Trying to show a table with 3 columns that are prices that need to be displayed. the columns are differentiated by 'price_type' and there are 3 different price types.
Its probably something obvious im missing but something like:
Select price as 'current', price as '10min', price as '30min'
from table
where Price_Type(current) = 'current' AND Price_Type(10min) = '10min' AND
Price_Type(30min) = '30min'
Order by date desc
I'm not sure what the actual syntax would be, but any help is appreciated.
With conditional aggregation:
select date,
max(case when Price_Type = 'current' then price end) as [current],
max(case when Price_Type = '10min' then price end) as [10min],
max(case when Price_Type = '30min' then price end) as [30min]
from table
group by date
order by date desc
Related
I am struggling to work out combining a query that should give me 3 columns of Month, total_sold_products and drinks_sold_products
Query 1:
Select month(date), count(id) as total_sold_products
from Products
where date between '2022-01-01' and '2022-12-31'
Query 2
Select month(date), count(id) as drinks_sold_products
from Products where type = 'drinks' and date between '2022-01-01' and '2022-12-31'
I tried the union function but it summed count(id) twice and gave me only 2 columns
Many thanks!
Union is for attaching sets of data on top of each other. You need conditional aggregation or a join. See below.
SELECT MONTH(date),
COUNT(*) AS total_sold_products,
COUNT(CASE WHEN type = 'drinks' THEN 1 ELSE 0 END) AS drinks_sold_products,
FORMAT((CASE
WHEN COUNT(*) > 0 THEN
COUNT(CASE WHEN type = 'drinks' THEN 1 ELSE 0 END)/COUNT(*)
ELSE 0 END),
'P') AS Percentage
FROM Products
WHERE date BETWEEN'2022-01-01' AND '2022-12-31'
GROUP BY MONTH(date)
I have a raw data that has multiple dates per category, and I use code case when category = 'referral' then min(date) end as date_referral to get earliest dates of each category per id.
However, it will not return data in a row but create row per category, as such:
id date_entered date_referral date_reply date_final
-------------------------------------------------------------------------
1 2020-12-20 null null null
1 2020-12-20 2020-12-21 null null
1 2020-12-20 null 2020-12-21 null
1 2020-12-20 null null 2020-12-24
I tried enforcing single rows by using distinct or group by (separately and together):
select distinct id
, date_entered
, case when category = 'referral' then min(date) end as date_referral
, case when category = 'reply' then min(date) end as date_reply
, case when category = 'final' then min(date) end as date_final
from data
group by id
, date_entered
, category
but it will keep returning multiple rows, with each row being calculated earliest date per category. I also tried creating cte after this code to select distinct id, date_entered, date_referral, date_reply, date_final from table but that also still returns multiple rows..
How can I combine these rows and make it return one single row?
You should not group by category.
Use conditional aggregation like this:
select id, date_entered,
min(case when category = 'referral' then date end) as date_referral,
min(case when category = 'reply' then date end) as date_reply,
min(case when category = 'final' then date end) as date_final
from data
group by id, date_entered
I have list of line items from invoices with a field that indicates if a line was delivered or picked up. I need to find a percentage of delivered items from the total number of lines.
SALES_NBR | Total | Deliveryrate
1 = Delivered 0 = picked up from FULFILLMENT.
SELECT SALES_NBR,
COUNT (ITEMS) as Total,
SUM (case when FULFILLMENT = '1' then 1 else 0 end) as delivered,
(SELECT delivered/total) as Deliveryrate
FROM Invoice_table
WHERE STORE IN '0123'
And SALE_DATE >='2020-02-01'
And SALE_DATE <='2020-02-07'
Group By SALES_NBR, Deliveryrate;
My query executes but never finishes for some reason. Is there any easier way to do this? Fulfillment field does not contain any NULL values.
Any help would be appreciated.
I need to find a percentage of delivered items from the total number of lines.
The simplest method is to use avg():
select SALES_NBR,
avg(fulfillment) as delivered_ratio
from Invoice_table
where STORE = '0123' and
SALE_DATE >='2020-02-01' and
SALE_DATE <='2020-02-07'
group by SALES_NBR;
I'm not sure if the group by sales_nbr is needed.
If you want to get a "nice" query, you can use subqueries like this:
select
qry.*,
qry.delivered/qry.total as Deliveryrate
from (
select
SALES_NBR,
count(ITEMS) as Total,
sum(case when FULFILLMENT = '1' then 1 else 0 end) as delivered
from Invoice_table
where STORE IN '0123'
and SALE_DATE >='2020-02-01'
and SALE_DATE <='2020-02-07'
group by SALES_NBR
) qry;
But I think this one, even being ugglier, could perform faster:
select
SALES_NBR,
count(ITEMS) as Total,
sum(case when FULFILLMENT = '1' then 1 else 0 end) as delivered,
sum(case when FULFILLMENT = '1' then 1 else 0 end)/count(ITEMS) as Deliveryrate
from Invoice_table
where STORE IN '0123'
and SALE_DATE >='2020-02-01'
and SALE_DATE <='2020-02-07'
group by SALES_NBR
I'm trying to figure out if what I'm trying to do is possible. Instead of resorting to multiple queries on a table, I wanted to group the records by business date and id then group by the id and select one date for a field and another date for the other field.
SELECT
*
{AMOUNT FROM DATE}
{AMOUNT FROM OTHER DATE}
FROM (
SELECT
date,
id,
SUM(amount) AS amount
FROM
table
GROUP BY id, date
AS subquery
GROUP BY id
It seems that you're looking to do a pivot query. I usually use cross tabs for this. Based on the query you posted, it could look like:
SELECT
id,
SUM(CASE WHEN date = '20190901' THEN amount ELSE 0 END) AmountFromSept01,
SUM(CASE WHEN date = '20191001' THEN amount ELSE 0 END) AmountFromOct01
FROM (
SELECT
date,
id,
SUM(amount) AS amount
FROM
table
GROUP BY id, date
)AS subquery
GROUP BY id;
You could also use a CTE.
WITH CTE AS(
SELECT
date,
id,
SUM(amount) AS amount
FROM
table
GROUP BY id, date
)
SELECT
id,
SUM(CASE WHEN date = '20190901' THEN amount ELSE 0 END) AmountFromSept01,
SUM(CASE WHEN date = '20191001' THEN amount ELSE 0 END) AmountFromOct01
FROM CTE
GROUP BY id;
Or even be a rebel and do the operation directly.
SELECT
id,
SUM(CASE WHEN date = '20190901' THEN amount ELSE 0 END) AmountFromSept01,
SUM(CASE WHEN date = '20191001' THEN amount ELSE 0 END) AmountFromOct01
FROM CTE
GROUP BY id;
However, some people have tested for performance and found that pre-aggregating can improve performance.
If I understand you correctly, then you're just trying to pivot, but only with two particular dates:
select id,
date1 = sum(iif(date = '2000-01-01', amount, null)),
date2 = sum(iif(date = '2000-01-02', amount, null))
from [table]
group by id
i have table named source table with data like this :
And i want to do query that subtract row with status plus and minus to be like this group by product name :
How to do that in SQL query? thanks!
Group by the product and then use a conditional SUM()
select product,
sum(case when status = 'plus' then total else 0 end) -
sum(case when status = 'minus' then total else 0 end) as total,
sum(case when status = 'plus' then amount else 0 end) -
sum(case when status = 'minus' then amount else 0 end) as amount
from your_table
group by product
There is another method using join, which works for the particular data you have provided (which has one "plus" and one "minus" row per product):
select tplus.product, (tplus.total - tminus.total) as total,
(tplus.amount - tminus.amount) as amount
from t tplus join
t tminus
on tplus.product = tminus.product and
tplus.status = 'plus' and
tplus.status = 'minus';
Both this and the aggregation query work well for the data you have provided. In other words, there are multiple ways to solve this problem (each has its strengths).
you can query as below:
select product , sum (case when [status] = 'minus' then -Total else Total end) as Total
, sum (case when [status] = 'minus' then -Amount else Amount end) as SumAmount
from yourproduct
group by product