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)
)
Related
I have table in Teradata SQL like below:
ID trans_date
------------------------
123 | 2021-01-01
887 | 2021-01-15
123 | 2021-02-10
45 | 2021-03-11
789 | 2021-10-01
45 | 2021-09-02
And I need to calculate average monthly number of transactions made by customers in a period between 2021-01-01 and 2021-09-01, so client with "ID" = 789 will not be calculated because he made transaction later.
In the first month (01) were 2 transactions
In the second month was 1 transaction
In the third month was 1 transaction
In the nineth month was 1 transactions
So the result should be (2+1+1+1) / 4 = 1.25, isn't is ?
How can I calculate it in Teradata SQL? Of course I showed you sample of my data.
SELECT ID, AVG(txns) FROM
(SELECT ID, TRUNC(trans_date,'MON') as mth, COUNT(*) as txns
FROM mytable
-- WHERE condition matches the question but likely want to
-- use end date 2021-09-30 or use mth instead of trans_date
WHERE trans_date BETWEEN date'2021-01-01' and date'2021-09-01'
GROUP BY id, mth) mth_txn
GROUP BY id;
Your logic translated to SQL:
--(2+1+1+1) / 4
SELECT id, COUNT(*) / COUNT(DISTINCT TRUNC(trans_date,'MON')) AS avg_tx
FROM mytable
WHERE trans_date BETWEEN date'2021-01-01' and date'2021-09-01'
GROUP BY id;
You should compare to Fred's answer to see which is more efficent on your data.
every customer has different first-time purchase date, I want to COUNT the number of purchases they have between the following 10 months after the first purchase?
sample table
TransactionID Client_name PurchaseDate Revenue
11 John Lee 10/13/2014 327
12 John Lee 9/15/2015 873
13 John Lee 11/29/2015 1,938
14 Rebort Jo 8/18/2013 722
15 Rebort Jo 5/21/2014 525
16 Rebort Jo 2/4/2015 455
17 Rebort Jo 3/20/2016 599
18 Tina Pe 10/8/2014 213
19 Tina Pe 6/10/2016 3,494
20 Tina Pe 8/9/2016 411
my code below just use ROW_NUM function to identify the first purchase, but I don't know how to do the calculations or there's a better way to do it?
SELECT client_name,
purchasedate,
Dateadd(month, 10, purchasedate) TenMonth,
Row_number()
OVER (
partition BY client_name
ORDER BY client_name) RM
FROM mytable
You might try something like this - I assume you're using SQL Server from the presence of DATEADD() and the fact that you're using a window function (ROW_NUMBER()):
WITH myCTE AS (
SELECT TransactionID, Client_name, PurchaseDate, Revenue
, MIN(PurchaseDate) OVER ( PARTITION BY Client_name ) AS min_PurchaseDate
FROM myTable
)
SELECT Client_name, COUNT(*)
FROM myCTE
WHERE PurchaseDate <= DATEADD(month, 10, min_PurchaseDate)
GROUP BY Client_name
Here I'm creating a common table expression (CTE) with all the data, including the date of first purchase, then I grab a count of all the purchases within a 10-month timeframe.
Hope this helps.
Give this a whirl ... Subquery to get the min purchase date, then LEFT JOIN to the main table to have a WHERE clause for the ten month date range, then count.
SELECT Client_name, COUNT(mt.PurchaseDate) as PurchaseCountFirstTenMonths
FROM myTable mt
LEFT JOIN (
SELECT Client_name, MIN(PurchaseDate) as MinPurchaseDate GROUP BY Client_name) mtmin
ON mt.Client_name = mtmin.Client_name AND mt.PurchaseDate = mtmin.MinPurchaseDate
WHERE mt.PurchaseDate >= mtmin.MinPurchaseDate AND mt.PurchaseDate <= DATEADD(month, 10, mtmin.MinPurchaseDate)
GROUP BY Client_name
ORDER BY Client_name
btw I'm guessing there's some kind of ClientID involved, as nine character full name runs the risk of duplicates.
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
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 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.