select max(total), date
from (select sum(total) as total, date as date
from canteen
group by date) as max
i want to select the highest sale and the date of the highest sale from table.
With my query this error is shown.
Exception, Error code 8,120, SQLState S0001] Column 'max.date' is
invalid in the select list because it is not contained in either an
aggregate function or the GROUP BY clause.
You can use order by, set your data to order by sale descending and fetch first row.
If you wants result by date then you can use ROW_NUMBER()
select TOP(1) total, date
from
(
select sum(total) as total, date as date
from canteen
group by date
) as max
Order by todal desc
This will return all dates achieving the max sales total. Here is a demo.
; with x as (
select sum(total) as total, date from canteen group by date
)
, y as (
select dr = dense_rank() over(order by total desc), *
from x
)
select * from y where dr = 1
If you want to get all dates with max(total), here it is
;with temp as
(
select sum(total) as total, date as date
from canteen
group by date
)
SELECT TOP 1 WITH TIES *
FROM temp
ORDER BY temp.total desc
You have forgotten to add group by & order by in your outer query. I have modified it to display all the sales in descending order. So the highest sale will be at top.
select max(total) total, date
from (select sum(total) as total, date as date
from canteen
group by date) as max
group by date
order by total desc
Related
I have a table with snapshot data. It has productid and date and quantity columns. I need to find min value in the max date. Let's say, we have product X: X had the last snapshot at Y date but it has two snapshots at Y with 9 and 8 quantity values. I need to get
product_id | date | quantity
X Y 8
So far I came up with this.
select
productid
, max(snapshot_date) max_date
, min(quantity) min_quantity
from snapshot_table
group by 1
It works but I don't know why. Why this does not bring min value for each date?
I would use RANK here along with a scalar subquery:
WITH cte AS (
SELECT *, RANK() OVER (ORDER BY quantity) rnk
FROM snapshot_table
WHERE snapshot_date = (SELECT MAX(snapshot_date) FROM snapshot_table)
)
SELECT productid, snapshot_date, quantity
FROM cte
WHERE rnk = 1;
Note that this solution caters to the possibility that two or more records happened to be tied for having the lower quantity among those most recent records.
Edit: We could simplify by doing away with the CTE and instead using the QUALIFY clause for the restriction on the RANK:
SELECT productid, snapshot_date, quantity
FROM snapshot_table
WHERE snapshot_date = (SELECT MAX(snapshot_date) FROM snapshot_table)
QUALIFY RANK() OVER (ORDER BY quantity) = 1;
Consider also below approach
select distinct product_id,
max(snapshot_date) over product as max_date,
first_value(quantity) over(product order by snapshot_date desc, quantity) as min_quantity
from your_table
window product as (partition by product_id)
use row_number()
with cte as (select *,
row_number() over(partition by product_id order by date desc) rn
from table_name) select * from cte where rn=1
I have a table with columns month, name and transaction_id. I would like to count the number of transactions per month and name. However, for each month I want to have the top N names with the highest transaction counts.
The following query groups by month and name. However the LIMIT is applied to the complete result and not per month:
SELECT
month,
name,
COUNT(*) AS transaction_count
FROM my_table
GROUP BY month, name
ORDER BY month, transaction_count DESC
LIMIT N
Does anyone have an idea how I can get the top N results per month?
Use row_number():
SELECT month, name, transaction_count
FROM (SELECT month, name, COUNT(*) AS transaction_count,
ROW_NUMBER() OVER (PARTITION BY month ORDER BY COUNT(*) DESC) as seqnum
FROM my_table
GROUP BY month, name
) mn
WHERE seqnum <= N
ORDER BY month, transaction_count DESC
I have gotten all rows for the latest date like this:
SELECT date, quarter, sales_region, revenue
FROM regions
WHERE date = (SELECT MAX(date) FROM regions)
ORDER BY 1
So how would I get the rows for the second latest date?
I have tried but no luck:
SELECT MAX(date), quarter, sales_region, revenue
FROM regions
WHERE date < (SELECT MAX(date) FROM regions)
ORDER BY 1
Here is one method:
SELECT date, quarter, sales_region, revenue
FROM regions
WHERE date = (SELECT DISTINCT date
FROM regions r2
ORDER BY date DESC
OFFSET 1 FETCH FIRST 1 ROW ONLY
)
ORDER BY 1;
Another method uses dense_rank():
select r.*
from (select r.*, dense_rank() over (order by date desc) as seqnum
from regions r
) r
where seqnum = 2;
Gordon answered your question precisely, but if you want to get the records for the last two dates in one query, you could use IN instead of =, and get the top two records with LIMIT 2:
SELECT date, quarter, sales_region, revenue
FROM regions
WHERE date IN (SELECT DISTINCT date
FROM regions r2
ORDER BY date DESC
LIMIT 2)
ORDER BY 1;
Starting with version 8.4, you can also use FETCH FIRST 2 ROW ONLY instead of LIMIT 2.
I have a question that is asking:
-List the max sales for each year?
I think I have the starter query but I can't figure out how to get all the years in my answer:
SELECT TO_CHAR(stockdate,'YYYY') AS year, sales
FROM sample_newbooks
WHERE sales = (SELECT MAX(sales) FROM sample_newbooks);
This query gives me the year with the max sales. I need max sales for EACH year. Thanks for your help!
Use group by and max if all you need is year and max sales of the year.
select
to_char(stockdate, 'yyyy') year,
max(sales) sales
from sample_newbooks
group by to_char(stockdate, 'yyyy')
If you need rows with all the columns with max sales for the year, you can use window function row_number:
select
*
from (
select
t.*,
row_number() over (partition by to_char(stockdate, 'yyyy') order by sales desc) rn
from sample_newbooks t
) t where rn = 1;
If you want to get the rows with ties on sales, use rank:
select
*
from (
select
t.*,
rank() over (partition by to_char(stockdate, 'yyyy') order by sales desc) rn
from sample_newbooks t
) t where rn = 1;
I have a database of transactions, accounts, profit/loss, and date. I need to find the dates which the largest profit occurs by account. I have already found a way to find these actually max/min values but I can't seem to be able to pull the actual date from it. My code so far is like this:
Select accountnum, min(ammount)
from table
where date > '02-Jan-13'
group by accountnum
order by accountnum
Ideally I would like to see account num, the min or max, and then the date which this occurred on.
Try something like this to get the min and max amount for each customer and the date it happened.
WITH max_amount as (
SELECT accountnum, max(amount) amount, date
FROM TABLE
GROUP BY accountnum, date
),
min_amount as (
SELECT accountnum, min(amount) amount, date
FROM TABLE
GROUP BY accountnum, date
)
SELECT t.accountnum, ma.amount, ma.date, mi.amount, ma.date
FROM table t
JOIN max_amount ma
ON ma.accountnum = t.accountnum
JOIN min_amount mi
ON mi.accountnum = t.accountnum
If you want the data for just this year you could add a where clause to the end of the statement
WHERE t.date > '02-Jan-13'
The easiest way to do this is using window/analytic functions. These are ANSI standard and most databases support them (MySQL and Access being two notable exceptions).
Here is one way:
select t.accountnum, min_amount, max_amount,
min(case when amount = min_amount then date end) as min_amount_date,
min(case when amount = min_amount then date end) as max_amount_date,
from (Select t.*,
min(amount) over (partition by accountnum) as min_amount,
max(amount) over (partition by accountnum) as max_amount
from table t
where date > '02-Jan-13'
) t
group by accountnum, min_amount, max_amount;
order by accountnum
The subquery calculates the minimum and maximum amount for each account, using min() as a window function. The outer query selects these values. It then uses conditional aggregation to get the first date when each of those values occurred.
;with cte as
(
select accountnum, ammount, date,
row_number() over (partition by accountnum order by ammount desc) rn,
max(ammount) over (partition by accountnum) maxamount,
min(ammount) over (partition by accountnum) minamount
from table
where date > '20130102'
)
select accountnum,
ammount as amount,
date as date_of_max_amount,
minamount,
maxamount
from cte where rn = 1