There are two tables:
Clients (id, name)
Order (id, id_client, name), where id_client - foreign key.
Write a query that selects the identifier and name of the first table and the number of records in the second table, associated with them. The result should be sorted by surname in descending order.
I've tried
SELECT
Clients.id, Clients.name, count(id)
FROM clients
INNER JOIN Order ON Clients.id = Order.id_client
GROUP BY
Clients.id, Clients.name
ORDER BY
Clients.name DESC
But it doesn't work. What is wrong?
SELECT
c.ID,
c.Name,
COUNT(o.ID)
FROM
Clients c
LEFT JOIN [Order] o
ON
o.id_client = c.id
GROUP BY
c.ID,
c.Name
ORDER BY
c.Name DESC
SELECT Clients.id, Clients.name, count(client.id) FROM clients INNER JOIN Order on Clients.id=Order.id_client GROUP BY Clients.id, Clients.name ORDER BY Clients.name DESC
Change count(id) to
count(Clients.id) or count(Order.id)
I don't know which table you need count(id) from. I hope you understand where the issue is.
SELECT
c.ID,
c.Name,
COUNT(o.ID)
FROM
Clients c,
Order o
WHERE o.id_client = c.id
GROUP BY
c.ID
c.Name
Related
I have a feeling I'll feel stupid when this is answered. I have a table of clients and a table of orders. I want a query that gives me a list of all clients, and their last order info if there is one, sorted by client name.
SELECT c.id, c.name, o.order_time, o.item_name
FROM clients AS c LEFT JOIN(
SELECT client_id, max(order_time) AS order_time
FROM orders GROUP BY client_id
) AS o
ON(c.id = o.client_id)
ORDER BY UPPER(c.name)"
My issue is I get the rows I want if I remove o.item_name but the query as written isn't valid because there's no way to get o.item_name without putting it in the GROUP BY. That, of course, causes it to return multiple rows per client. Hopefully my intent is clear.
You can do this using a window function:
SELECT c.id, c.name, o.order_time, o.item_name
FROM clients AS c
LEFT JOIN (
SELECT client_id,
item_name,
order_time,
row_number() over (partition by client_id order by order_time desc) as rn
FROM orders
) AS o ON c.id = o.client_id and o.rn = 1
ORDER BY UPPER(c.name);
another option is to use Postgres' distinct on() operator which is usually faster than a solution using window functions:
SELECT c.id, c.name, o.order_time, o.item_name
FROM clients AS c
LEFT JOIN (
SELECT distinct on (client_id) client_id,
item_name,
order_time
FROM orders
order by client_id, order_time desc
) AS o ON c.id = o.client_id
ORDER BY UPPER(c.name);
In Postgres, you can use distinct on:
SELECT DISTINCT ON (c.name) c.id, c.name, o.order_time, o.item_name
FROM clients c LEFT JOIN
orders o
ON c.id = o.client_id
ORDER BY UPPER(c.name), o.order_time DESC;
I seem to be running into a query syntax error but cannot seem to isolate it. I am using MS Access and when I run the queries I get a syntax error in FROM clause.
I have two tables and they are in a one to many relationship:
Table 1 called (customer) with the following fields:
ID
FirstName
Table 2 called (tblservice) with the following fields:
serviceID
Timing
Total
customerID <-Foreign Key
First Query:
select c.id, c.firstname, avg(s.Total) / (select count(id) from customer) as LifetimeValue
from tblservice as s join customer as c on s.id = c.id
group by s.id
Second Query(30 day span):
select c.id, c.firstname, avg(s.Total) / (select count(id) from customer) as LifetimeValue
from tblservice as s join customer as c on s.id = c.id
where (s.Timing)>=DateAdd("d",-30,Date())
group by s.id
Try this:
select c.id, c.firstname, avg(s.Total) / count(c.id) as LifetimeValue
from tblservice as s inner join customer as c on s.id = c.id
group by c.id, c.firstname
and
select c.id, c.firstname, avg(s.Total) / count(c.id) as LifetimeValue
from tblservice as s inner join customer as c on s.id = c.id
where (s.Timing)>=DateAdd("d",-30,Date())
group by c.id, c.firstname
You cannot select c.id and c.firstname unless you group by them. And you can use count(c.id) since you are already grouping by c.id. I have not used SQL in MS Access though. So I am not 100% sure. Try it out.
MS Access may require you to use INNER JOIN instead of just JOIN.
I'm trying to do something like:
SELECT c.id, c.name, COUNT(orders.id)
FROM customers c
JOIN orders o ON o.customerId = c.id
However, SQL will not allow the COUNT function. The error given at execution is that c.Id is not valid in the select list because it isn't in the group by clause or isn't aggregated.
I think I know the problem, COUNT just counts all the rows in the orders table. How can I make a count for each customer?
EDIT
Full query, but it's in dutch... This is what I tried:
select k.ID,
Naam,
Voornaam,
Adres,
Postcode,
Gemeente,
Land,
Emailadres,
Telefoonnummer,
count(*) over (partition by k.id) as 'Aantal bestellingen',
Kredietbedrag,
Gebruikersnaam,
k.LeverAdres,
k.LeverPostnummer,
k.LeverGemeente,
k.LeverLand
from klanten k
join bestellingen on bestellingen.klantId = k.id
No errors but no results either..
When using an aggregate function like that, you need to group by any columns that aren't aggregates:
SELECT c.id, c.name, COUNT(orders.id)
FROM customers c
JOIN orders o ON o.customerId = c.id
GROUP BY c.id, c.name
If you really want to be able to select all of the columns in Customers without specifying the names (please read this blog post in full for reasons to avoid this, and easy workarounds), then you can do this lazy shorthand instead:
;WITH o AS
(
SELECT CustomerID, CustomerCount = COUNT(*)
FROM dbo.Orders GROUP BY CustomerID
)
SELECT c.*, o.OrderCount
FROM dbo.Customers AS c
INNER JOIN dbo.Orders AS o
ON c.id = o.CustomerID;
EDIT for your real query
SELECT
k.ID,
k.Naam,
k.Voornaam,
k.Adres,
k.Postcode,
k.Gemeente,
k.Land,
k.Emailadres,
k.Telefoonnummer,
[Aantal bestellingen] = o.klantCount,
k.Kredietbedrag,
k.Gebruikersnaam,
k.LeverAdres,
k.LeverPostnummer,
k.LeverGemeente,
k.LeverLand
FROM klanten AS k
INNER JOIN
(
SELECT klantId, klanCount = COUNT(*)
FROM dbo.bestellingen
GROUP BY klantId
) AS o
ON k.id = o.klantId;
I think this solution is much cleaner than grouping by all of the columns. Grouping on the orders table first and then joining once to each customer row is likely to be much more efficient than joining first and then grouping.
The following will count the orders per customer without the need to group the overall query by customer.id. But this also means that for customers with more than one order, that count will repeated for each order.
SELECT c.id, c.name, COUNT(orders.id) over (partition by c.id)
FROM customers c
JOIN orders ON o.customerId = c.id
I need a help constructung an sql query for mysql database. 2 Table as follows:
tblcities (id,name)
tblmembers(id,name,city_id)
Now I want to retrieve the 'city' details that has maximum number of 'members'.
Regards
SELECT tblcities.id, tblcities.name, COUNT(tblmembers.id) AS member_count
FROM tblcities
LEFT JOIN tblmembers ON tblcities.id = tblmembers.city_id
GROUP BY tblcities.id
ORDER BY member_count DESC
LIMIT 1
Basically: retrieve all cities and count how many members each has, sort by that member count in descending order, making the highest count first - then show only that first city.
Terrible, but that's a way of doing it:
SELECT * FROM tblcities WHERE id IN (
SELECT city_id
FROM tblMembers
GROUP BY city_id
HAVING COUNT(*) = (
SELECT MAX(TOTAL)
FROM (
SELECT COUNT(*) AS TOTAL
FROM tblMembers
GROUP BY city_id
) AS AUX
)
)
That way, if there is a tie, still you'll get all cities with the maximum number of members...
Select ...
From tblCities As C
Join (
Select city_id, Count(*) As MemberCount
From tblMembers
Order By Count(*) Desc
Limit 1
) As MostMembers
On MostMembers.city_id = C.id
select top 1 c.id, c.name, count(*)
from tblcities c, tblmembers m
where c.id = m.city_id
group by c.id, c.name
order by count(*) desc
Say I have 2 tables: Customers and Orders. A Customer can have many Orders.
Now, I need to show any Customers with his latest Order. This means if a Customer has more than one Orders, show only the Order with the latest Entry Time.
This is how far I managed on my own:
SELECT a.*, b.Id
FROM Customer a INNER JOIN Order b ON b.CustomerID = a.Id
ORDER BY b.EntryTime DESC
This of course returns all Customers with one or more Orders, showing the latest Order first for each Customer, which is not what I wanted. My mind was stuck in a rut at this point, so I hope someone can point me in the right direction.
For some reason, I think I need to use the MAX syntax somewhere, but it just escapes me right now.
UPDATE: After going through a few answers here (there's a lot!), I realized I made a mistake: I meant any Customer with his latest record. That means if he does not have an Order, then I do not need to list him.
UPDATE2: Fixed my own SQL statement, which probably caused no end of confusion to others.
I don't think you do want to use MAX() as you don't want to group the OrderID. What you need is an ordered sub query with a SELECT TOP 1.
select *
from Customers
inner join Orders
on Customers.CustomerID = Orders.CustomerID
and OrderID = (
SELECT TOP 1 subOrders.OrderID
FROM Orders subOrders
WHERE subOrders.CustomerID = Orders.CustomerID
ORDER BY subOrders.OrderDate DESC
)
Something like this should do it:
SELECT X.*, Y.LatestOrderId
FROM Customer X
LEFT JOIN (
SELECT A.Customer, MAX(A.OrderID) LatestOrderId
FROM Order A
JOIN (
SELECT Customer, MAX(EntryTime) MaxEntryTime FROM Order GROUP BY Customer
) B ON A.Customer = B.Customer AND A.EntryTime = B.MaxEntryTime
GROUP BY Customer
) Y ON X.Customer = Y.Customer
This assumes that two orders for the same customer may have the same EntryTime, which is why MAX(OrderID) is used in subquery Y to ensure that it only occurs once per customer. The LEFT JOIN is used because you stated you wanted to show all customers - if they haven't got any orders, then the LatestOrderId will be NULL.
Hope this helps!
--
UPDATE :-) This shows only customers with orders:
SELECT A.Customer, MAX(A.OrderID) LatestOrderId
FROM Order A
JOIN (
SELECT Customer, MAX(EntryTime) MaxEntryTime FROM Order GROUP BY Customer
) B ON A.Customer = B.Customer AND A.EntryTime = B.MaxEntryTime
GROUP BY Customer
While I see that you've already accepted an answer, I think this one is a bit more intuitive:
select a.*
,b.Id
from customer a
inner join Order b
on b.CustomerID = a.Id
where b.EntryTime = ( select max(EntryTime)
from Order
where a.Id = b.CustomerId
);
a.Id = b.CustomerId because you want the max EntryTime of all orders (in b) for the customer (a.Id).
I would have to run something like this through an execution plan to see the difference in execution, but where the TOP function is done after-the-fact and that using order by can be expensive, I believe that using max(EntryTime) would be the best way to run this.
You can use a window function.
SELECT *
FROM (SELECT a.*, b.*,
ROW_NUMBER () OVER (PARTITION BY a.ID ORDER BY b.orderdate DESC,
b.ID DESC) rn
FROM customer a, ORDER b
WHERE a.ID = b.custid)
WHERE rn = 1
For each customer (a.id) it sorts all orders and discards everything but the latest.
ORDER BY clause includes both order date and entry id, in case there are multiple orders on the same date.
Generally, window functions are much faster than any look-ups using MAX() on large number of records.
This query is much faster than the accepted answer :
SELECT c.id as customer_id,
(SELECT co.id FROM customer_order co WHERE
co.customer_id=c.id
ORDER BY some_date_column DESC limit 1) as last_order_id
FROM customer c
SELECT Cust.*, Ord.*
FROM Customers cust INNER JOIN Orders ord ON cust.ID = ord.CustID
WHERE ord.OrderID =
(SELECT MAX(OrderID) FROM Orders WHERE Orders.CustID = cust.ID)
Something like:
SELECT
a.*
FROM
Customer a
INNER JOIN Order b
ON a.OrderID = b.Id
INNER JOIN (SELECT Id, max(EntryTime) as EntryTime FROM Order b GROUP BY Id) met
ON
b.EntryTime = met.EntryTime and b.Id = met.Id
One approach that I haven't seen above yet:
SELECT
C.*,
O1.ID
FROM
dbo.Customers C
INNER JOIN dbo.Orders O1 ON
O1.CustomerID = C.ID
LEFT OUTER JOIN dbo.Orders O2 ON
O2.CustomerID = C.ID AND
O2.EntryTime > O1.EntryTime
WHERE
O2.ID IS NULL
This (as well as the other solutions I believe) assumes that no two orders for the same customer can have the exact same entry time. If that's a concern then you would have to make a choice as to what determines which one is the "latest". If that's a concern post a comment and I can expand the query if needed to account for that.
The general approach of the query is to find the order for a customer where there is not another order for the same customer with a later date. It is then the latest order by definition. This approach often gives better performance then the use of derived tables or subqueries.
A simple max and "group by" is sufficient.
select c.customer_id, max(o.order_date)
from customers c
inner join orders o on o.customer_id = c.customer_id
group by c.customer_id;
No subselect needed, which slows things down.