I have this query
SELECT table_article.articleID, table_article_to_date.sellDate, MIN(table_article.price) AS minPrice
FROM table_article table_article
LEFT JOIN table_article_to_date table_article_to_date ON (table_article.ord_no=table_article_to_date.ord_no)
WHERE table_article.price > 0 AND table_article_to_date.sellDate BETWEEN_TWO_DATES
GROUP BY table_article.articleID, table_article_to_date.sellDate, table_article.price
For the sell_date I use a time range. My problem is, that i get more than one entrie each articleID.
I wish to have the lowest price of each articleID in a specified time range. DISTINCT is not woking with MIN
Givs a change to make this with a query?
The problem here is that you are adding the sell date to the GROUP BY clause. TO solve the issue, you need to take it out and make use of subqueries to get it back. To achieve this, you can do an inner join of the query and a query with the id, sell date and prize on the id and prize.
Also, no need for the prize in the group by, since it's already in the MIN.
SELECT articleData.articleId, articleData.sellDate, articleData.price FROM
(
SELECT articleId, MIN(price)
FROM table
[...]
GROUP BY articleId
) AS minPrice
INNER JOIN
(
SELECT articleId, sellDate, price
FROM table
) AS articleData
ON minPrice.price = articleData.price AND minPrice.articleId = articleData.articleId
Related
I am trying to pull the most recent sale amount for each salesperson. The salespeople have made a sale on multiple days, I only want the most recent one.
My attempt below:
SELECT salesperson, amount
FROM table
WHERE date = (SELECT MAX(date) FROM table);
Use correlated subquery :
SELECT t.salesperson, t.amount
FROM table t
WHERE t.date = (SELECT MAX(t1.date)
FROM table t1
WHERE t1.salesperson = t.salesperson -- for each salesperson
);
If you are using PostgreSQL, you can take advantage of DISTINCT ON:
SELECT DISTINCT ON (salesperson) salesperson, amount
FROM table t
ORDER BY salesperson, date DESC
This will return only one row for each salesperson. The ORDER BY clause says to return the one with the largest date for that salesperson.
Unfortunately, DISTINCT ON is not supported by other databases.
I have a very simplified table / view like below to illustrate the issue:
The stock column represents the current stock quantity of the style at the retailer. The reason the stock column is included is to avoid joins for reporting. (the table is created for reporting only)
I want to query the table to get what is currently in stock, grouped by stylenumber (across retailers). Like:
select stylenumber,sum(sold) as sold,Max(stock) as stockcount
from MGTest
I Expect to get Stylenumber, Total Sold, Most Recent Stock Total:
A, 6, 15
B, 1, 6
But using ...Max(Stock) I get 10, and with (Sum) I get 25....
I have tried with over(partition.....) also without any luck...
How do I solve this?
I would answer this using window functions:
SELECT Stylenumber, Date, TotalStock
FROM (SELECT M.Stylenumber, M.Date, SUM(M.Stock) as TotalStock,
ROW_NUMBER() OVER (PARTITION BY M.Stylenumber ORDER BY M.Date DESC) as seqnum
FROM MGTest M
GROUP BY M.Stylenumber, M.Date
) m
WHERE seqnum = 1;
The query is a bit tricky since you want a cumulative total of the Sold column, but only the total of the Stock column for the most recent date. I didn't actually try running this, but something like the query below should work. However, because of the shape of your schema this isn't the most performant query in the world since it is scanning your table multiple times to join all of the data together:
SELECT MDate.Stylenumber, MDate.TotalSold, MStock.TotalStock
FROM (SELECT M.Stylenumber, MAX(M.Date) MostRecentDate, SUM(M.Sold) TotalSold
FROM [MGTest] M
GROUP BY M.Stylenumber) MDate
INNER JOIN (SELECT M.Stylenumber, M.Date, SUM(M.Stock) TotalStock
FROM [MGTest] M
GROUP BY M.Stylenumber, M.Date) MStock ON MDate.Stylenumber = MStock.Stylenumber AND MDate.MostRecentDate = MStock.Date
You can do something like this
SELECT B.Stylenumber,SUM(B.Sold),SUM(B.Stock) FROM
(SELECT Stylenumber AS 'Stylenumber',SUM(Sold) AS 'Sold',MAX(Stock) AS 'Stock'
FROM MGTest A
GROUP BY RetailerId,Stylenumber) B
GROUP BY B.Stylenumber
if you don't want to use joins
My solution, like that of Gordon Linoff, will use the window functions. But in my case, everything will turn around the RANK window function.
SELECT stylenumber, sold, SUM(stock) totalstock
FROM (
SELECT
stylenumber,
SUM(sold) OVER(PARTITION BY stylenumber) sold,
RANK() OVER(PARTITION BY stylenumber ORDER BY [Date] DESC) r,
stock
FROM MGTest
) T
WHERE r = 1
GROUP BY stylenumber, sold
I need help on a course question for my school. So I am supposed to get two tables Seller & Item, and I need to return the most active seller based on the most items offered. I have the tables as links below.
How could I just return one record with the sellers ID# and Name? Do I need to do a subquery? Thank you so much in advance.
Actually this is a way to accomplish it via subquery. I don't know that any teacher would anticipate students using >= all though:
select s.sellerid, min(s.name) as name
from seller s inner join item i on i.sellerid = s.sellerid
group by s.sellerid
having count(*) >= all (
select count(*)
from item
group by sellerid
)
You can also do it doubly=nested without even needing aliases!
select * from seller where sellerid in
(
select sellerid from item group by sellerid
having count(*) >= all (select count(*) from item group by sellerid)
)
I have a Select query to extract Customer Names and Purchase Dates from a table. My goal is to select only those names and dates for customers who have ordered on more than one distinct date. My code is as follows:
SELECT Customer, PurchDate
FROM (SELECT DISTINCT PurchDate, Customer
FROM (SELECT CDate(FORMAT(DateAdd("h",-7,Mid([purchase-date],1,10)+""+Mid([purchase-date],12,8)), "Short Date")) AS PurchDate,
[buyer-name] AS Customer
FROM RawImport
WHERE sku ALIKE "%RE%"))
GROUP BY Customer, PurchDate
HAVING COUNT(PurchDate)>1
ORDER BY PurchDate
This returns no results, even though there are many customers with more than one Purchase Date. The inner two Selects work perfectly and return a set of distinct dates for each customer, so I believe there is some problem in my GROUP/HAVING/ORDER clauses.
Thanks in advance for any help!
You are doing in the inner select
SELECT DISTINCT PurchDate, Customer
and in the outter select
GROUP BY Customer, PurchDate
That mean all are
having count(*) = 1
I cant give you the exact sintaxis in access but you need something like this
I will use YourTable as a replacement of your inner derivated table to make it easy to read
SELECT DISTINCT Customer, PurchDate
FROM YourTable
WHERE Customer IN (
SELECT Customer
FROM (SELECT DISTINCT Customer, PurchDate
FROM YourTable)
GROUP BY Customer
HAVING COUNT(*) > 1
)
inner select will give you which customer order on more than one day.
outside select will bring you those customer on all those days.
.
Maybe you can try something simple to get the list of customer who brought in more than one day like this
SELECT [buyer-name]
FROM RawImport
WHERE sku ALIKE "%RE%"
GROUP BY [buyer-name]
HAVING Format(MAX(purchase-date,"DD/MM/YYYY")) <>
Format(MIN(purchase-date,"DD/MM/YYYY"))
I wish to get the latest date per product ID:
SELECT ProductID, Max(Date) as Date FROM MyTable GROUP BY ProductID
I was always under the impression I had to do a nested join, but this appears to work!
EDIT: Thanks for all replies
That's what aggregates do!
For each unique value of ProductId, return the maximum Date value. Not nested JOIN is necessary.
It is correct. It returns the biggest value of the column Date of all rows with the same ProductID.
Yes, per ProductId being exact. If it represents Product and there exists Date for this product - then completely YES
It is correct. You would need to use a nested JOIN if you were wanting to retrieve other columns in addition to the ProductID columns from the table. E.g:
SELECT ProductID, Quantity, ...
FROM MyTable
JOIN
(
SELECT ProductID, Max(Date) AS Date
FROM MyTable
GROUP BY ProductID
) T1 ON MyTable.ProductID = T1.ProductID AND MyTable.Date = T1.Date;