I have the following sales table that displays the customer ID, their name, the order amount, and the order date.
ID
Name
Order
Date
1
A
25
11/10/2006
1
A
10
5/25/2010
1
A
10
6/18/2018
2
B
20
3/31/2008
2
B
15
11/15/2010
3
C
35
1/1/2019
3
C
20
4/12/2007
3
C
10
3/20/2010
3
C
5
10/19/2012
4
D
15
12/12/2013
4
D
15
2/18/2010
5
E
25
12/11/2006
6
F
10
5/1/2016
I am trying to group the data so that for each customer it would only show me their most recent order and the amount, as per below:
ID
Name
Order
Date
1
A
10
6/18/2018
2
B
15
11/15/2010
3
C
35
1/1/2019
4
D
15
12/12/2013
5
E
25
12/11/2006
6
F
10
5/1/2016
So far I've only been able to group by ID and Name, because adding the Order column would also group by that column as well.
SELECT
ID,
Name,
MAX(Date) 'Most recent date'
FROM Table
GROUP BY Customer, Customer
How can I also add the order amount for each Customer?
SELECT ID, Name, Order, Date FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Name ORDER BY Date DESC) AS sn
FROM your_table_name
) A WHERE sn = 1;
You could use a subqoery for max date
SELECT
ID,
Name,
MAX(Date) 'Most recent date'
FROM Table
GROUP BY Customer, Customer
select a.ID, a.Name, b.max_date
from Table a
inner join (
select name, max(Date) max_date
from Table
group by name
) b on a. name = b.name and a.date = b.max_date
You can use this query to get the expected result:
SELECT S.*
FROM Sales S
CROSS APPLY
(
SELECT ID, Max(Date) MaxDate
FROM Sales
GROUP BY ID
)T
WHERE S.ID = T.ID
AND S.Date = T.MaxDate
ORDER BY S.ID
Related
I have the following query
select S.id, X.id, 15,15,1 from schema_1.tbl_2638 S
JOIN schema_1.tbl_2634_customid X on S.field_1=x.fullname
That returns the following results, where you can see the first column is duplicated on matches to the 2nd table.
1 1 15 15 1
2 3 15 15 1
2 2 15 15 1
3 5 15 15 1
3 4 15 15 1
I'm trying to get a query that would just give me a single row per 1st ID, and the min value from 2nd ID. So I want a result that would be:
1 1 15 15 1
2 2 15 15 1
3 4 15 15 1
I'm a little rust on my SQL skills, how would I write the query to provide the above result?
From your result you can do,this to achieve your result, for much more compicated structures, you can always take a look at window fucntions
select S.id, MIN(X.id) x_id, 15,15,1 from schema_1.tbl_2638 S
JOIN schema_1.tbl_2634_customid X on S.field_1=x.fullname
GROUP BY 1,3,4,5
window function can be used, need always a outer SELECT
SELECT
s_id,x_idm a,b,c
FROM
(select S.id as s_id, X.id as x_id, 15 a ,15 b,1 c
, ROW_NUMBER() OVER (PARTITION BY S.id ORDER BY X.id ASC) rn
from schema_1.tbl_2638 S
JOIN schema_1.tbl_2634_customid X on S.field_1=x.fullname)
WHERE rn = 1
Or as CTE
WITH CTE as (select S.id as s_id, X.id as x_id, 15 a ,15 b,1 c
, ROW_NUMBER() OVER (PARTITION BY S.id ORDER BY X.id ASC) rn
from schema_1.tbl_2638 S
JOIN schema_1.tbl_2634_customid X on S.field_1=x.fullname)
SELECT s_id,x_id,a,b,c FROM CTE WHERE rn = 1
I need to make a PIVOT table from Source like this table
FactID UserID Date Product QTY
1 11 01/01/2020 A 600
2 11 02/01/2020 A 400
3 11 03/01/2020 B 500
4 11 04/01/2020 B 200
6 22 06/01/2020 A 1000
7 22 07/01/2020 A 200
8 22 08/01/2020 B 300
9 22 09/01/2020 B 100
Need Pivot Like this where Product QTY is QTY by Last Date
UserID A B
11 400 200
22 200 100
My try PostgreSQL
Select
UserID,
MAX(CASE WHEN Product='A' THEN 'QTY' END) AS 'A',
MAX(CASE WHEN Product='B' THEN 'QTY' END) AS 'B'
FROM table
GROUP BY UserID
And Result
UserID A B
11 600 500
22 1000 300
I mean I get a result by the maximum QTY and not by the maximum date!
What do I need to add to get results by the maximum (last) date ??
Postgres doesn't have "first" and "last" aggregation functions. One method for doing this (without a subquery) uses arrays:
select userid,
(array_agg(qty order by date desc) filter (where product = 'A'))[1] as a,
(array_agg(qty order by date desc) filter (where product = 'B'))[1] as b
from tab
group by userid;
Another method uses select distinct with first_value():
select distinct userid,
first_value(qty) over (partition by userid order by product = 'A' desc, date desc) as a,
first_value(qty) over (partition by userid order by product = 'B' desc, date desc) as b
from tab;
With the appropriate indexes, though, distinct on might be the fastest approach:
select userid,
max(qty) filter (where product = 'A') as a,
max(qty) filter (where product = 'B') as b
from (select distinct on (userid, product) t.*
from tab t
order by userid, product, date desc
) t
group by userid;
In particular, this can use an index on userid, product, date desc). The improvement in performance will be most notable if there are many dates for a given user.
You can use DENSE_RANK() window function in order to filter by the last date per each product and UserID before applying conditional aggregation such as
SELECT UserID,
MAX(CASE WHEN Product='A' THEN QTY END) AS "A",
MAX(CASE WHEN Product='B' THEN QTY END) AS "B"
FROM
(
SELECT t.*, DENSE_RANK() OVER (PARTITION BY Product,UserID ORDER BY Date DESC) AS rn
FROM tab t
) q
WHERE rn = 1
GROUP BY UserID
Demo
presuming all date values are distinct(no ties occur for dates)
I have 2 tables, one is credit and other one is creditdetails.
Creditdetails creates new row every day for each of credit.
ID Amount ref_id date
1 2 1 16.03
2 3 1 17.03
3 4 1 18.03
4 1 2 16.03
5 2 2 17.03
6 0 2 18.03
I want to sum up amount of every row with the unique id and last date. So the output should be 4 + 0.
You can use ROW_NUMBER to filter on the latest amount per ref_id.
Then SUM it.
SELECT SUM(q.Amount) AS TotalLatestAmount
FROM
(
SELECT
cd.ref_id,
cd.Amount,
ROW_NUMBER() OVER (PARTITION BY cd.ref_id ORDER BY cd.date DESC) AS rn
FROM Creditdetails cd
) q
WHERE q.rn = 1;
A test on db<>fiddle here
With this query:
select ref_id, max(date) maxdate
from creditdetails
group by ref_id
you get all the last dates for each ref_id, so you can join it to the table creditdetails and sum over amount:
select sum(amount) total
from creditdetails c inner join (
select ref_id, max(date) maxdate
from creditdetails
group by ref_id
) g
on g.ref_id = c.ref_id and g.maxdate = c.date
I think you want something like this,
select sum(amount)
from table
where date = ( select max(date) from table);
with the understanding that your date column doesn't appear to be in a standard format so I can't tell if it needs to be formatted in the query to work properly.
Order_ID
=========
id price
A 10
A 10
B 20
B 20
C 30
C 30
D 40
D 40
Client
==================
Client Name id
1 ClientInc. A
1 ClientInc. A
1 ClientInc. B
1 ClientInc. B
1 ClientInc. C
1 ClientInc. C
1 ClientInc. D
1 ClientInc. D
I have two tables that I need to join (Order_ID and Client) and want to sum the price by distinct order_ID and create the report below:
Desired Solution
========================
id Name Sum(Price)
1 ClientInc. 100
This is the current query I am using:
SELECT merchant,
name,
SUM(price)
FROM order_id a
JOIN client b
ON a.id = b.id
GROUP BY merchant, name
It is displaying the following output by summarizing every order_id, but the problem is that I want to SUM a distinct order ID:
Current Wrong Report
======================
id Name Sum(Price)
1 ClientInc. 200
SELECT merchant,
name,
SUM(price)
FROM ( SELECT DISTINCT id,price
FROM order_id
) a
JOIN client b
ON a.id = b.id
GROUP BY merchant, name;
Consider Table like this.
I will call it Test
Id A B C D
1 1 1 8 25
2 1 2 5 35
3 1 3 2 75
4 2 2 2 45
5 3 2 5 26
Now I want rows with max 'Id' Grouped by 'A'
Id A B C D
3 1 3 2 75
4 2 2 2 45
5 3 2 5 26
-
--Work, but I do not want
SELECT MAX(Id), A FROM Test GROUP BY A
--I want but do not work
SELECT MAX(Id), A, B, C, D FROM Test GROUP BY A
--Work but I do not want
SELECT MAX(Id), A, B, C, D FROM Test GROUP BY A, B, C, D
--Work and I want
SELECT old.Id, old.A, new.B, new.C, new.D
FROM(
SELECT
MAX(Id) AS Id, A
FROM
Test GROUP BY A
)old
JOIN Test new
ON old.Id = new.Id
Is there a better way to write last query without join
Most databases support window functions:
select *
from (
select *, row_number() over (partition by a order by id desc) rn
from test
) t
where rn = 1
Most DBMS now support Common Table Expressions (CTE). You can use one.
;with maxa as (
select row_number() over(partition by a order by id desc) rn,
id,a,b,c,d from test
)
select id,a,b,c,d
from maxa
where rn=1