division problem and attaching a % symbol in Query.
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Tuesday, September 18, 2012 7:24 AM
@AVD AS AvailableDays,(@AVD - Cast(SUM(T.ActualWorkBillable) as int)) as AfterHolidays,
((@AVD - Cast(SUM(T.ActualWorkBillable) as int)) / @AVD) as Percentage
These three fields which I am using in my query
I want to divide "AfterHolidays" with "AvailableDays" and show the value in Percentage column with two decimal places and with '%' sign.
I am doing this but there is no decimal places and no '%' symbol.
How to I do this.
All Replies
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Tuesday, September 18, 2012 7:27 AM
pls try
((@AVD*1.0) - SUM(T.ActualWorkBillable)) as AfterHolidays,
cast((((@AVD*1.0) - SUM(T.ActualWorkBillable) ) / (@AVD*1.0)) as varchar(20) )+'%' PercentageVT
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- Edited by v.vtMicrosoft Community Contributor Tuesday, September 18, 2012 7:29 AM
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Tuesday, September 18, 2012 7:29 AM
Pleas Try
Convert(VARCHAR,((@AVD - Cast(SUM(T.ActualWorkBillable) as Float) / CAST(@AVD as Float))+'%' as Percentage
Thanks, Sachin Surve
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Tuesday, September 18, 2012 7:30 AM
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Tuesday, September 18, 2012 7:37 AM
try
cast(cast((((@AVD*1.0) - SUM(T.ActualWorkBillable) ) / (@AVD*1.0)) as decimal(10,2))as varchar(20) )+'%' Percentage
vt
Please mark answered if I've answered your question and vote for it as helpful to help other user's find a solution quicker
- Marked As Answer by John.Eddie Wednesday, September 19, 2012 8:47 AM
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Tuesday, September 18, 2012 7:45 AM
Replace varchar(15) with varchar(Max) or Varchar(4000)
sample code
declare @AVG int, @Val int set @AVG=123456 set @Val=10000000 select convert(varchar,cast(cast(@Val AS decimal(19,2))/cast(@AVG AS decimal(19,2)) AS decimal(10,2)))
Thanks, Sachin Surve
- Marked As Answer by John.Eddie Wednesday, September 19, 2012 8:47 AM
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Tuesday, September 18, 2012 8:26 AM
CAST
(@AVD as Decimal(19,2)) - CONVERT(varchar,CAST(CAST(SUM(T.ActualWorkBillable) as decimal(19,2)))) / CAST(@AVD as decimal(19,2)) as Percentage
- Edited by John.Eddie Tuesday, September 18, 2012 8:26 AM
- Marked As Answer by John.Eddie Wednesday, September 19, 2012 8:47 AM
- Unmarked As Answer by Naomi NMicrosoft Community Contributor, Moderator Thursday, September 20, 2012 2:12 AM
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Tuesday, September 18, 2012 8:33 AM
what is the datatype of T.ActualWorkBillable as well as @AVD? are they of type of INT?
CONVERT( NVARCHAR(100), ( CONVERT( INT, @AVD ) - ( SUM ( CONVERT( INT, T.ActualWorkBillable) ) / CONVERT( INT, @AVD ) ) ) + '%' as Percentage
regards
joon
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Tuesday, September 18, 2012 8:45 AM
This is working correct and getting percentage .
but tell me how to I format this to percentage upto 2 decimal and need to show % symbol in collumn.
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Tuesday, September 18, 2012 9:08 AM
CONVERT(VARCHAR,CAST(CAST(@ADV - SUM(T.ActualWorkBillable) as Decimal(19,2))/CAST(@AVD As Decimal(19,2))*100 AS DECIMAL(19,2)))+'%'This is working correct and getting percentage .
but tell me how to I format this to percentage upto 2 decimal and need to show % symbol in collumn.
Thanks, Sachin Surve
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Tuesday, September 18, 2012 10:26 AM
CONVERT( NVARCHAR(100), ( CONVERT( decimal (35,2) , @AVD ) - ( SUM ( CONVERT( decimal (35,2), T.ActualWorkBillable) ) / CONVERT( decimal (35,2), @AVD ) ) ) + '%' as Percentage
you can decide, where you want to apply the decimal points, either on the sum part or for whole calculation.
regards
joon
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Tuesday, September 18, 2012 11:20 AM
Netiquette in an SQL forum is to post DDL. Rows are not fields. Display formatting is done in the presentation layer. Puttign that percent sign on this is the worst way to write SQL. You do not know about INTEGER math in SQL; if you use integers, you will an integer results. Here is a guess at what you meant.
DECLARE @in_available_day_cnt INTEGER;
DECLARE @after_holidays_day_cnt INTEGER;
SET @after_holidays_day_cnt
= (@in_available_day_cnt
- SUM(T.actual_work_billable);
DECLARE @after_holidays_day_percent DECIMAL (12,3);
SET @after_holidays_day_percentage
= (@after_holidays_day_cnt
- SUM(T.Actual_work_billable)
/ @after_holidays_day_cnt;
--CELKO-- Books in Celko Series for Morgan-Kaufmann Publishing: Analytics and OLAP in SQL / Data and Databases: Concepts in Practice Data / Measurements and Standards in SQL SQL for Smarties / SQL Programming Style / SQL Puzzles and Answers / Thinking in Sets / Trees and Hierarchies in SQL
- Proposed As Answer by Naomi NMicrosoft Community Contributor, Moderator Thursday, September 20, 2012 2:12 AM

