I have one table with following data..
saleId amount date
-------------------------
1 2000 10/10/2012
2 3000 12/10/2012
3 2000 11/12/2012
2 3000 12/10/2012
1 4000 11/10/2012
4 6000 10/10/2012
From my table I want result with max of sum amount between dates 10/10/2012 and 12/10/2012 which for the data above will be:
saleId amount
---------------
1 6000
2 6000
4 6000
Here 6000 is the max of the sums (by saleId) so I want ids 1, 2 and 4.
You have to use Sub-queries like this:
SELECT saleId , SUM(amount) AS Amount
FROM Table1
GROUP BY saleId
HAVING SUM(amount) =
(
SELECT MAX(AMOUNT) FROM
(
SELECT SUM(amount) AS AMOUNT FROM Table1
WHERE date BETWEEN '10/10/2012' AND '12/10/2012'
GROUP BY saleId
) AS A
)
See this SQLFiddle
This query goes through the table only once and is fairly optimised.
select top(1) with ties saleid, amount
from (
select saleid, sum(amount) amount
from tbl
where date between '20121010' and '20121210'
group by saleid
) x
order by amount desc;
You can produce the SUM with the WHERE clause as a derived table, then SELECT TOP(1) in the query using WITH TIES to show all the ones with the same (MAX) amount.
When presenting dates to SQL Server, try to always use the format YYYYMMDD for robustness.
Related
The below Product table, Product ID - 100 as duplicated twice, and also there are negative profits are needs to Substract while calculating the Profit wise Total.
PID | Pname | Profit
100 AB 20
100 AB 20
101 BC 30
102 CD -10
103 DE -10
Expected Result: 30
Please provide the SQL query to get this result. Thanks in advance!!!
Is this what you want?
select sum(profit)
from (select distinct t.*
from t
) t
WITH CTE AS (
SELECT ROW_NUMBER() OVER (PARTITION BY PID ORDER BY PID ) AS rn,
PID,Pname,Profit FROM TableName
)
SELECT CAST(SUM(Profit) AS INT) AS Profit FROM CTE
WHERE rn=1
Note:- First you to get the DISTINCT Record then..use sum function...
In a firebird database with a table "Sales", I need to select the first sale of all customers. See below a sample that show the table and desired result of query.
---------------------------------------
SALES
---------------------------------------
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
3 25 05/04/16 08:10
4 31 07/03/16 10:22
5 22 01/02/16 12:30
6 22 10/01/16 08:45
Result: only first sale, based on sale date.
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
4 31 07/03/16 10:22
6 22 10/01/16 08:45
I've already tested following code "Select first row in each GROUP BY group?", but it did not work.
In Firebird 2.5 you can do this with the following query; this is a minor modification of the second part of the accepted answer of the question you linked to tailored to your schema and requirements:
select x.id,
x.customerid,
x.dthrsale
from sales x
join (select customerid,
min(dthrsale) as first_sale
from sales
group by customerid) p on p.customerid = x.customerid
and p.first_sale = x.dthrsale
order by x.id
The order by is not necessary, I just added it to make it give the order as shown in your question.
With Firebird 3 you can use the window function ROW_NUMBER which is also described in the linked answer. The linked answer incorrectly said the first solution would work on Firebird 2.1 and higher. I have now edited it.
Search for the sales with no earlier sales:
SELECT S1.*
FROM SALES S1
LEFT JOIN SALES S2 ON S2.CUSTOMERID = S1.CUSTOMERID AND S2.DTHRSALE < S1.DTHRSALE
WHERE S2.ID IS NULL
Define an index over (customerid, dthrsale) to make it fast.
in Firebird 3 , get first row foreach customer by min sales_date :
SELECT id, customer_id, total, sales_date
FROM (
SELECT id, customer_id, total, sales_date
, row_number() OVER(PARTITION BY customer_id ORDER BY sales_date ASC ) AS rn
FROM SALES
) sub
WHERE rn = 1;
İf you want to get other related columns, This is where your self-answer fails.
select customer_id , min(sales_date)
, id, total --what about other colums
from SALES
group by customer_id
So simple as:
select CUSTOMERID min(DTHRSALE) from SALES group by CUSTOMERID
I have a table SalePrices in SQL server and data same as below:
SPID ProductID Price Date
001 Pro01 10 2016-03-10
002 Pro01 20 2016-03-11
003 Pro02 10 2016-03-13
004 Pro02 20 2016-03-15
What I want is create a view that show only one ProductID and Price that I have modified at the last time. So what I want is same as the result below:
ProductID Price Date
Pro01 20 2016-03-11
Pro02 20 2016-03-15
There're few different approaches for this, for example, using row_number():
;with cte as (
select
ProductID, Price, Date,
row_number() over(partition by ProductID order by Date desc) as rn
from <Table>
)
select
ProductID, Price, Date
from cte
where
rn = 1
sql fiddle demo
Another version with windowing functions, this one with FIRST_VALUE();
SELECT ProductID, price, date
FROM products
WHERE spid IN (
SELECT FIRST_VALUE(spid) OVER (PARTITION BY ProductID ORDER BY date DESC) spid
FROM products
)
An SQLfiddle to test with.
Note that Roman's version with ROW_NUMBER should work from SQL Server 2005 and newer, while this will only work for SQL Server 2012 and newer.
TRY THIS:
SELECT
ProductID
, Price
, Date FROM tablename AS A
JOIN (SELECT ProductID,MAX(Date) AS DATE FROM tablename
GROUP BY ProductID
) AS B ON A.Date=B.DATE AND A.ProductID=B.ProductID
one more approach...
select productid,price,date
from
table t1
where date=(select max(date) from table t2 where t1.productid=t2.productid)
Your last record will have the highest SPID:
select
ProductId, Price, Date
from
SalePrices sap
where
sap.spid =(
select
max(sap2.spid)
from
SalePrices sap2
where
sap2.productId = sap.productId)
This query will give u desired result:
ProductID Price Date
Pro01 20 2016-03-11
Pro02 20 2016-03-15
My table looks something like this:
group date cash checks
1 1/1/2013 0 0
2 1/1/2013 0 800
1 1/3/2013 0 700
3 1/1/2013 0 600
1 1/2/2013 0 400
3 1/5/2013 0 200
-- Do not need cash just demonstrating that table has more information in it
I want to get the each unique group where date is max and checks is greater than 0. So the return would look something like:
group date checks
2 1/1/2013 800
1 1/3/2013 700
3 1/5/2013 200
attempted code:
SELECT group,MAX(date),checks
FROM table
WHERE checks>0
GROUP BY group
ORDER BY group DESC
problem with that though is it gives me all the dates and checks rather than just the max date row.
using ms sql server 2005
SELECT group,MAX(date) as max_date
FROM table
WHERE checks>0
GROUP BY group
That works to get the max date..join it back to your data to get the other columns:
Select group,max_date,checks
from table t
inner join
(SELECT group,MAX(date) as max_date
FROM table
WHERE checks>0
GROUP BY group)a
on a.group = t.group and a.max_date = date
Inner join functions as the filter to get the max record only.
FYI, your column names are horrid, don't use reserved words for columns (group, date, table).
You can use a window MAX() like this:
SELECT
*,
max_date = MAX(date) OVER (PARTITION BY group)
FROM table
to get max dates per group alongside other data:
group date cash checks max_date
----- -------- ---- ------ --------
1 1/1/2013 0 0 1/3/2013
2 1/1/2013 0 800 1/1/2013
1 1/3/2013 0 700 1/3/2013
3 1/1/2013 0 600 1/5/2013
1 1/2/2013 0 400 1/3/2013
3 1/5/2013 0 200 1/5/2013
Using the above output as a derived table, you can then get only rows where date matches max_date:
SELECT
group,
date,
checks
FROM (
SELECT
*,
max_date = MAX(date) OVER (PARTITION BY group)
FROM table
) AS s
WHERE date = max_date
;
to get the desired result.
Basically, this is similar to #Twelfth's suggestion but avoids a join and may thus be more efficient.
You can try the method at SQL Fiddle.
Using an in can have a performance impact. Joining two subqueries will not have the same performance impact and can be accomplished like this:
SELECT *
FROM (SELECT msisdn
,callid
,Change_color
,play_file_name
,date_played
FROM insert_log
WHERE play_file_name NOT IN('Prompt1','Conclusion_Prompt_1','silent')
ORDER BY callid ASC) t1
JOIN (SELECT MAX(date_played) AS date_played
FROM insert_log GROUP BY callid) t2
ON t1.date_played = t2.date_played
SELECT distinct
group,
max_date = MAX(date) OVER (PARTITION BY group), checks
FROM table
Should work.
I know it's just late in the day and my brain is just fried....
Using Teradata, I need to COUNT DISTINCT MEMBERS that haven't had a TRANS in the past six months and also COUNT the number of TRANS they had historically (prior to the six months). We can just assume the cutoff date to be 01/01/2012. All table is contained in a single table.
For example:
Member | Tran Date
123 | 01/01/2011
789 | 06/01/2011
123 |10/31/2011
678 | 04/03/2011
789 | 06/01/2012
So 2 members had a total of 3 transactions dated prior to 1/1/2012 with no transactions later than 1/1/2012.
In this example, my result would be:
MEMBERS | TRANS
2 | 3
Try this solution:
SELECT
COUNT(DISTINCT member_id) AS MEMBERS,
COUNT(*) AS TRANS
FROM
tbl
WHERE
member_id NOT IN
(
SELECT DISTINCT member_id
FROM tbl
WHERE trans_date > '2012-01-01'
)
You can't do it in one SQL statement. Use subqueries. This is TSQL coz I am unfamiliar with Teradata.
DECLARE #CUTOFF DATETIME = DATEADD(MO,-6,GETDATE()) --6MTHS AGO
SELECT COUNT(MEMBERID) AS MEMBERS, SUM(TRANSCOUNT) AS TRANS FROM (
SELECT DISTINCT
MEMBERID,
(SELECT COUNT(*) TRANSDATE WHERE TRANSDATA.MEMBERID = MEMBER.MEMBERIF) AS TRANSCOUNT
FROM MEMBER WHERE NOT EXISTS
(SELECT * FROM TRANSDATA, MEMBER WHERE
TRANSDATA.MEMBERID = MEMBER.MEMBERIF
AND TRANDATE > #CUTOFF)
)