It seems a dataset I've been given has multiple customer IDs with the same Order Numbers. How would I go about identifying those customer who have been effected?
I did try the usual but then realised some Order Numbers are duplicated but with the same Cus ID?
SELECT Cus ID, Ord Num, COUNT(*)
FROM OUTPUT
GROUP BY Cus ID, Order Num, Amount
HAVING COUNT(*) > 1
You can find the orders using a subquery:
SELECT OrdNum, COUNT(DISTINCT CustId)
FROM OUTPUT
GROUP BY OrdNum
HAVING COUNT(DISTINCT CustId) > 1
You can get the details by joining the information back in:
SELECT o.*
FROM (SELECT OrdNum, COUNT(DISTINCT CustId)
FROM OUTPUT
GROUP BY OrdNum
HAVING COUNT(DISTINCT CustId) > 1
) as oc INNER JOIN
OUTPUT as o
on oc.OrdNum = o.OrdNum
EDIT:
In Access, you can do:
SELECT o.*
FROM (SELECT OrdNum
FROM OUTPUT
GROUP BY OrdNum
HAVING MIN(CustId) <> MAX(CustId)
) as oc INNER JOIN
OUTPUT as o
on oc.OrdNum = o.OrdNum;
Related
I need to pull the second last record in a date column called OrderDate. However, I need to bring only one date (I am making the search into a table with all the purchases orders, dates and costs, in which a have to bring only the second last and its cost). The way its query is written today (and working) is pulling me the the newest date.
select distinct
a.PurchaseNum, a.ItemID, a.SupplierNum, a.Location, a.OrderDate, a.Cost
from
PurchaseOrder a
inner join
(select
l.SupplierNum, l.ItemID, l.Location, maxdate = max(l.OrderDate)
from
PurchaseOrder l
where
l.Cost <> 0
group by
l.SupplierNum, l.itemid, l.Location) l on a.SupplierNum = l.SupplierNumand a.itemid = l.itemid
and l.Location = a.Location
and a.OrderDate = l.maxdate
I have tried to use lag(), offset (but with limitations once is within a join, forcing me to use the order by and include the dateOrder column which is not what I want because we need only one date)
A bit of context: I have a report in which I need to show the last and second last cost of a purchase order for each supplier. Bring the last cost of an order is easy, the problem is go back to the second last... and it is where I am stuck right now.
Any thought?
If I'm understanding you correctly, here's one option using row_number to return the 2 highest orderdate records:
select *
from (
select *,
row_number() over (partition by SupplierNum, ItemID, Location
order by OrderDate desc) rn
from PurchaseOrder
where cost <> 0
) t
where rn <= 2
Inner query does order by desc and outside query does order by asc.
select distinct top 1 a.*
from PurchaseOrder a
inner join
(
select Top 2 l.*
from PurchaseOrder l
where
l.Cost <> 0
group by l.SupplierNum, l.itemid, l.Location order by orderdate desc) l
on a.SupplierNum= l.SupplierNumand a.itemid = l.itemid and l.Location=a.Location and a.OrderDate = l.Orderdate
order by a.orderdate
or
SELECT TOP 1 * FROM (SELECT * FROM PurchaseOrder a
EXCEPT SELECT TOP (SELECT (COUNT(*)-2) FROM PurchaseOrder a where
l.Cost <> 0
group by l.SupplierNum, l.itemid, l.Location) * FROM PurchaseOrder) A
or
SELECT *
FROM PurchaseOrder a
WHERE OrderDate = ( SELECT MAX(OrderDate)
FROM PurchaseOrder
WHERE Orderdate < ( SELECT MAX(OrderDate)
FROM PurchaseOrder l where
l.Cost <> 0
group by l.SupplierNum, l.itemid, l.Location
)
) ;
or
SELECT TOP (1) *
FROM PurchaseOrder
WHERE OrderDate < ( SELECT MAX(OrderDate)
FROM PurchaseOrder where ....
)
ORDER BY OrderDate DESC ;
I'm struggling to right a SQL command to get the top 10 names from the following (using standard SQL, cant use TOP) for the following 2 relations:
Orders (customer_email, item_id, date)
Items(id, name, store, price)
Any advice on how to do this? I think I would need to group them, but then what do I do to get the top 10 groupings based on count?
select *
from (select x.*, row_number() over(order by num_orders desc) as rn
from (select i.name, count(*) as num_orders
from orders o
join items i
on o.item_id = i.id
group by i.name) x) x
where rn <= 10
SELECT
COUNT(*) count_per_item
, i.id
, i.name
FROM
Orders o
JOIN
Items i
ON (o.item_id = i.id)
GROUP BY
i.id
, i.name
ORDER BY
count_per_item DESC
LIMIT 10;
I have a table that is having 2 duplicate rows (total of 3 rows), so I used the code below to get the duplicate value in the column
SELECT CustNo, COUNT(*) TotalCount
FROM Rental
GROUP BY CustNo
HAVING COUNT(*) > 1
ORDER BY COUNT(*) DESC
So once I get the repeated value, I need to get the CustNo derived as duplicate from the customer table. How do I go about taking this value and using it in the select statment all in the same query.
I also have the select statement prepared like this.
Select * from Customer where CustNo = 'T0002';
Thanks.
Select * from Customer
where CustNo IN
(
SELECT CustNo
FROM Rental
GROUP BY CustNo
HAVING COUNT(*) > 1
)
You can use join:
SELECT c.*
FROM (SELECT CustNo, COUNT(*) TotalCount
FROM Rental
GROUP BY CustNo
HAVING COUNT(*) > 1
) cc JOIN
Customer c
on cc.CustNo = c.CustNo;
Select C.* from Customer C RIGHT JOIN (
SELECT CustNo
FROM Rental
GROUP BY CustNo
HAVING COUNT(*) > 1) D
ON C.CustNo = D.CustNo
You can also try this,
With tblDups as(
select CustNo,count(CustNo) as TotalCount from a_rental
Group by CustNo
Having count(CustNo) >1)
select b.* from a_rental b
inner join tblDups a on a.CustNo = b.Custno
I need to list columns from customer table, the date from first order and all data from last one, in a 1:N relationship between customer and order tables. I'm using Oracle 10g.
How the best way to do that?
TABLE CUSTOMER
---------------
id NUMBER
name VARCHAR2(200)
subscribe_date DATE
TABLE ORDER
---------------
id NUMBER
id_order NUMBER
purchase_date DATE
purchase_value NUMBER
Here is one way of doing it, using the row_number function, one join, and on aggregation:
select c.*,
min(o.purchase_date) as FirstPurchaseDate,
min(case when seqnum = 1 then o.id_order end) as Last_IdOrder,
min(case when seqnum = 1 then o.purchase_date end) as Last_PurchaseDate,
min(case when seqnum = 1 then o.purchase_value end) as Last_PurchaseValue
from Customer c join
(select o.*,
row_number() over (partition by o.id order by purchase_date desc) as seqnum
from orders o
) o
on c.customer_id = o.order_id
group by c.customer_id, c.name, c.subscribe_date
It's not obvious how to join the customer table to the orders table (order is a reserved word in Oracle so your table can't be named order). If we assume that the id_order in orders joins to the id in customer
SELECT c.id customer_id,
c.name name,
c.subscribe_date,
o.first_purchase_date,
o.id last_order_id,
o.purchase_date last_order_purchase_date,
o.purchase_value last_order_purchase_value
FROM customer c
JOIN (SELECT o.*,
min(o.purchase_date) over (partition by id_order) first_purchase_date,
rank() over (partition by id_order order by purchase_date desc) rnk
FROM orders o) o ON (c.id = o.id_order)
WHERE rnk = 1
I'm confused by your field names, but I'm going to assume that ORDER.id is the id in the CUSTOMER table.
The earliest order date is easy.
select CUSTOMER.*, min(ORDER.purchase_date)
from CUSTOMER
inner join ORDER on CUSTOMER.id = ORDER.id
group by CUSTOMER.*
To get the last order data, join this to the ORDER table again.
select CUSTOMER.*, min(ORD_FIRST.purchase_date), ORD_LAST.*
from CUSTOMER
inner join ORDER ORD_FIRST on CUSTOMER.id = ORD_FIRST.id
inner join ORDER ORD_LAST on CUSTOMER.id = ORD_LAST.id
group by CUSTOMER.*, ORD_LAST.*
having ORD_LAST.purchase_date = max(ORD_FIRST.purchase_date)
Maybe something like this assuming the ID field in the Order table is actually the Customer ID:
SELECT C.*, O1.*, O2.purchase_Date as FirstPurchaseDate
FROM Customer C
LEFT JOIN
(
SELECT Max(purchase_date) as pdate, id
FROM Orders
GROUP BY id
) MaxPurchaseOrder
ON C.Id = MaxPurchaseOrder.Id
LEFT JOIN Orders O1
ON MaxPurchaseOrder.pdate = O1.purchase_date
AND MaxPurchaseOrder.id = O1.id
LEFT JOIN
(
SELECT Min(purchase_date) as pdate, id
FROM Orders
GROUP BY id
) MinPurchaseOrder
ON C.Id = MinPurchaseOrder.Id
LEFT JOIN Orders O2
ON MinPurchaseOrder.pdate = O2.purchase_date
AND MinPurchaseOrder.id = O2.id
And the sql fiddle.
If I turn this sub-query which selects sales persons and their highest price paid for any item they sell:
select *,
(select top 1 highestProductPrice
from orders o
where o.salespersonid = s.id
order by highestProductPrice desc ) as highestProductPrice
from salespersons s
in to this join in order to improve efficiency:
select *, highestProductPrice
from salespersons s join (
select salespersonid, highestProductPrice, row_number(
partition by salespersonid
order by salespersonid, highestProductPrice) as rank
from orders ) o on s.id = o.salespersonid
It still touches every order record (it enumerates the entire table before filtering by salespersonid it seems.) However you cannot do this:
select *, highestProductPrice
from salespersons s join (
select salespersonid, highestProductPrice, row_number(
partition by salespersonid
order by salespersonid, highestProductPrice) as rank
from orders
where orders.salepersonid = s.id) o on s.id = o.salespersonid
The where clause in the join causes a `multi-part identifier "s.id" could not be bound.
Is there any way to join the top 1 out of each order group with a join but without touching each record in orders?
Try
SELECT
S.*,
T.HighestProductPrice
FROM
SalesPersons S
CROSS APPLY
(
SELECT TOP 1 O.HighestProductPrice
FROM Orders O
WHERE O.SalesPersonid = S.Id
ORDER BY O.SalesPersonid, O.HighestProductPrice DESC
) T
would
select s.*, max(highestProductPrice)
from salespersons s
join orders o on o.salespersonid = s.id
group by s.*
or
select s.*, highestProductPrice
from salespersons s join (select salepersonid,
max(highestProductPrice) as highestProductPrice
from orders o) as o on o.salespersonid = s.id
work?