I want to sum up qty*price where type is 'in' and then subtract the sum of qty*price where type is 'out'. Something like this.
SELECT cid, name, SUM(paid_amt), (SELECT SUM(qty*price) WHERE type = 'in' - SELECT SUM(qty*price) WHERE type = 'out'), type FROMtransactionsGROUP BY cid
Here's is my SQL query.
You can solve your problem by using two different sum expressions, each with a slightly different case statement inside of it. The idea is that your case statement only returns a value to the aggregate sum expression if it's of the correct typing.
select
cid
, sum(case when type = "in" then qty*price else 0 end)
- sum(case when type = "out" then qty*price else 0 end)
from
your_table
group by
cid
Related
SQL-Noob here.
Having a bunch of items which are either bought or sold.
I would like to Group by item-name and get a sum of purchases and sales.
Also i would like to calculate the stock of items.
I tried a subquery:
SELECT NAME,
(
SELECT SUM(quantity)
FROM transactions
WHERE `type` = "Buy") AS "purchases",
(
SELECT SUM(quantity)
FROM transactions
WHERE `type` = "Sell") AS "sales"
FROM transactions
GROUP BY NAME
which results in wrong values:
After that i tried CASE-Statements
SELECT NAME,
SUM(CASE WHEN `type` = "Buy" THEN quantity END) AS "purchases",
SUM(CASE WHEN `type` = "Sell" THEN quantity END) AS "sales"
FROM transactions
GROUP BY name
which works:
Question 1: Is there a better way to achieve the result?
Question 2: How to calculate the Stock of the items?
Something like purchases - sales AS "Stock" doesn't seem to work.
Thanks alot
Use conditional aggregation:
select name,
sum(case when type = 'Buy' then quantity else 0 end) as purchases,
sum(case when type = 'Sell' then quantity else 0 end) as sales,
sum(case when type = 'Buy' then quantity else - quantity end) as stock
from transactions
where type in ('Buy', 'Sell')
group by name
The where clause is not necessary of there are just those types in the table.
Side notes:
using single quotes for literal strings rather than double quotes; this is standard SQL, that all databases understand
in MySQL, use backticks if you need to quote an identifier rather than double quotes; or better yet, use identifiers that do not require quoting
I've looked over similar questions and I just can't seem to get this right.
I have a table with three columns: ID, Date, and Method. None are unique.
I want to be able to see for any given date, how many rows match a certain pattern on Method.
So, for example, if the table has 100 rows, and 8 of them have the date "01-01-2020" and of those 8, two of them have a method of "A", I would want a return row that says "01-01-2020", "8", "2", and "25%".
My SQL is pretty rudimentary. I have been able to make a query to get me the count of each method by date:
select Date, count(*) from mytable WHERE Method="A" group by Date;
But I haven't been able to figure out how to put together the results that I am needing. Can someone help me out?
You could perform a count over a case expression for that method, and then divide the two counts:
SELECT date,
COUNT(*),
COUNT(CASE method WHEN 'A' THEN 1 END),
COUNT(CASE method WHEN 'A' THEN 1 END) / COUNT(*) * 100
FROM mytable
GROUP BY date
I'm assuming you're interested in all methods rather than just 'A', so you could do the following:
with ptotals as
(
SELECT
thedate,
count(*) as NumRows
FROM
mytable
group by
thedate
)
select
mytable.thedate,
mytable.themethod,
count(*) as method_count,
100 * count(*) / max(ptotals.NumRows) as Pct
from
mytable
inner join
ptotals
on
mytable.thedate = ptotals.thedate
group by
mytable.thedate,
mytable.themethod
You can use AVG() for the ratio/percentage:
SELECT date, COUNT(*),
SUM(CASE WHEN method = 'A' THEN 1 ELSE 0 END),
AVG(CASE WHEN method = 'A' THEN 100.0 ELSE 0 END)
FROM t
GROUP BY date;
I am unable to derive a SQL query for the following table content.
When i tried below query i am getting above said output. Can someone help me to give the required query for it.
select Name, count(Status) from mytable where Status='Open' group by mytable union
select Name, count(Status) from mytable where Status='Cleared' group by mytable
Use case expressions in the select list to do conditional aggregation.
select Name,
count(case when Status = 'Open' then 1 end) as opencnt,
count(case when Status = 'Cleared' then 1 end) as clearedcnt
from mytable
where Status in ('Open', 'Cleared')
group by Name
COUNT() counts non-null values. The case expressions above returns null when the conditions aren't fulfilled.
I want to count the number of rows where a particular field = 'Q1'.
I usually use count(particular_field), but this does not allow me to count only when that field = 'Q1'.
Does the query SUM(particular_field = 'Q1') work for this matter? Or am I able to do count(particular_field = 'Q1')?
you can either do
select count(*)
from table
where particular_field = 'Q1'
or
select sum(case when particular_field ='Q1' then 1 else 0 end)
from table
You should be able to use a CASE statement with your SUM() (See SQL Fiddle)
SELECT SUM(CASE WHEN particular_field = 'Q1' THEN 1 ELSE 0 END) yourCount
FROM yourTable
Or (See SQL Fiddle) - this will give you a list of the count and each field. If you only want the one value, then use a WHERE clause to filter:
select count(*), particular_field
from yourTable
group by particular_field
I have about the following query:
SELECT *
FROM Product
ORDER BY (SELECT CASE
(SELECT MIN(date) FROM SomethingElse WHERE productId = Product.id) AS here
WHEN NULL
THEN 0
ELSE here
) DESC
However, AS in CASE doesn't work. I need a way to save the value in a variable and use it again.
So you are trying to do CASE when expression IS NULL THEN 0 ELSE expression END?
In this case you can just use COALESCE(expression, 0)