simple SQL subquery - sql

It seems im a complete idiot when it comes to SQL....
All i need is get one value from other table, but there is multiple rows with same customerId on second table.. and i would need to get one with highest timestamp
CREATE OR REPLACE VIEW CUS_SETTINGS as
SELECT
c.id as id,
c.LANG as Language,
c.ALLOWEMAIL as AllowEmail,
l.CONFIRMED as confirmed
FROM cus.CUSTOMER c
????? something with l
/
LEFT JOIN will bring every row so i have multiple duplicate id's
What i need is propably subquery, but i cant get it to work...
(SELECT CONFIRMED FROM settings WHERE ?? c.id == l.id ?? AND MAX(TIMESTAMP) )
i've tried many many variations of joins and subqueries.. but for some reason.. SQL is just
too confusing....

You can use a ROW_NUMBER() in the subquery:
SELECT c.id as id, c.LANG as Language, c.ALLOWEMAIL as AllowEmail,
l.CONFIRMED as confirmed
FROM cus.CUSTOMER c JOIN
(SELECT s.*,
ROW_NUMBER() OVER (PARTITION BY s.id ORDER BY s.timestamp DESC) as seqnum
FROM settings s
) s
ON s.id = c.id AND s.seqnum = 1;
Note: You might want a LEFT JOIN if you want to keep all customers, even those with no settings.

You can use analytic functions. MAX() OVER(PARTITION BY) clause can give you max timestamped id.
Analytic Functions Docs 11gR2
SELECT SECONDD.CONFIRMED
FROM CUSTOMER CU
INNER JOIN
(SELECT
*
FROM (SELECT SECONDD.*,
MAX (S.TIMESTAMP) OVER (PARTITION BY S.ID)
AS MAXTIMESTAMP
FROM SETTINGS SECONDD)
WHERE TIMESTAMP = MAXTIMESTAMP) SECONDD
ON SECONDD.ID = CU.ID

Don't worry; while this sounds very basic, it isn't :-)
The easiest way to get the CONFIRMED for the latest TIMESTAMP in Oracle is KEEP LAST. E.g.:
SELECT customer_id, MAX(confirmed) KEEP (DENSE_RANK LAST ORDER BY timestamp)
FROM settings
GROUP BY customer_id;
The related CREATE VIEW statement:
CREATE OR REPLACE VIEW cus_settings as
SELECT
c.id AS id,
c.lang AS language,
c.allowemail AS allowemail,
l.last_confirmed AS confirmed
FROM cus.customer c
LEFT JOIN
(
SELECT
customer_id,
MAX(confirmed) KEEP (DENSE_RANK LAST ORDER BY timestamp) AS last_confirmed
FROM settings
GROUP BY customer_id;
) l ON l.customer_id = c.id;
Or:
CREATE OR REPLACE VIEW cus_settings as
SELECT
c.id AS id,
c.lang AS language,
c.allowemail AS allowemail,
(
SELECT MAX(confirmed) KEEP (DENSE_RANK LAST ORDER BY timestamp)
FROM settings s
WHERE s.customer_id = c.id
) AS confirmed
FROM cus.customer c;

Related

How to use DISTINCT ON but ORDER BY another expression?

The model Subscription has_many SubscriptionCart.
A SubscriptionCart has a status and an authorized_at date.
I need to pick the cart with the oldest authorized_at date from all the carts associated to a Subscription, and then I have to order all the returned Subscription results by this subscription_carts.authorized_at column.
The query below is working but I can't figure out how to select DISTINCT ON subscription.id to avoid duplicates but ORDER BY subscription_carts.authorized_at .
raw sql query so far:
select distinct on (s.id) s.id as subscription_id, subscription_carts.authorized_at, s.*
from subscriptions s
join subscription_carts subscription_carts on subscription_carts.subscription_id = s.id
and subscription_carts.plan_id = s.plan_id
where subscription_carts.status = 'processed'
and s.status IN ('authorized','in_trial', 'paused')
order by s.id, subscription_carts.authorized_at
If I try to ORDER BY subscription_carts.authorized_at first, I get an error because the DISTINCT ON and ORDER BY expressions must be in the same order.
The solutions I've found seem too complicated for what I need and I've failed to implement them because I don't understand them fully.
Would it be better to GROUP BY subscription_id and then pick from that group instead of using DISTINCT ON? Any help appreciated.
This requirement is necessary to make DISTINCT ON work; to change the final order, you can add an outer query with another ORDER BY clause:
SELECT *
FROM (SELECT DISTINCT ON (s.id)
s.id as subscription_id, subscription_carts.authorized_at, s.*
FROM subscriptions s
JOIN ...
WHERE ...
ORDER BY s.id, subscription_carts.authorized_at
) AS subq
ORDER BY authorized_at;
You don't have to use DISTINCT ON. While it is occasionally useful, I personally find window function based approaches much more clear:
-- Optionally, list all columns explicitly, to remove the rn column again
SELECT *
FROM (
SELECT
s.id AS subscription_id,
c.authorized_at,
s.*,
ROW_NUMBER () OVER (PARTITION BY s.id ORDER BY c.authorized_at) rn
FROM subscriptions s
JOIN subscription_carts c
ON c.subscription_id = s.id
AND c.plan_id = s.plan_id
WHERE c.status = 'processed'
AND s.status IN ('authorized', 'in_trial', 'paused')
) t
WHERE rn = 1
ORDER BY subscription_id, authorized_at

Sub query in a inner join

Well, in order to get the last entry by date i've done that query :
select cle
from ( select cle, clepersonnel, datedebut, row_number()
over(partition by clepersonnel order by datedebut desc) as rn
from periodeoccupation) as T
where rn = 1
This one is working and give me the last entry by date, so far i'm ok.
But its the first time i work with subquery and i have to make my query way more complex with multiple joins. But i cannot figure out how to make a subquery in a inner join.
This is what i try :
select personnel.prenom, personnel.nom
from personnel
inner join
( select cle, clepersonnel, datedebut, row_number()
over(partition by clepersonnel order by datedebut desc) as rn
from periodeoccupation) as T
ON personnel.cle = periodeoccupation.clepersonnel
where rn = 1
but its not working !
If you have any idea or tips...Thank you !
Simply change
ON personnel.cle = periodeoccupation.clepersonnel
to
ON personnel.cle = T.clepersonnel
The query join has been aliased with T and you have reference the alias as the table in the aliased query is out of scope in your outer statement.

Subquery Order by Oracle

My Database has 3 Tables, Device,Postal_Address and Company.
Every Device has an Company_ID, and for every company there is an address. The problem is, the old address is still there if i change it, its everytime making a new entry for this. So if i select all devices with address i get some entrys doubled.
select d.device,
(
select postalcode_address from
(
select
pa.postalcode_address,
row_number() over (order by pa.last_update desc) r,
pa.company_id
from
postal_address pa
)
where company_id=c.company_id
AND r=1
) as Postcode
from device d,company c,
where d.company_id = c.company_id(+)
I have tried it with order the address withlast_update, but its getting me nothing back, where is my error? so i need the newest entry in the postal_address for my company_id.
Your row_number() analytic call is getting the row number across all companies, so you'll only get a match for the most-recently-changed company; no other company will get r = 1. You need to partition by the company ID:
row_number() over (partition by pa.company_id order by pa.last_update desc) r,
Or move the where company_id=c.company_id inside that nested subquery; but that doesn't work in some versions of Oracle (the c alias might not be visible).
I wouldn't use a subquery to get the column value within the select list at all here though; you could use the same row number check as an inline view and join to it:
select d.device, pa.postalcode_address
from device d
left join company c on c.company_id = d.company_id
left join (
select company_id, postalcode_address,
row_number() over (partition by company_id order by last_update desc) rn
from postal_address
) pa on pa.company_id = c.company_id and pa.rn = 1;

Oracle: earliest date of stay (two columns)

i have three columns id_user, date_stay_start, date_stay_stop and I need to return complete date of the first stay (the oldest).
I am currently have something like this:
SELECT users.id_user, Min(orders.date_stay_start), Min(orders.date_stay_stop)
FROM users
JOIN orders ON users.id_user=orders.id_user
WHERE orders.date_stay_start IS NOT NULL
GROUP BY users.id_user;
But there is a problem. If I have two orders from one customer like this:
first order 02/Jun/2014 (date_stay_start) 10/Jun/2014
(date_stay_stop) and
second order 04/Jun/2014 (date_stay_start) 8/Jun/2014
(date_stay_stop).
The returned result will contain 02/Jun/2014 (date_stay_start) and 8/Jun/2014 (date_stay_stop).
But it should be 02/Jun/2014 (date_stay_start) 10/Jun/2014 (date_stay_stop).
I am using Oracle 11g. Thank you for any advice.
Something like this may work...
SELECT users.id_user, orders.date_stay_start, orders.date_stay_stop
FROM users
JOIN orders
ON users.id_user=orders.id_user
INNER JOIN (Select Min(Orders.date_Stay_start) DSS, id_user from orders group by ID_user) o
ON O.DSS = orders.date_stay_start
and O.id_user = orders.Id_user
WHERE orders.date_stay_start IS NOT NULL
GROUP BY users.id_user;
Or you can do something like this
select * from (
SELECT u.id_user
,o.date_stay_start
,o.date_stay_stop
,row_number() over (partition by u.id_user order by o.date_stay_start) rn
FROM users u
JOIN orders o
ON u.id_user = o.id_user
WHERE o.date_stay_start IS NOT NULL
) where rn = 1;

How to reference a custom field in SQL

I am using mssql and am having trouble using a subquery. The real query is quite complicated, but it has the same structure as this:
select
customerName,
customerId,
(
select count(*)
from Purchases
where Purchases.customerId=customerData.customerId
) as numberTransactions
from customerData
And what I want to do is order the table by the number of transactions, but when I use
order by numberTransactions
It tells me there is no such field. Is it possible to do this? Should I be using some sort of special keyword, such as this, or self?
use the field number, in this case:
order by 3
Sometimes you have to wrestle with SQL's syntax (expected scope of clauses)
SELECT *
FROM
(
select
customerName,
customerId,
(
select count(*)
from Purchases
where Purchases.customerId=customerData.customerId
) as numberTransactions
from customerData
) as sub
order by sub.numberTransactions
Also, a solution using JOIN is correct. Look at the query plan, SQL Server should give identical plans for both solutions.
Do an inner join. It's much easier and more readable.
select
customerName,
customerID,
count(*) as numberTransactions
from
customerdata c inner join purchases p on c.customerID = p.customerID
group by customerName,customerID
order by numberTransactions
EDIT: Hey Nathan,
You realize you can inner join this whole table as a sub right?
Select T.*, T2.*
From T inner join
(select
customerName,
customerID,
count(*) as numberTransactions
from
customerdata c inner join purchases p on c.customerID = p.customerID
group by customerName,customerID
) T2 on T.CustomerID = T2.CustomerID
order by T2.numberTransactions
Or if that's no good you can construct your queries using temporary tables (#T1 etc)
There are better ways to get your result but just from your example query this will work on SQL2000 or better.
If you wrap your alias in single ticks 'numberTransactions' and then call ORDER BY 'numberTransactions'
select
customerName,
customerId,
(
select count(*)
from Purchases
where Purchases.customerId=customerData.customerId
) as 'numberTransactions'
from customerData
ORDER BY 'numberTransactions'
The same thing could be achieved by using GROUP BY and a JOIN, and you'll be rid of the subquery. This might be faster too.
I think you can do this in SQL2005, but not SQL2000.
You need to duplicate your logic. SQL Server isn't very smart at columns that you've named but aren't part of the dataset in your FROM statement.
So use
select
customerName,
customerId,
(
select count(*)
from Purchases p
where p.customerId = c.customerId
) as numberTransactions
from customerData c
order by (select count(*) from purchases p where p.customerID = c.customerid)
Also, use aliases, they make your code easier to read and maintain. ;)