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Formatting Decimal Places In A Query In MS SQL

Hey - I have a quick question and know that it is probably pretty simple, but I am stumped. I have a query where I need to make a colum a number that looks like a percent with 2 significant digits:

SELECT tblNumericCovert.number1, tblNumericCovert.number2, [number1]/[number2] AS testDiv
FROM tblNumericCovert

where testDiv needs to spit out results like this ###.##

I am totally lost, if anyone can help, I would appreciate it.

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from tblDataHeader
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where fkDataHeaderID = @DataHeaderID
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set @ErrorMessage = null
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SET @SQL_Update = 'UPDATE ' + @server + '.' + @ASWTableName + ' set '
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SET @SQL_Where = ' WHERE '
SET @whereCounter = 0

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from tblASWField
inner join tblASWData on tblASWField.ASWFieldID = tblASWData.fkASWFieldID
where fkASWTableID = @ASWTableID and fkDataHeaderID = @DataHeaderID
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select getdate(), @ASWFieldName, 'testposition 3'
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if @DataTypeIsNumeric = 0
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set @SQL_Insert = @SQL_Insert + @ASWFieldName + ', '
set @SQL_InsertValues = @SQL_InsertValues + @Data + ', '
IF @AllowASWUpdate = 1
set @SQL_Update = @SQL_Update + @ASWFieldName + ' = ' + @Data + ', '
set @updateCounter = @updateCounter + 1
IF @IsPrimaryKey = 1
set @SQL_Where = @SQL_Where + @ASWFieldName + ' = ' + @Data + ' and '
SET @whereCounter = @whereCounter + 1

FETCH NEXT FROM ASWField_Cursor INTO @ASWFieldName, @AllowASWUpdate, @IsPrimaryKey, @DataTypeIsNumeric, @Data
select getdate(), 'testposition 4'
CLOSE ASWField_Cursor

SET @SQL_Where = LEFT(@SQL_Where, LEN(@SQL_Where) - 4)
SET @SQL_Insert = LEFT(@SQL_Insert, LEN(@SQL_Insert) - 1) + ') ' + LEFT(@SQL_InsertValues, LEN(@SQL_InsertValues) - 1) + ')'
SET @SQL_Update = LEFT(@SQL_Update, LEN(@SQL_Update) - 1) + @SQL_Where
SET @SQL_CheckIfAlreadyExist = 'Update #tmptblUpdateASW set ASWRowAlreadyExists = ' +
'(SELECT * from OPENQUERY(' + @shortServer + ','' SELECT count(*) FROM ' + @ASWTableName + ' ' + replace(@SQL_Where,char(39), char(39) + char(39)) + ' ''))'
select getdate(), 'testposition 4'
select getdate(), 'testposition 5'
select getdate(), 'testposition 6'

IF @whereCounter = 0
insert into tblASWUpdateLog(LogTime, fkDataHeaderID, fkASWTableID, ASWAction, ErrorMessage)
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update tblDataHeader set isSentToASW = 1 where DataHeaderID = @DataHeaderID
ELSE IF (select ASWRowAlreadyExists from #tmptblUpdateASW) = 0
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update tblDataHeader set isSentToASW = 1 where DataHeaderID = @DataHeaderID


FETCH NEXT FROM ASWTable_Cursor INTO @ASWTableID, @ASWTableName, @RuleWhen
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FETCH NEXT FROM Header_Cursor INTO @DataHeaderID
CLOSE Header_Cursor
DEALLOCATE Header_Cursor

UPDATE tblDataBatch set DateToASW = getDate() where DataBatchID = @DataBatchID

FETCH NEXT FROM Batch_Cursor INTO @DataBatchID
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DROP Table #tmptblUpdateASW


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SQL 6.5 Query Formatting Help
I am trying to run a query for the company that I work for and I am getting the right data but more than needed. the query is like this

select distinct(companyname),max(calldate) from tablename

what I want to see is the latest call made to a customer for each company
but what I get back is the company listed several times with the dates of each call. I only want the latest. Any help?

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SQL Query Formatting

Im new to sql server what i wanted to do was as follows:

i got a table with application_name and app_version

it looks like as follows

Application_name App_ver
Oracle 8.1.0
Oracle 8.5
Oracle 9.1
Oracle 10.1
Sql Server 6.5
Sql Server 7
Sql Server 2000
Sql Server 2005

i want to get query the table and produce an out put as follows:

Application_name App_vers
Oracle 8.1.0, 8.5, 9.1, 10.1
Sql Server 6.5, 7, 2000, 2005

please note i need the commas in place
i have written several queries to try and do it but no luck

im using sql server 2000

i will be greatful if any one can help...

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Formatting Results Of The Query
my current query returns data as follows:

stopID reason count

h1 e 5
h1 v 2
h1 a 1
h2 v 2
h2 m 8
h3 t 6
h3 v 1

Is it even possible to sort/format this query in SQL to
return the same data but in this type of format? Using
the values of the 'reason' column as column names?

stopID e v a m t

h1 5 2 1 0 0
h2 0 2 0 8 0
h3 0 1 0 0 6



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