How to get count of a particular row - sql

I have table that contain Id,Date and Status i.e open/close
i just want a result in sql that contain month wise open,close and total count of Id's
e.g In Jan open count 15,close count 5 and total count 20

Use RollUp() and Group By as below:
;WITH T AS
(
SELECT
Id,
DATENAME(MONTH,[Date]) AS [MonthName],
Status
FROM #tblTest
)
SELECT
[MonthName],
[Status],
StatusCount
FROM
(
SELECT
MonthName,
CASE ISNULL(Status,'') WHEN '' THEN 'Total' ELSE Status END AS Status,
Count(Status) AS StatusCount
FROM T
GROUP BY ROLLUP([MonthName],[Status])
)X
WHERE X.MonthName IS NOT NULL
ORDER BY X.[MonthName],X.[Status]
Output:
Note: If required data in single row by month then apply PIVOT

select year(date), month(date),
sum(case when status = 'open' then 1 else 0 end) as open_count,
sum(case when status = 'closed' then 1 else 0 end) as closed_count,
count(*) as total_count
from your_table
group by year(date), month(date)

Related

Combine 2 queries together

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)

CASE WHEN condition with MAX() function

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.)

MSSQL Group by and Select rows from grouping

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

Get the Highest Value in different Select SUM

I want to get the highest value in my query
Select SUM(CASE WHEN Day='Monday' THEN 1 END) AS'Total Monday',
SUM(CASE WHEN Day='Tuesday' THEN 1 END) AS'Total Tuesday'
FROM tbl_sched
WHERE teacherID='2014279384'
The Output would be TotalMonday ='1' and TotalTuesday ='2'
I need to get the highest value from the outputs which in this case is TotalTuesday=2
select max(daycnt) from
(Select SUM(CASE WHEN Day='Monday' THEN 1 END) AS daycnt
from tbl_sched WHERE teacherID='2014279384'
union all
Select SUM(CASE WHEN Day='Tuesday' THEN 1 END) AS daycnt
from tbl_sched WHERE teacherID='2014279384')
If you need the max between many columns:
Something interesting in SQLServer 2008 and above
SELECT (SELECT Max(v)
FROM (VALUES ([Total Monday]), ([Total Tuesday]), ...) AS value(v)) as [MaxDate]
From
(
Select SUM(CASE WHEN Day='Monday' THEN 1 END) AS'Total Monday',
SUM(CASE WHEN Day='Tuesday' THEN 1 END) AS'Total Tuesday'
..........
FROM tbl_sched
WHERE teacherID='2014279384'
)a
Another option:
SELECT Case When [Total Monday] > [Total Tuesday] then [Total Monday] else [Total Tuesday] End as maxvalue
FROM
(
Select SUM(CASE WHEN Day='Monday' THEN 1 END) AS'Total Monday',
SUM(CASE WHEN Day='Tuesday' THEN 1 END) AS'Total Tuesday'
FROM tbl_sched
WHERE teacherID='2014279384'
)a
I'd say the query below is better in terms of performance and highlights the intention better, because basically we are just GROUPing by days and COUNTing the groups, we don't need CASE's or SUM's (in which case SQL Server will have to go over all the records of the selected teacher).
SELECT TOP 1 Day, COUNT(*) AS Total
FROM tbl_sched
WHERE teacherID='2014279384'
AND Day IN ('Monday','Tuesday')
GROUP BY Day
ORDER BY Total DESC
You can just group by Day, sort by COUNT(*) DESC and get the top count:
SELECT TOP (1)
TotalCount = COUNT(*)
FROM
dbo.tbl_sched
WHERE
teacherID = '2014279384'
GROUP BY
Day
ORDER BY
TotalCount DESC
;
You can also include Day into the output to return the day that had the topmost result.
You can achieve this by using Max Function
Select MAX(SUM(CASE WHEN Day='Monday' THEN 1 END)) AS 'Total Monday',
MAX(SUM(CASE WHEN Day='Tuesday' THEN 1 END)) AS 'Total Tuesday'
FROM tbl_sched
WHERE teacherID='2014279384'

Multiple sub queries

Is it possible to get the result as below from the same table date-wise records:
Enrolled Enrolled as Email Enrolled as Text Deals Redeemed
<First Date> 7 5 2 6
<Next Date> 9 3 6 14
Table structure look something like this:
Customer_id, field1, field2, responsecode, created_date
My current query is something like this:
Select
Created,
Enroll = (Select COUNT(*) from tblCustomer where field1 <> '' group by created),
Email = (Select COUNT(field1) from tblCustomer where field1 = 'E-mail' and field1 <> '' group by created),
Cell = (Select COUNT(*) from tblCustomer where field1 = 'Cell Phone' and field1 <> '' group by created)
from tblCustomer
group by created
order by created
You don't want a COUNT(), but instead, a SUM( CASE/WHEN )
Select
Created,
count(*) TotalCnt,
SUM( CASE WHEN Field1 = 'E-mail' then 1 else 0 END ) as EMailCnt,
SUM( CASE WHEN Field1 = 'Cell Phone' then 1 else 0 END ) as CellCnt,
SUM( CASE WHEN RedeamedCondition then 1 else 0 END ) as RedeamCnt
from
tblCustomer
group by
created
order by
created
Note... if created is a date/time you will need to have the group by based on the date portion ONLY of the "created", otherwise you would get different counts for every second... From another post, the following gets only the date portion of a date time by basically removing the hours:minutes:seconds portion
DATEADD(dd, 0, DATEDIFF(dd, 0, created))
or if that doesn't make sense, you could just do it by
datepart( yy, created) as GrpYear,
datepart( mm, created) as GrpMonth,
datepart( dd, created) as GrpDay, ... rest of columns.....
Try:
select created_date,
count(field1) Enrolled,
count(case field1 when 'E-mail' then 1 end) Enrolled_as_Email,
count(case field1 when 'Cell Phone' then 1 end) Enrolled_as_Cell,
(Select COUNT(*)
from tbl_TransactionDishout d
where d.created = c.created_date and
d.DishoutResponseCode = '0000') Deals_Redeemed
from tblCustomer c
group by created_date
order by created_date