,'ABC123',1 UNION ALL
SELECT 3,2,NULL ,0 UNION ALL
SELECT 4,3,'UVW789',1 UNION ALL
SELECT 5,3,'UVW789',1
)
SELECT ID,UserId,LAG(LicensePlate,1,NULL) OVER(PARTITION BY UserId ORDER BY LicensePlate),HasCar FROM License

Case statements: SQL Server

id date value
------------------
1 1 null
1 2 a
1 3 b
1 4 null
2 1 null
2 2 null
2 3 null
2 4 null
2 5 null
If value is null in all id's then max of date of that id and if we have value then max of date with value id.
Required output is:
id date value
-----------------
1 3 b
2 5 null
Typical method for this type of problem is row_number(). You can create a CASE expression to define a priority:
select id,
date,
value
from (
select id,
date,
value,
row_number() over (partition by id order by case when value is not null then 1 else 2 end asc, date desc) rn
from UnnamedTable
) t1
where t1.rn = 1
Sql Fiddle Demo
WITH cte as (
SELECT id,
[date],
[value],
ROW_NUMBER() OVER (PARTITION BY [ID] ORDER BY [value] DESC, [date] DESC) as rn
FROM Table1
)
SELECT *
FROM cte
WHERE rn = 1

Select and aggregate last records base on order

I have different versions of the charges in a table. I want to grab and sum the last charge grouped by Type.
So I want to add 9.87, 9.63, 1.65.
I want the Parent ID , sum(9.87 + 9.63 + 1.65) as the results of this query.
We use MSSQL
ID ORDER CHARGES TYPE PARENT ID
1 1 6.45 1 1
2 2 1.25 1 1
3 3 9.87 1 1
4 1 6.54 2 1
5 2 5.64 2 1
6 3 0.84 2 1
7 4 9.63 2 1
8 1 7.33 3 1
9 2 5.65 3 1
10 3 8.65 3 1
11 4 5.14 3 1
12 5 1.65 3 1
WITH recordsList
AS
(
SELECT Type, Charges,
ROW_NUMBER() OVER (PArtition BY TYPE
ORDER BY [ORDER] DESC) rn
FROM tableName
)
SELECT SUM(Charges) totalCharge
FROM recordsLIst
WHERE rn = 1
SQLFiddle Demo
Use row_number() to identify the rows to be summed, and then sum them:
select SUM(charges)
from (select t.*,
ROW_NUMBER() over (PARTITION by type order by id desc) as seqnum
from t
) t
where seqnum = 1
Alternatively you could use a window aggregate MAX():
SELECT SUM(Charges)
FROM (
SELECT
[ORDER],
Charges,
MaxOrder = MAX([ORDER]) OVER (PARTITION BY [TYPE])
FROM atable
) s
WHERE [ORDER] = MaxOrder
;
SELECT t.PARENT_ID, SUM(t.CHARGES)
FROM dbo.test73 t
WHERE EXISTS (
SELECT 1
FROM dbo.test73
WHERE [TYPE] = t.[TYPE]
HAVING MAX([ORDER]) = t.[ORDER]
)
GROUP BY t.PARENT_ID
Demo on SQLFiddle