I have here a sample table data which i want to use pivot and make my DedCode data into column.
the AMount example is already computed to its total but how can i count the number of employee if it is same with branch,deptcode and emptype.
my sample table data.
expected output:
You can use conditional aggregation:
select branch, deptcode, emptype, sum(empcount) as empcount,
sum(case when dedcode = 'PHIC' then amount else 0 end) as phic,
sum(case when dedcode = 'SLOAN' then amount else 0 end) as sloan,
sum(case when dedcode = 'VLOAN' then amount else 0 end) as vloan
from t
group by branch, deptcode, emptype;
Related
Below is the my base table
I need to create view like below format based on my base table.
Here Deposit_Acc_Count is combination of (Type -> SB,RD, FD) and Loan_Acc_Count is (Type ->LO) same as for amount.
Can anyone help me to achieve this.
You can use conditional aggregation while grouping by br_code and cif columns such as
SELECT br_code, cif,
COUNT(CASE WHEN type IN ('SB','RD', 'FD') THEN Amt END) AS deposit_acc_count,
SUM(CASE WHEN type IN ('SB','RD', 'FD') THEN Amt ELSE 0 END) AS deposit_acc_amt,
COUNT(CASE WHEN type = 'LO' THEN Amt END) AS loan_acc_count,
SUM(CASE WHEN type = 'LO' THEN Amt ELSE 0 END) AS loan_acc_amt,
COUNT(*) AS total_count,
SUM(amt) AS total_amt
FROM t
GROUP BY br_code, cif
Demo
I was wondering how I can SUM the values of a column based on another column's values being distinct like below. I tried the following two ways, each giving errors due to the aggregate function. I am trying to get NonDistinctTotals with the queries below.
SELECT SUM(InvoiceSaleAmt) AS NonDistinctTotals, SUM(case when count(*) over (partition by InvoiceNo) = 1 then InvoiceSaleAmt else 0 END) as DistinctTotals, SUM(CASE WHEN PaymentType= 'CASH' THEN CashTotal ELSE 0 END) AS CashTotal
FROM #InvoiceTable a
group by LocationId, InvoiceNo
Error: Windowed functions cannot be used in the context of another windowed function or aggregate.
SELECT SUM(InvoiceSaleAmt) AS NonDistinctTotals, SUM(CASE WHEN InvoiceNoin (SELECT DISTINCT InvoiceNofrom #InvoiceTable) THEN InvoiceSaleAmt else 0 END) as DistinctTotals, SUM(CASE WHEN PaymentType= 'CASH' THEN CashTotal ELSE 0 END) AS CashTotal
FROM #InvoiceTable a
group by LocationId, InvoiceNo
Error: Cannot perform an aggregate function on an expression containing an aggregate or a subquery.
Use a subquery:
SELECT SUM(InvoiceSaleAmt) AS NonDistinctTotals,
SUM(case when cnt = 1 then InvoiceSaleAmt else 0 END) as DistinctTotals,
SUM(CASE WHEN PaymentType = 'CASH' THEN CashTotal ELSE 0 END) AS CashTotal
FROM (SELECT it.*, COUNT(*) over (partition by InvoiceNo) as cnt
FROM #InvoiceTable it
) it
GROUP BY LocationId, InvoiceNo
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
I have a table of Accounts, having columns:
Acc_Id, Transaction_TypeId, Amount
I want to get result as When Transaction_TypeId = 1 then Sum of Amount as 'Total Advance Payments'.
Else when Transaction_typeId = 2 then Sum of Amount as 'Total Reciepts'
Here is my SQL query:
SELECT SUM(Amount) AS 'Sum' , Transaction_TypeId INTO #temp1 FROM AccountDetailTable WHERE Account_MasterId = 1 GROUP BY Transaction_TypeId
SELECT Sum as 'Total Advance' from #temp1 WHERE #temp1.Transaction_TypeId = 1;
SELECT Sum as 'Total Cash Receipts' FROM #temp1 WHERE #temp1.Transaction_TypeId = 2;
DROP TABLE #temp1;
but this query returns me two different result sets. How can i get the values in same result sets?
Use a CASE expression:
SELECT SUM(CASE WHEN Transaction_TypeId = 1 THEN somecolumn END) as [Total Advance],
SUM(CASE WHEN Transaction_TypeId = 2 THEN somecolumn END) as [Total Cash Receipts]
FROM #temp1;
You should use CASE EXPRESSION like this:
SELECT
sum(case when #temp1.Transaction_TypeId = 1 then amount else 0 end) as 'Total Advance',
sum(case when #temp1.Transaction_TypeId = 2 then amount else 0 end) as 'Total Cash Receipts'
FROM #temp1
I have the following data:
In SQL Server How can I have group by weekdate so I have only one row for each weekdate, example for the weekdate 2015-11-14:
Any clue?
Use conditional aggregation.
select cast(weekdate as date),
sum(case when permittype = 0 then total else 0 end) as permittype0,
sum(case when permittype = 1 then total else 0 end) as permittype1
from tablename
group by cast(weekdate as date)
I would do this using conditional aggregation:
select weekdate,
sum(case when permittype = 0 then total else 0 end) as permitttype0,
sum(case when permittype = 1 then total else 0 end) as permitttype1
from followingdata t
group by weekdate
order by weekdate;
You can also use pivot syntax, if you prefer.