I have two tables
Customer table
+--------+---------+
| refno | deposit |
+--------+---------+
| 1/13 | -10 |
| 10/13 | 500 |
| 100/13 | 0 |
| 101/13 | 250 |
| 102/13 | 1000 |
+--------+---------+
Ledger Table
+--------+----------+------+----------+
| refno | quantity | rate | recieved |
+--------+----------+------+----------+
| 1/13 | 2 | 70 | 0 |
| 10/13 | 3 | 80 | 0 |
| 100/13 | 3 | 60 | 0 |
| 101/13 | 4 | 60 | 0 |
| 102/13 | 10 | 65 | 0 |
+--------+----------+------+----------+
I want to customer column(deposit) added in ledger column (total)
I don't want to create another table.
i want
refno | total = customer.deposit+(ledger.quantity*ledger.rate-ledger.received)
1/13 | -200
10/13 | 4210
100/13| 625
101/13| 280
102/13| 1000
Regards,
select c.refno, c.deposit + l.total
from customer c join ledger l on c.refno = l.refno
I think author means update other than select from directly, but not sure which table used as basic table. For example, if we want to add value in Customer to that in Ledger, it should look like this:
update Ledger l inner join Cusomter c on l.refno = c.refno set l.total = l.total + c.deposit.
If you have duplicate refno in Ledger, you can try this way :
select b.refno, b.total + IFNULL(a.deposit, 0) as total
from
(select refno, sum(total) as total
from Ledger
group by refno) b
inner join Customer a on a.refno = b.refno
SQLFiddle demo
Is this what you want?
select l.refno, l.total, c.deposit, l.total + coalesce(c.deposit, 0) as TotalWithDeposit
from ledger l left outer join
customer c
on l.refno = c.refno;
Your question mentions aggregate functions, but I don't see any aggregations going on.
EDIT:
select l.refno, l.total, c.deposit,
c.deposit + (l.quantity * l.rate - l.received) as TotalWithDeposit
from ledger l left outer join
customer c
on l.refno = c.refno;
Related
Blog table:
| bid | btitle |
| 29 | ...... |
| 38 | ...... |
likes table:
| lid | bid |
| 1 | 29 |
| 2 | 29 |
| 3 | 29 |
| 4 | 38 |
| 5 | 38 |
comment table
| commid | bid |
| 1 | 29 |
| 2 | 29 |
| 3 | 38 |
I had tried the following query but that will not work for me:
SELECT blog.bid,blog.btitle,COUNT(likes.lid) AS likecnt,COUNT(comment.comid) AS commentcnt FROM blog,likes,comment WHERE blog.bid=likes.bid AND blog.bid=comment.bid GROUP BY blog.bid
i want output like:
| bid | btitle | likecnt | commentcnt |
| 29 | ...... | 3 | 2 |
| 38 | ...... | 2 | 1 |
You can do left join with separate aggregation :
select b.bid, b.btitle,
coalesce(l.likecnt, 0) as likecnt,
coalesce(c.commentcnt, 0) as commentcnt
from blog b left join
(select l.bid, count(*) as likecnt
from likes l
group by l.bid
) l
on l.bid = b.bid left join
(select c.bid, count(*) as commentcnt
from comment c
group by c.bid
) c
on c.bid = l.bid;
If you want only matching bids the use INNER JOIN instead of LEFT JOIN & remove COALESCE().
Under many circumstances, correlated subqueries may be the fastest solution:
select b.bid, b.btitle,
(select count(*) from likes l where l.bid = b.bid) as num_likes,
(select count(*) from comment c where c.bid = b.bid) as num_comments
from blog b;
When is this a win performance wise. First, you want indexes on likes(bid) and comments(bid). With those indexes, it might be the fastest approach for your query.
It is particularly better if you have a where clause filtering the blogs in the outer query. It only has to do the counts for the blogs in the result set.
Use proper joins and count DISTINCT values because multiple joins increase the number of returned rows:
SELECT b.bid, b.btitle,
COUNT(DISTINCT l.lid) AS likecnt,
COUNT(DISTINCT c.comid) AS commentcnt
FROM blog b
LEFT JOIN likes l ON b.bid = l.bid
LEFT JOIN comment c ON b.bid = c.bid
GROUP BY b.bid, b.btitle
See the demo.
I use LEFT joins just in case there are no comments or likes for a post.
Results:
| bid | btitle | likecnt | commentcnt |
| --- | ------ | ------- | ---------- |
| 29 | ...... | 3 | 2 |
| 38 | ...... | 2 | 1 |
For every ID_Number, there is a bill_date and then two types of bills that happen. I want to return the latest date (max date) for each ID number and then add together the two types of bill amounts. So, based on the table below, it should return:
| 1 | 201604 | 10.00 | |
| 2 | 201701 | 28.00 | |
tbl_charges
+-----------+-----------+-----------+--------+
| ID_Number | Bill_Date | Bill_Type | Amount |
+-----------+-----------+-----------+--------+
| 1 | 201601 | A | 5.00 |
| 1 | 201601 | B | 7.00 |
| 1 | 201604 | A | 4.00 |
| 1 | 201604 | B | 6.00 |
| 2 | 201701 | A | 15.00 |
| 2 | 201701 | B | 13.00 |
+-----------+-----------+-----------+--------+
Then, if possible, I want to be able to do this in a join in another query, using ID_Number as the column for the join. Would that change the query here?
Note: I am initially only wanting to run the query for about 200 distinct ID_Numbers out of about 10 million. I will be adding an 'IN' clause for those IDs. When I do the join for the final product, I will need to know how to get those latest dates out of all the other join possibilities. (ie, how do I get ID_Number 1 to join with 201604 and not 201601?)
I would use NOT EXISTS and GROUP BY
select, t1.id_number, max(t1.bill_date), sum(t1.amount)
from tbl_charges t1
where not exists (
select 1
from tbl_charges t2
where t1.id_number = t2.id_number and
t1.bill_date < t2.bill_date
)
group by t1.id_number
the NOT EXISTS filter out the irrelevant rows and GROUP BY do the sum.
I would be inclined to filter in the where:
select id_number, sum(c.amount)
from tbl_charges c
where c.date = (select max(c2.date)
from tbl_charges c2
where c2.id_number = c.id_number and c2.bill_type = c.bill_type
)
group by id_number;
Or, another fun way is to use in with tuples:
select id_number, sum(c.amount)
from tbl_charges c
where (c.id_number, c.bill_type, c.date) in
(select c2.id_number, c2.bill_type, max(c2.date)
from tbl_charges c2
group by c2.id_number, c2.bill_type
)
group by id_number;
I'm having a table with main invoice data, and two table with invoice items:
items which are based on hourly work, with an hourly rate and an amount of hours
items which are products, with a unit count an unit price
For the invoice overview page, I'd like to retrieve all invoices and their total amounts with one query.
A simplified schema
invoices_main
| invoice_id |
| 1 |
| 2 |
| 3 |
invoices_items_products
| item_id | invoice_id | item_count | item_unit_price |
| 1 | 1 | 1 | 999.95 |
| 2 | 1 | 20 | 49.50 |
| 3 | 2 | 3 | 15.00 |
| 4 | 2 | 5 | 5.00 |
| 5 | 3 | 2 | 150.00 |
invoices_items_hourly
| item_id | invoice_id | item_hours | item_hourly_rate |
| 1 | 1 | 3.50 | 90.00 |
| 2 | 1 | 1.00 | 140.00 |
| 3 | 2 | 12.00 | 90.00 |
| 4 | 3 | 1.50 | 90.00 |
With the help of this question, I've constructed the following query:
SELECT
I.invoice_id,
IFNULL(
SUM(ROUND(P.item_unit_price * P.item_count, 2)),
0
) + IFNULL(
SUM(ROUND(H.item_hourly_rate * H.item_hours, 2)),
0
) AS invoice_total_amount
FROM
invoices_main I
LEFT JOIN invoices_items_products P ON I.invoice_id = P.invoice_id
LEFT JOIN invoices_items_hours H ON I.invoice_id = H.invoice_id
GROUP BY
I.invoice_id
It works kind of, but if an invoice has both products and hourly items, with at least multiple entries for one of both, items are duplicated due to the joins and the total amount becomes way too high.
Thus, in the above example schema, it goes wrong with invoice_id 1 and 2, but work with 3.
How can I retrieve a list of invoices with their respective total amounts, in a way that works even if an invoice has multiple products and multiple hourly items?
Try putting both left join's into a subquery instead.
SELECT
I.invoice_id,
IFNULL
(
(
SELECT SUM(ROUND(H.item_hourly_rate * H.item_hours, 2))
FROM invoices_items_hours AS H
WHERE H.invoice_id = I.invoice_id
)
, 0
) +
IFNULL
(
(
SELECT SUM(ROUND(P.item_unit_price * P.item_count, 2))
FROM invoices_items_products AS P
WHERE P.invoice_id = I.invoice_id
)
, 0
) AS invoice_total_amount
FROM invoices_main AS I
GROUP BY I.invoice_id
As mentioned in the comments, you should sum up the revenue in each table per invoice_id before doing the join. If you're looking to get the revenue from both of these places then you can add (B.unit_revenue + C.hourly_revenue) total_revenue to the first SELECT statement below.
SELECT A.invoice_id, B.unit_revenue, C.hourly_revenue FROM
invoices_main AS A
JOIN (
SELECT invoice_id, SUM(item_count * item_unit_price) unit_revenue
FROM invoices_items_products GROUP BY invoice_id
) B
ON
A.invoice_id = B.invoice_id
JOIN (
SELECT invoice_id, SUM(item_hours * item_hourly_rate) hourly_revenue FROM
invoices_items_hours GROUP BY invoice_id
) C
ON
A.invoice_id = C.invoice_id
This query runs on an invoices table to help me decide who I need to pay
Here's the base table:
The users table
+---------+--------+
| user_id | name |
+---------+--------+
| 1 | Peter |
| 2 | Lois |
| 3 | Stewie |
+---------+--------+
The invoices table:
+------------+---------+----------+--------+---------------+---------+
| invoice_id | user_id | currency | amount | description | is_paid |
+------------+---------+----------+--------+---------------+---------+
| 1 | 1 | usd | 140 | Cow hoof | 0 |
| 2 | 1 | usd | 45 | Cow tail | 0 |
| 3 | 1 | gbp | 1 | Cow nostril | 0 |
| 4 | 2 | gbp | 1500 | Cow nose hair | 0 |
| 5 | 2 | cad | 1 | eyelash | 1 |
+------------+---------+----------+--------+---------------+---------+
I want a resulting table that looks like this:
+---------+-------+----------+-------------+
| user_id | name | currency | SUM(amount) |
+---------+-------+----------+-------------+
| 1 | Peter | usd | 185 |
| 2 | Lois | gbp | 1500 |
+---------+-------+----------+-------------+
The conditions are:
Only consider invoices that have not been paid, so where is_paid = 0
Group them by user_id, by currency
If the SUM(amount) < $100 for the user_id, currency pair then don't bother showing the result, since we don't pay invoices that are less than $100 (or equivalent, based on a fixed exchange rate).
Here's what I've got so far (not working -- which I guess is because I'm filtering by a GROUP'ed parameter):
SELECT
users.user_id, users.name,
invoices.currency, SUM(invoices.amount)
FROM
mydb.users,
mydb.invoices
WHERE
users.user_id = invoices.user_id AND
invoices.is_paid != true AND
SUM(invoices.amount) >=
CASE
WHEN invoices.currency = 'usd' THEN 100
WHEN invoices.currency = 'gbp' THEN 155
WHEN invoices.currency = 'cad' THEN 117
END
GROUP BY
invoices.currency, users.user_id
ORDER BY
users.name, invoices.currency;
Help?
You can't use SUM in a WHERE. Use HAVING instead.
Use HAVING clause instead of SUM in WHERE condition
Try this:
SELECT u.user_id, u.name, i.currency, SUM(i.amount) invoiceAmount
FROM mydb.users u
INNER JOIN mydb.invoices i ON u.user_id = i.user_id
WHERE i.is_paid = 0
GROUP BY u.user_id, i.currency
HAVING SUM(i.amount) >= (CASE i.currency WHEN 'usd' THEN 100 WHEN 'gbp' THEN 155 WHEN 'cad' THEN 117 END)
ORDER BY u.name, i.currency;
Try something like this:
SELECT
user_id, name, currency, sum(amount) due
FROM
invoice i
JOIN users u ON i.user_id=u.user_id
WHERE
is_paid = 0 AND
GROUP BY user_id, currency
having due >= 100
do you store exchange rates? Multiply rates with amount to get actual amount with respect to base currency.
sum(amount*ex_rate) due
I have scheduled payments, so I have these tables:
customer
+id
paymentSchedule
+id
+customer_id
+amount //total price
+dueDate //date to be paid
payments
+id
+date
+customer_id
+paymentSchedule_id
+amount //amount paid, it can be a partial payment
How do I write a query to get Today's due amount by customer.
I mean I need to join the tables (thats my main problem) and then substract the
sum of the payments.mount minus the sum of the scheduledPaymens.amount
but.. how?
Thanks in advance
This is probably not 100%, but should be pretty solid to help you tweak:
SELECT customer_id, (due.amount - paid.amount) as amountDue
FROM
(SELECT customer_id, SUM(amount) as amount
FROM paymentSchedule
WHERE dateDate <= getDate()
and customer_id = #custid) as due
LEFT JOIN
(SELECT customer_id, SUM(amount) as amount
FROM payments
WHERE customer_id = #custid) as paid ON paid.customer_id = due.customer_id
Ok, this is how I understood the problem. I simplified the tables because they where just complicating things, and adding dates is just straight forward.
PaymentSchedule
+----+-------------+-----------------+
| id | customer_id | original_amount |
+----+-------------+-----------------+
| 1 | Tom | 100 |
| 2 | Tom | 200 |
| 3 | Tom | 300 |
| 4 | Moe | 400 |
+----+-------------+-----------------+
Payments
+----+--------------------+-------------+
| id | paymentSchedule_id | paid_amount |
+----+--------------------+-------------+
| 1 | 1 | 70 |
| 2 | 2 | 150 |
| 3 | 2 | 50 |
| 4 | 4 | 300 |
| 5 | 4 | 25 |
+----+--------------------+-------------+
Result of query
+-------------+-------------------+-----------------+-----------+----------------+
| CUSTOMER_ID | PAYMENTSCHEDULEID | ORIGINAL_AMOUNT | TOTALPAID | PENDINGPAYMENT |
+-------------+-------------------+-----------------+-----------+----------------+
| Tom | 1 | 100 | 70 | 30 |
| Tom | 2 | 200 | 200 | 0 |
| Tom | 3 | 300 | 0 | 300 |
| Moe | 4 | 400 | 325 | 75 |
+-------------+-------------------+-----------------+-----------+----------------+
Query with double select
select *, s.original_amount - s.TotalPaid as PendingPayment from (
select
ps.customer_id, ps.id as PaymentScheduleId, ps.original_amount,
coalesce(sum(p.paid_amount), 0) as TotalPaid
from paymentSchedule ps
left join payments p on p.paymentSchedule_id = ps.id
group by ps.customer_id, PaymentScheduleId, ps.original_amount
) as S
Query with single select
select
ps.customer_id, ps.id as PaymentScheduleId, ps.original_amount,
coalesce(sum(p.paid_amount), 0) as TotalPaid,
ps.original_amount - coalesce(sum(p.paid_amount), 0) as PendingPayment
from paymentSchedule ps
left join payments p on p.paymentSchedule_id = ps.id
group by ps.customer_id, PaymentScheduleId, ps.original_amount
The result of both queries is the same. I just wonder which one runs faster. You can try both and tell us :)
Let me know if this this is the result you expected
Something like this should be a good starting point for you to tweak.
SELECT c.*
FROM customer c
INNER JOIN paymentSchedule ps
ON c.id = ps.customer_id
LEFT JOIN payments p
ON ps.id = p.paymentSchedule_id
WHERE ps.dueDate = 'This depends on how you store dueDate'
AND ps.amount - p.amount > 0