I have got two tables, first one called card:
| card_id | card_name |
And table withdrawal:
| withdrawal_id | card_id | transaction_date |
I want to get the last four transaction_dates for each card_id.
I tried this code but it wouldn't give the last 4 dates:
SELECT a.card_id, b.transaction_date
FROM card AS a
JOIN withdrawal AS b ON a.card_id = b.card_id
GROUP BY a.card_id, b.transaction_date
ORDER BY a.card_id, b.transaction_date
What do I need to change?
It seems like you want to get the first N rows per group, which can be done using something like this:
SELECT a.card_id, b.transaction_date
FROM card a
JOIN withdrawal b ON a.card_id = b.card_id
WHERE(
SELECT COUNT(*)
FROM card c
JOIN withdrawal w ON w.card_id = c.card_id
WHERE c.card_id = a.card_id AND w.transaction_date <= b.transaction_date
) <= 4
ORDER BY a.card_id, b.transaction_date
See this question for more info on getting rows per group.
Here is another way of achieving it
select
w.withdrawal_id,
w.card_id,
w.transaction_date
from card c
left join
(
select
r1.*
from withdrawal r1
where
(
select count(*)
from withdrawal r2
where r1.card_id = r2.card_id
AND r1.transaction_date <= r2.transaction_date
) <=4
order by r1.transaction_date desc
)w
on w.card_id = c.card_id
order by c.card_id
demo
Related
i need a query that should first look the oldest order which has status 0 (zero). and retrieves all the similar orders of that kind(matches exact total qty, itemSku and number of distinct items ordered).
***OrdersTable***
ID OrderNumber CustomerId Status created_at
1 123456 1 0 2018-01-01
2 234567 1 0 2018-01-02
3 345678 1 0 2018-01-03
4 456789 1 0 2018-01-04
***PurchasedProductsTable***
OrderId itemSku Qty
1 1000001 1
1 1000002 2
2 1000001 3
3 1000001 1
3 1000002 2
4 1000001 3
In the above table the query should first look at the oldest (created_at ASC) order (i.e with Id 1) having status 0 (in order table). and along with that order it should retrieves all the other orders that matches the same itemSku, qty and total distinct items count (in purchasedProducts table).
here order 1 and 3 matches the same itemSKu (1000001 and 1000002) and qty ( 1 and 2) and both have (2) distinct items count respectively so order 1 and 3 should be retrived at first.and when i marked order 1 and 3 as shipped (i.e chang status to 2).
and if i run query again it should retrive similar oders. now order 2 and 4 as order 2 and 4 are similar orders. (have same itemSkus (1000001, Qty (3) and distinct items count (1)).
please help thanks
You have to go trough your tables two times :)
Something like this :
SELECT DISTINCT O2.ID
FROM OrdersTable O1
INNER JOIN PurchasedProductsTable P1 ON O1.ID = P1.OrderId
INNER JOIN PurchasedProductsTable P2 ON P1.itemSku = P2.itemSku
AND P1.Qty = P2.Qty
INNER JOIN OrdersTable O2 ON O2.ID = P2.OrderId
WHERE O1.ID =
(SELECT ID FROM OrdersTable WHERE Status = 0
ORDER BY created_at ASC LIMIT 1)
AND (SELECT COUNT(*) FROM PurchasedProductsTable WHERE OrderId = O1.ID)
= (SELECT COUNT(*) FROM PurchasedProductsTable WHERE OrderId = O2.ID)
ORDER BY O2.ID ASC;
https://www.db-fiddle.com/f/65t9GgSfqMpzNVgnrJp2TR/2
You can get the earliest order via a limit and ordered by the date.
Then you can left join to get that order and any other order that at least has the same items.
Then once you have those order id's from the sub-query result, you can get the order details.
SELECT o.*
FROM
(
SELECT DISTINCT ord2.ID as OrderId
FROM
(
SELECT ID, CustomerId, Status
FROM OrdersTable
WHERE Status = 0
ORDER BY created_at
LIMIT 1
) AS ord1
JOIN PurchasedProductsTable AS pprod1
ON pprod1.OrderId = ord1.ID
LEFT JOIN OrdersTable ord2
ON ord2.CustomerId = ord1.CustomerId
AND ord2.Status = ord1.Status
LEFT JOIN PurchasedProductsTable pprod2
ON pprod2.OrderId = ord2.ID
AND pprod2.itemSku = pprod1.itemSku
AND pprod2.Qty = pprod1.Qty
GROUP BY ord1.CustomerId, ord1.ID, ord2.ID
HAVING COUNT(pprod1.itemSku) = COUNT(pprod2.itemSku)
) q
JOIN OrdersTable AS o ON o.ID = q.OrderId;
Test on RexTester here
I have the following tables, for example:
invoices
ID Name
1 A
2 B
3 C
4 D
5 E
transactions
ID Invoice_ID User_ID
1 1 10
2 1 10
3 1 10
4 2 30
5 3 20
6 3 40
7 2 30
8 2 30
9 4 40
10 3 50
Now I want to make a select that will pull the invoices and the user_id from the related transactions, but of course if I do that I won't get all the ids, since they may be distinct but there will be only one column for that. What I want to do is that if there are distinct User_ids, I will display a pre-defined text in the column instead of the actual result.
select invoices.id, invoices.name, transactions.user_id(if there are distinct user_ids -> return null)
from invoices
left join transactions on invoices.id = transactions.invoice_id
and then this would be the result
ID Name User_ID
1 A 10
2 B 30
3 C null
4 D 40
5 E null
Is this possible?
You can do the following :
select
invoices.id,
invoices.name,
IF (
(SELECT COUNT(DISTINCT user_id) FROM transactions WHERE transactions.invoice_id = invoices.id) = 1,
(SELECT MAX(user_id) FROM transactions WHERE transactions.invoice_id = invoices.id),
null
) AS user_id
from invoices
Or, alternatively, you can use the GROUP_CONCAT function to output a comma-separated list of users for each invoice. It is not exactly what you asked, but maybe in fact it will be more useful :
select
invoices.id,
invoices.name,
GROUP_CONCAT(DISTINCT transactions.user_id SEPARATOR ',') AS user_ids
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
Try somethingh like:
select i.id, i.name, t.user_id
from invoices i left join
(
select invoice_ID, User_ID
from transactions
group by invoice_ID
having count(invoice_ID)=1
) t on i.id=t.invoice_id
SQL fiddle
You could list all the transactions that have multiple user ids, like this:
select invoices.id, invoices.name, null
from invoices
left join transactions on invoices.id = transactions.invoice_id having count(distinct transactions.user_id) > 1
Also, I think this CASE might suit your needs here:
select invoices.id, invoices.name,
case when count(distinct transactions.user_id) > 1 then null else transactions.user_id end
from invoices
left join transactions on invoices.id = transactions.invoice_id
group by invoices.id
although, I'm not sure this is syntactically correct
I have a couple of tables:
Invoice
-----------------
ID total
1 500.00
2 100.00
3 10.00
Payment
---------------------------------------
ID invoiceId Amount Method
1 1 400 CASH
2 2 60 CASH
3 2 40 CREDIT
I need a query that gets all invoices where at least one payment.method is CREDIT and the sum of all payments for that invoice is greater than the total of the invoice.
And I need it to be fast.
How can I do this?
SELECT a.ID InvoiceID,
a.Total TotalInvoice,
b.TotalPayment
FROM Invoice a
INNER JOIN
(
SELECT InvoiceID, SUM(Amount) TotalPayment
FROM Payment
GROUP BY InvoiceID
HAVING SUM(Method = 'CREDIT') > 0
) b ON a.ID = b.InvoiceID AND
a.Total < b.TotalPayment
SQLFiddle Demo
Another method:
SELECT
i.`id`,
i.`total` AS `total_invoiced`,
SUM(p.`amount`) AS `total_payments`,
SUM(IF(p.`method`='credit', 1, 0)) AS `count_credit`
FROM `invoices` i
LEFT JOIN `payments` p ON (p.`invoice_id`=i.`id`)
WHERE 1=1
GROUP BY i.`id`
HAVING (`total_payments` > i.`total`) AND (`count_credit` > 0)
I changed some table/field names. Sorry for the inconvenience.
http://www.sqlfiddle.com/#!2/7402d/1/0
try this one:
select iagg.ID
from (
select i.ID
, sum (p.amount) tl_paid
, min (i.total) tl
from invoice i
join payment p on ( p.invoiceID = i.ID )
group by i.ID
) iagg
where exists (
select 1
from payment p2
where p2.invoiceId = iagg.ID
and p2.method = 'CREDIT'
)
and iagg.tl_paid > iagg.tl
;
the minimum operator over the total attribute is mandated by the grouping operator, the minimum is actually taken from a set of identical values.
I've got a budget table:
user_id product_id budget created
-----------------------------------------------------------------
1 1 300 2011-12-01
2 1 400 2011-12-01
1 1 500 2011-12-03
2 2 400 2011-12-04
I've also got a manager_user table, joining a manager with the user
user_id manager_id product_id
------------------------------------
1 5 1
1 9 2
2 5 1
2 5 2
3 5 1
What I'd like to do is grab each of the user that's assigned to Manager #5, and also get their 'budgets'... but only the most recent one.
Right now my statement looks like this:
SELECT * FROM manager_user mu
LEFT JOIN budget b
ON b.user_id = mu.user_id AND b.product_id = mu.product_id
WHERE mu.manager_id = 5
GROUP BY mu.user_id, mu.product_id
ORDER BY b.created DESC;
The problem is it doesn't pull the most recent budget. Any suggestions? Thanks!
To accomplish your task you can do as follows:
select b1.user_id,
b1.budget
from budget b1 inner join (
select b.user_id,
b.product_id,
max(created) lastdate
from budget b
group by b.user_id, b.product_id ) q
on b1.user_id=q.user_id and
b1.product_id=q.product_id and
b1.created=q.lastdate
where b1.user_id in
(select user_id from manager_user where manager_id = 5);
I'm assuming here that your (user_id, product_id, created) combination is unique.
For what it's worth, here's the code that returned what I was looking for:
SELECT DISTINCT(b1.id),mu.user_id,mu.product_id,b1.budget,b1.created
FROM budget b1
INNER JOIN (
SELECT b.user_id, b.product_id, MAX(created) lastdate
FROM budget b
GROUP BY b.user_id, b.product_id) q
ON b1.user_id=q.user_id AND
b1.product_id=q.product_id AND
b1.created=q.lastdate
RIGHT JOIN manager_user mu
ON mu.user_id = b1.user_id AND
mu.product_id = b1.product_id
WHERE mu.manager_id = 5;
Thanks for the help Andrea!
I posted something similar to this yesterday, but now I'd like something a little different from my query-
I'm trying to query a database to retrieve the number of one-time users who have visited a website over time. The data looks something like this:
Day | UserID
1 | A
1 | B
2 | B
3 | A
4 | B
4 | C
5 | D
I'd like the query result to look this this
Time Span | COUNT(DISTINCT UserID)
Day 1 to Day 1 | 2
Day 1 to Day 2 | 1
Day 1 to Day 3 | 0
Day 1 to Day 4 | 1
Day 1 to Day 5 | 2
The result is 2,1,0,1,2 because, at the end of those days, there are X number of users who have visited a single time. e.g. for day 5, at the end of day 5, users c and d have visited only once each.
I think I'm looking for a query similar to this:
select d.day, (select count(distinct userid) from visits where day<=d.day)
from (select distinct day from visits) d
The difference between the query above and what I'm looking for is that I'd like this new query to consider only one-time users for each time span, and not repeat users.
Thanks
This subquery should work for the clarified requirements.
select d.day, count(distinct case when b.userid is null then a.userid end)
from (select day from visits group by day) d
inner join
(
select a.day, a.userid, count(*) c
from visits a
join visits b on a.userid=b.userid and b.day <= a.day
group by a.day, a.userid
having count(*) = 1
) a on a.day <= d.day
left join
(
select a.day, a.userid, count(*) c
from visits a
join visits b on a.userid=b.userid and b.day <= a.day
group by a.day, a.userid
having count(*) > 1
) b on a.userid = b.userid and b.day <= d.day
group by d.day
Original
You must have taken the idea from SQL Server - it is the only RDBMS (IIRC) that will allow you to reference a twice removed (nesting) query. Please indicate what you want and we can rewrite the query.
For the exact query shown, you don't need 2 levels of subquery
SELECT
C.col_c1 AS Data,
(
SELECT count(col_b1)
FROM tbl
WHERE col_b2 <= C.col_c1
) A
FROM (
SELECT col_c1 # subquery to get distinct c1
FROM tbl
GROUP BY col_c1) C;