my query as below will get result as in the picture
SELECT Dept,
Item,
Month,
OpenQty,
OpenVal
FROM StockTransaction
WHERE year = 2016 and
month between 1 and 10
actually the result i want are only the minimum month for each dept and item will hold the openqty and openval value. Other than that will hold zero value.
how to write query to get the result as above picture?
You can use window functions:
SELECT Dept, Item, Month,
(CASE WHEN row_number() over (partition by dept, item order by month) = 1
THEN OpenQty ELSE 0
END) as OpenQty,
(CASE WHEN row_number() over (partition by dept, item order by month) = 1
THEN OpenVal ELSE 0
END) as OpenVal
FROM StockTransaction
WHERE year = 2016 and
month between 1 and 10;
Related
There are a lot questions on CASE WHEN topic, but the closest my question is related to this How to use CASE WHEN condition with MAX() function query which has not been resolved.
Here is some of my sample data:
date
debet
2022-07-15
57190.33
2022-07-14
815616516.00
2022-07-15
40866.67
2022-07-14
1221510.00
So, I want to all records for the last two dates and three additional columns: sum(sales) for the previous day, sum for the current day and the difference between them:
SELECT
[debet],
[date] ,
SUM( CASE WHEN [date] = MAX(date) THEN [debet] ELSE 0 END ) AS sum_act,
SUM( CASE WHEN [date] = MAX(date) - 1 THEN [debet] ELSE 0 END ) AS sum_prev ,
(
SUM( CASE WHEN [date] = MAX(date) THEN [debet] ELSE 0 END )
-
SUM( CASE WHEN [date] = MAX(date) - 1 THEN [debet] ELSE 0 END )
) AS diff
FROM
Table
WHERE
[date] = ( SELECT MAX(date) FROM Table WHERE date < ( SELECT MAX(date) FROM Table) )
OR
[date] = ( SELECT MAX(date) FROM Table WHERE date = ( SELECT MAX(date) FROM Table ) )
GROUP BY
[date],
[debet]
Further, of course, it informs that I can't use the aggregate function inside CASE WHEN. Now I use this combination: sum(CASE WHEN [date] = dateadd(dd,-3,cast(getdate() as date)) THEN [debet] ELSE 0 END). But here every time I need to make an adjustment for weekends and holidays. The question is, is there any other way than using 'getdate' in 'case when' Statement to get max date?
Expected result:
date
sum_act
sum_prev
diff
2022-07-15
97190.33
0.00
97190.33
2022-07-14
0.00
508769.96
-508769.96
You can use dense_rank() to filter the last 2 dates in your table. After that you can use either conditional case expression with sum() to calculate the required value
select [date],
sum_act = sum(case when rn = 1 then [debet] else 0 end),
sum_prev = sum(case when rn = 2 then [debet] else 0 end),
diff = sum(case when rn = 1 then [debet] else 0 end)
- sum(case when rn = 2 then [debet] else 0 end)
from
(
select *, rn = dense_rank() over (order by [date] desc)
from tbl
) t
where rn <= 2
group by [date]
db<>fiddle demo
Two steps:
Get the sums for the last three dates
Show the results for the last two dates.
Well, we could also get all daily sums in step 1, but we just need the last three in order to calculate the sums for the last two days, so why aggregate more data than necessary?
Here is the query. You may have to put the date column name in brackets in SQL Server, as date is a keyword in SQL.
select top(2)
date,
sum_debit_current,
sum_debit_previous,
sum_debit_current - sum_debit_previous as diff
(
select
date,
sum(debet) as sum_debit_current,
lag(sum(debet)) over (order by date) as sum_debit_previous
from table
where date in (select distinct top(3) date from table order by date desc)
group by date
)
order by date desc;
(SQL Server uses TOP(n) instead of standard SQL FETCH FIRST 3 ROWS and while SELECT DISTINCT TOP(3) date looks like "get the top 3 rows, then apply distinct on their date", it is really "apply distinct on the dates, then get the top 3" like in standard SQL.)
I've got the following table:
and I was wondering if there is an SQL query, which would give me the begin and end Calender week (CW), where the value is greater than 0.
So in the case of the table above, a result like below:
Thanks in advance!
You can assign a group by counting the number of zeros and then aggregating:
select article_nr, min(year), max(year)
from (select t.*,
sum(case when amount = 0 then 1 else 0 end) over (partition by article_nr order by year) as grp
from t
) t
where amount > 0
group by article_nr, grp;
select Atricle_Nr, min(Year&CW) as 'Begin(Year&CW)',max(Year&CW) as 'End(Year&CW)'
from table where Amount>0 group by Atricle_Nr;
I have a table with 4 columns and need to create a newcolumn, which represents that if for the following year's quarter exists at least one data='colour', so for this quarter all values are changed to 'colour. How to do it?
You can use condition and get count of no for each quarter as follows:
select t.*,
case when count(case when data = 'no' then 1 end)
over (partition by year, quarter) > 0
then 'nocolour'
else data
end as new_column
from your_table t
You can use window functions:
select t.*,
min(data) over(partition by year, quarter) as new_column
from mytable t
If there is any 'no' for a given quarter, new_column takes value 'no' for all rows of the quarter. This works because string-wise, 'no' < 'yes'.
Assuming that you really mean "has data = 'yes' and the next quarter and you want to update the table, you can use:
update t
set newcolumn = (case when next_max_data = 'yes' then 'colour' else 'no colour')
from t join
(select year, quarter,
lead(max(data)) over (order by year, quarter) as next_max_data
from t
group by year, quarter
) tt
on t.year = tt.year and t.quater = tt.quarter;
If you want the the quarter a year later, you would use lead(max(data), 4) in the subquery.
I want to select the first non-null row with the minimum date. I'll like to use a CASE WHEN that condition is met, then 1 ELSE 0.
So more like CASE WHEN row IS NOT and DATE is minimum DATE then 1 ELSE 0. I just need to select ONLY one row.
Another option (for BigQuery Standard SQL)
#standardSQL
SELECT *, 0 AS marker FROM `project.dataset.table` WHERE item_count IS NULL
UNION ALL
SELECT *, IF(1 = ROW_NUMBER() OVER(PARTITION BY user ORDER BY date), 1, 0)
FROM `project.dataset.table` WHERE NOT item_count IS NULL
ORDER BY user, date
Consider:
select
t.*
case when date = min(case when itemcount is not null then date end) over(partition by user order by date)
then 1
else 0
end as marker
from mytable t
I am unsure whether BigQuery supports minif() as a window function:
select
t.*
case when date = minif(date, itemcount is not null) over(partition by user order by date)
then 1
else 0
end as marker
from mytable
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