multiple resultsets: multiply

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Friday, March 23, 2012

Multiply Rows with T - SQL

I need to multiply rows of a table with each other. I found a function like this

exp(sum(ln(floatfield1))).

But I can not find the ln function. Is there any other possibility ?

Thanks for any help.

Uwe

in function ? is this for MS SQL Server ?

Post your table structure, some sample data and the expected result. We should be able to help you to come up with the query|||

Uwe, LOG is the natural logarithm function in SQL Server. Also, see some of the previous discussions of this question: http://groups.google.com/groups/search?q=aggregate+product+kass+sqlserver Steve Kass Drew University Uwe Helmer@.discussions.microsoft.com wrote:
> I need to multiply rows of a table with each other. I found a function
> like this
>
> exp(sum(ln(floatfield1))).
>
> But I can not find the ln function. Is there any other possibility ?
>
>
>
> Thanks for any help.
>
> Uwe
>
>

|||

In your SQL statment

exp(sum(LOG(floatfield1))).

In SQL

Constant e (2.71828182845905…) based Natural log is set as LOG and 10 based is written as LOG10.

Some examples:

LOG ( e ) = 1.0

EXP( LOG( n ) ) = n.

10 ^ LOG10(n) = n

This seelect statement

SELECT LOG (EXP (5))

returns 5.

Multiply record fields with same mainID.

Hi everybody,
What's the most efficient way to get the following result:
I have a number of records with a MainID and a Value fields.
eg:
SubID MainID Value
1 1 1
2 1 1
3 1 2
4 1 3
5 2 1
6 2 1
7 3 2
8 3 2
I need the product of the Value field for each MainID.
The result has to be like this:
MainID PrValue
1 6 (1*1*2*3)
2 1 (1*1)
3 4 (2*2)
TIA,
Martin.select MainID, sum(value) as PrValue
from table
group by MainID
"martin" <kashaan007@.hotmail.com> wrote in message
news:Of%23nDPnRFHA.3144@.tk2msftngp13.phx.gbl...
> Hi everybody,
> What's the most efficient way to get the following result:
> I have a number of records with a MainID and a Value fields.
> eg:
> SubID MainID Value
> 1 1 1
> 2 1 1
> 3 1 2
> 4 1 3
> 5 2 1
> 6 2 1
> 7 3 2
> 8 3 2
> I need the product of the Value field for each MainID.
> The result has to be like this:
> MainID PrValue
> 1 6 (1*1*2*3)
> 2 1 (1*1)
> 3 4 (2*2)
> TIA,
> Martin.
>
>|||Try,
use northwind
go
create table t (
SubID int,
MainID int,
Value int
)
go
insert into t values(1, 1, 1)
insert into t values(2, 1, 1)
insert into t values(3, 1, 2)
insert into t values(4, 1, 3)
insert into t values(5, 2, 1)
insert into t values(6, 2, 1)
insert into t values(7, 3, 2)
insert into t values(8, 3, 2)
go
select
MainID,
POWER(10, SUM(LOG10(Value)))
from
t
group by
MainID
drop table t
go
I took the idea from:
The T-SQL Banker
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/html/TheT-SQLBanker.asp
AMB
"martin" wrote:
> Hi everybody,
> What's the most efficient way to get the following result:
> I have a number of records with a MainID and a Value fields.
> eg:
> SubID MainID Value
> 1 1 1
> 2 1 1
> 3 1 2
> 4 1 3
> 5 2 1
> 6 2 1
> 7 3 2
> 8 3 2
> I need the product of the Value field for each MainID.
> The result has to be like this:
> MainID PrValue
> 1 6 (1*1*2*3)
> 2 1 (1*1)
> 3 4 (2*2)
> TIA,
> Martin.
>
>|||Hi,AMB
It gives me a wrong output for the col1=2
CREATE TABLE #Test
(
col1 INT NOT NULL,
col2 INT NOT NULL
)
INSERT INTO #Test VALUES (1,10)
INSERT INTO #Test VALUES (1,2)
INSERT INTO #Test VALUES (1,3)
INSERT INTO #Test VALUES (2,4)
INSERT INTO #Test VALUES (2,2)
SELECT col1,POWER(10, SUM(LOG10(col2)))
FROM #Test GROUP BY col1
"Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in message
news:2DA4CABB-2DBD-43CA-B0DE-995B8B644F14@.microsoft.com...
> Try,
> use northwind
> go
> create table t (
> SubID int,
> MainID int,
> Value int
> )
> go
> insert into t values(1, 1, 1)
> insert into t values(2, 1, 1)
> insert into t values(3, 1, 2)
> insert into t values(4, 1, 3)
> insert into t values(5, 2, 1)
> insert into t values(6, 2, 1)
> insert into t values(7, 3, 2)
> insert into t values(8, 3, 2)
> go
> select
> MainID,
> POWER(10, SUM(LOG10(Value)))
> from
> t
> group by
> MainID
> drop table t
> go
> I took the idea from:
> The T-SQL Banker
>
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/html/TheT-SQLBanker.asp
>
> AMB
> "martin" wrote:
> > Hi everybody,
> >
> > What's the most efficient way to get the following result:
> >
> > I have a number of records with a MainID and a Value fields.
> >
> > eg:
> >
> > SubID MainID Value
> > 1 1 1
> > 2 1 1
> > 3 1 2
> > 4 1 3
> > 5 2 1
> > 6 2 1
> > 7 3 2
> > 8 3 2
> >
> > I need the product of the Value field for each MainID.
> >
> > The result has to be like this:
> >
> > MainID PrValue
> > 1 6 (1*1*2*3)
> > 2 1 (1*1)
> > 3 4 (2*2)
> >
> > TIA,
> >
> > Martin.
> >
> >
> >
> >|||Alejandro's solution is good for values greater than zero. Here's a
more general solution for all integers (this one adapted from Celko and
others):
SELECT mainid,
CAST(ROUND(
COALESCE(EXP(SUM(LOG(ABS(NULLIF(value,0))))),0)
* SIGN(MIN(ABS(value)))
* (COUNT(NULLIF(SIGN(value),1))%2*-2+1)
,0) AS INTEGER) AS product
FROM T
GROUP BY mainid
--
David Portas
SQL Server MVP
--|||Thanks a lot.
Works great.
Martin.
"Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in message
news:2DA4CABB-2DBD-43CA-B0DE-995B8B644F14@.microsoft.com...
> Try,
> use northwind
> go
> create table t (
> SubID int,
> MainID int,
> Value int
> )
> go
> insert into t values(1, 1, 1)
> insert into t values(2, 1, 1)
> insert into t values(3, 1, 2)
> insert into t values(4, 1, 3)
> insert into t values(5, 2, 1)
> insert into t values(6, 2, 1)
> insert into t values(7, 3, 2)
> insert into t values(8, 3, 2)
> go
> select
> MainID,
> POWER(10, SUM(LOG10(Value)))
> from
> t
> group by
> MainID
> drop table t
> go
> I took the idea from:
> The T-SQL Banker
>
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/
html/TheT-SQLBanker.asp
>
> AMB
> "martin" wrote:
> > Hi everybody,
> >
> > What's the most efficient way to get the following result:
> >
> > I have a number of records with a MainID and a Value fields.
> >
> > eg:
> >
> > SubID MainID Value
> > 1 1 1
> > 2 1 1
> > 3 1 2
> > 4 1 3
> > 5 2 1
> > 6 2 1
> > 7 3 2
> > 8 3 2
> >
> > I need the product of the Value field for each MainID.
> >
> > The result has to be like this:
> >
> > MainID PrValue
> > 1 6 (1*1*2*3)
> > 2 1 (1*1)
> > 3 4 (2*2)
> >
> > TIA,
> >
> > Martin.
> >
> >
> >
> >|||If we add +0.001 it works fine
SELECT col1,POWER(10, SUM(LOG10(col2))+0.001)
FROM #Test GROUP BY col1
"Uri Dimant" <urid@.iscar.co.il> wrote in message
news:uYVS5cnRFHA.1172@.TK2MSFTNGP12.phx.gbl...
> Hi,AMB
> It gives me a wrong output for the col1=2
> CREATE TABLE #Test
> (
> col1 INT NOT NULL,
> col2 INT NOT NULL
> )
> INSERT INTO #Test VALUES (1,10)
> INSERT INTO #Test VALUES (1,2)
> INSERT INTO #Test VALUES (1,3)
> INSERT INTO #Test VALUES (2,4)
> INSERT INTO #Test VALUES (2,2)
>
> SELECT col1,POWER(10, SUM(LOG10(col2)))
> FROM #Test GROUP BY col1
>
>
>
> "Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in
message
> news:2DA4CABB-2DBD-43CA-B0DE-995B8B644F14@.microsoft.com...
> > Try,
> >
> > use northwind
> > go
> >
> > create table t (
> > SubID int,
> > MainID int,
> > Value int
> > )
> > go
> >
> > insert into t values(1, 1, 1)
> > insert into t values(2, 1, 1)
> > insert into t values(3, 1, 2)
> > insert into t values(4, 1, 3)
> > insert into t values(5, 2, 1)
> > insert into t values(6, 2, 1)
> > insert into t values(7, 3, 2)
> > insert into t values(8, 3, 2)
> > go
> >
> > select
> > MainID,
> > POWER(10, SUM(LOG10(Value)))
> > from
> > t
> > group by
> > MainID
> >
> > drop table t
> > go
> >
> > I took the idea from:
> >
> > The T-SQL Banker
> >
>
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/html/TheT-SQLBanker.asp
> >
> >
> > AMB
> >
> > "martin" wrote:
> >
> > > Hi everybody,
> > >
> > > What's the most efficient way to get the following result:
> > >
> > > I have a number of records with a MainID and a Value fields.
> > >
> > > eg:
> > >
> > > SubID MainID Value
> > > 1 1 1
> > > 2 1 1
> > > 3 1 2
> > > 4 1 3
> > > 5 2 1
> > > 6 2 1
> > > 7 3 2
> > > 8 3 2
> > >
> > > I need the product of the Value field for each MainID.
> > >
> > > The result has to be like this:
> > >
> > > MainID PrValue
> > > 1 6 (1*1*2*3)
> > > 2 1 (1*1)
> > > 3 4 (2*2)
> > >
> > > TIA,
> > >
> > > Martin.
> > >
> > >
> > >
> > >
>|||Cast 10 to float.
SELECT
col1,
cast(POWER(cast(10 as float), SUM(LOG10(col2))) as decimal(5))
FROM #Test GROUP BY col1
go
AMB
"Uri Dimant" wrote:
> Hi,AMB
> It gives me a wrong output for the col1=2
> CREATE TABLE #Test
> (
> col1 INT NOT NULL,
> col2 INT NOT NULL
> )
> INSERT INTO #Test VALUES (1,10)
> INSERT INTO #Test VALUES (1,2)
> INSERT INTO #Test VALUES (1,3)
> INSERT INTO #Test VALUES (2,4)
> INSERT INTO #Test VALUES (2,2)
>
> SELECT col1,POWER(10, SUM(LOG10(col2)))
> FROM #Test GROUP BY col1
>
>
>
> "Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in message
> news:2DA4CABB-2DBD-43CA-B0DE-995B8B644F14@.microsoft.com...
> > Try,
> >
> > use northwind
> > go
> >
> > create table t (
> > SubID int,
> > MainID int,
> > Value int
> > )
> > go
> >
> > insert into t values(1, 1, 1)
> > insert into t values(2, 1, 1)
> > insert into t values(3, 1, 2)
> > insert into t values(4, 1, 3)
> > insert into t values(5, 2, 1)
> > insert into t values(6, 2, 1)
> > insert into t values(7, 3, 2)
> > insert into t values(8, 3, 2)
> > go
> >
> > select
> > MainID,
> > POWER(10, SUM(LOG10(Value)))
> > from
> > t
> > group by
> > MainID
> >
> > drop table t
> > go
> >
> > I took the idea from:
> >
> > The T-SQL Banker
> >
> http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/html/TheT-SQLBanker.asp
> >
> >
> > AMB
> >
> > "martin" wrote:
> >
> > > Hi everybody,
> > >
> > > What's the most efficient way to get the following result:
> > >
> > > I have a number of records with a MainID and a Value fields.
> > >
> > > eg:
> > >
> > > SubID MainID Value
> > > 1 1 1
> > > 2 1 1
> > > 3 1 2
> > > 4 1 3
> > > 5 2 1
> > > 6 2 1
> > > 7 3 2
> > > 8 3 2
> > >
> > > I need the product of the Value field for each MainID.
> > >
> > > The result has to be like this:
> > >
> > > MainID PrValue
> > > 1 6 (1*1*2*3)
> > > 2 1 (1*1)
> > > 3 4 (2*2)
> > >
> > > TIA,
> > >
> > > Martin.
> > >
> > >
> > >
> > >
>
>|||Good catch!!!
AMB
"Uri Dimant" wrote:
> Hi,AMB
> It gives me a wrong output for the col1=2
> CREATE TABLE #Test
> (
> col1 INT NOT NULL,
> col2 INT NOT NULL
> )
> INSERT INTO #Test VALUES (1,10)
> INSERT INTO #Test VALUES (1,2)
> INSERT INTO #Test VALUES (1,3)
> INSERT INTO #Test VALUES (2,4)
> INSERT INTO #Test VALUES (2,2)
>
> SELECT col1,POWER(10, SUM(LOG10(col2)))
> FROM #Test GROUP BY col1
>
>
>
> "Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in message
> news:2DA4CABB-2DBD-43CA-B0DE-995B8B644F14@.microsoft.com...
> > Try,
> >
> > use northwind
> > go
> >
> > create table t (
> > SubID int,
> > MainID int,
> > Value int
> > )
> > go
> >
> > insert into t values(1, 1, 1)
> > insert into t values(2, 1, 1)
> > insert into t values(3, 1, 2)
> > insert into t values(4, 1, 3)
> > insert into t values(5, 2, 1)
> > insert into t values(6, 2, 1)
> > insert into t values(7, 3, 2)
> > insert into t values(8, 3, 2)
> > go
> >
> > select
> > MainID,
> > POWER(10, SUM(LOG10(Value)))
> > from
> > t
> > group by
> > MainID
> >
> > drop table t
> > go
> >
> > I took the idea from:
> >
> > The T-SQL Banker
> >
> http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/html/TheT-SQLBanker.asp
> >
> >
> > AMB
> >
> > "martin" wrote:
> >
> > > Hi everybody,
> > >
> > > What's the most efficient way to get the following result:
> > >
> > > I have a number of records with a MainID and a Value fields.
> > >
> > > eg:
> > >
> > > SubID MainID Value
> > > 1 1 1
> > > 2 1 1
> > > 3 1 2
> > > 4 1 3
> > > 5 2 1
> > > 6 2 1
> > > 7 3 2
> > > 8 3 2
> > >
> > > I need the product of the Value field for each MainID.
> > >
> > > The result has to be like this:
> > >
> > > MainID PrValue
> > > 1 6 (1*1*2*3)
> > > 2 1 (1*1)
> > > 3 4 (2*2)
> > >
> > > TIA,
> > >
> > > Martin.
> > >
> > >
> > >
> > >
>
>|||David,
This is really a good one.
AMB
"David Portas" wrote:
> Alejandro's solution is good for values greater than zero. Here's a
> more general solution for all integers (this one adapted from Celko and
> others):
> SELECT mainid,
> CAST(ROUND(
> COALESCE(EXP(SUM(LOG(ABS(NULLIF(value,0))))),0)
> * SIGN(MIN(ABS(value)))
> * (COUNT(NULLIF(SIGN(value),1))%2*-2+1)
> ,0) AS INTEGER) AS product
> FROM T
> GROUP BY mainid
> --
> David Portas
> SQL Server MVP
> --
>|||Use better the one posted by David Portas, and if you decide to use this one,
read the correction in my answer to Uri.
AMB
"martin" wrote:
> Thanks a lot.
> Works great.
> Martin.
>
> "Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in message
> news:2DA4CABB-2DBD-43CA-B0DE-995B8B644F14@.microsoft.com...
> > Try,
> >
> > use northwind
> > go
> >
> > create table t (
> > SubID int,
> > MainID int,
> > Value int
> > )
> > go
> >
> > insert into t values(1, 1, 1)
> > insert into t values(2, 1, 1)
> > insert into t values(3, 1, 2)
> > insert into t values(4, 1, 3)
> > insert into t values(5, 2, 1)
> > insert into t values(6, 2, 1)
> > insert into t values(7, 3, 2)
> > insert into t values(8, 3, 2)
> > go
> >
> > select
> > MainID,
> > POWER(10, SUM(LOG10(Value)))
> > from
> > t
> > group by
> > MainID
> >
> > drop table t
> > go
> >
> > I took the idea from:
> >
> > The T-SQL Banker
> >
> http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/
> html/TheT-SQLBanker.asp
> >
> >
> > AMB
> >
> > "martin" wrote:
> >
> > > Hi everybody,
> > >
> > > What's the most efficient way to get the following result:
> > >
> > > I have a number of records with a MainID and a Value fields.
> > >
> > > eg:
> > >
> > > SubID MainID Value
> > > 1 1 1
> > > 2 1 1
> > > 3 1 2
> > > 4 1 3
> > > 5 2 1
> > > 6 2 1
> > > 7 3 2
> > > 8 3 2
> > >
> > > I need the product of the Value field for each MainID.
> > >
> > > The result has to be like this:
> > >
> > > MainID PrValue
> > > 1 6 (1*1*2*3)
> > > 2 1 (1*1)
> > > 3 4 (2*2)
> > >
> > > TIA,
> > >
> > > Martin.
> > >
> > >
> > >
> > >
>
>|||Ok thanks.
Martin.
"Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in message
news:210DC41A-BF9B-4211-9E16-995A554C0374@.microsoft.com...
> Use better the one posted by David Portas, and if you decide to use this
one,
> read the correction in my answer to Uri.
>
> AMB
> "martin" wrote:
> > Thanks a lot.
> >
> > Works great.
> >
> > Martin.
> >
> >
> > "Alejandro Mesa" <AlejandroMesa@.discussions.microsoft.com> wrote in
message
> > news:2DA4CABB-2DBD-43CA-B0DE-995B8B644F14@.microsoft.com...
> > > Try,
> > >
> > > use northwind
> > > go
> > >
> > > create table t (
> > > SubID int,
> > > MainID int,
> > > Value int
> > > )
> > > go
> > >
> > > insert into t values(1, 1, 1)
> > > insert into t values(2, 1, 1)
> > > insert into t values(3, 1, 2)
> > > insert into t values(4, 1, 3)
> > > insert into t values(5, 2, 1)
> > > insert into t values(6, 2, 1)
> > > insert into t values(7, 3, 2)
> > > insert into t values(8, 3, 2)
> > > go
> > >
> > > select
> > > MainID,
> > > POWER(10, SUM(LOG10(Value)))
> > > from
> > > t
> > > group by
> > > MainID
> > >
> > > drop table t
> > > go
> > >
> > > I took the idea from:
> > >
> > > The T-SQL Banker
> > >
> >
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnsqlmag02/
> > html/TheT-SQLBanker.asp
> > >
> > >
> > > AMB
> > >
> > > "martin" wrote:
> > >
> > > > Hi everybody,
> > > >
> > > > What's the most efficient way to get the following result:
> > > >
> > > > I have a number of records with a MainID and a Value fields.
> > > >
> > > > eg:
> > > >
> > > > SubID MainID Value
> > > > 1 1 1
> > > > 2 1 1
> > > > 3 1 2
> > > > 4 1 3
> > > > 5 2 1
> > > > 6 2 1
> > > > 7 3 2
> > > > 8 3 2
> > > >
> > > > I need the product of the Value field for each MainID.
> > > >
> > > > The result has to be like this:
> > > >
> > > > MainID PrValue
> > > > 1 6 (1*1*2*3)
> > > > 2 1 (1*1)
> > > > 3 4 (2*2)
> > > >
> > > > TIA,
> > > >
> > > > Martin.
> > > >
> > > >
> > > >
> > > >
> >
> >
> >

Multiply record fields with same mainID.

multiply by 1/4

Hi,
What's wrong with this:
DECLARE @.a as float
DECLARE @.b as float
SET @.a = (0.25) * 100
SET @.b = (1/4) * 100
print @.a
print @.b
Returns:
25
0
Why is the following returning 0 and not 25?
SET @.b = (1/4) * 100
Something similar is coded somewhere in one of our apps and is causing
incorrect reporting.
Thanks!gracie wrote:
> Hi,
> What's wrong with this:
> DECLARE @.a as float
> DECLARE @.b as float
> SET @.a = (0.25) * 100
> SET @.b = (1/4) * 100
> print @.a
> print @.b
> Returns:
> 25
> 0
> Why is the following returning 0 and not 25?
> SET @.b = (1/4) * 100
> Something similar is coded somewhere in one of our apps and is causing
> incorrect reporting.
> Thanks!
Do:
SET @.b = (1/4.0) * 100 ;
Otherwise you get an integer division, which equals 0.
Better still, specify a datatype for numeric literals:
SET @.b = (CAST(1 AS FLOAT)/CAST(4 AS FLOAT)) * 100 ;
--
David Portas, SQL Server MVP
Whenever possible please post enough code to reproduce your problem.
Including CREATE TABLE and INSERT statements usually helps.
State what version of SQL Server you are using and specify the content
of any error messages.
SQL Server Books Online:
http://msdn2.microsoft.com/library/ms130214(en-US,SQL.90).aspx
--|||Perfect. Thanks.
Since they both return the correct result, why is specifying the datatype
for numeric literals better?
"gracie" wrote:
> Hi,
> What's wrong with this:
> DECLARE @.a as float
> DECLARE @.b as float
> SET @.a = (0.25) * 100
> SET @.b = (1/4) * 100
> print @.a
> print @.b
> Returns:
> 25
> 0
> Why is the following returning 0 and not 25?
> SET @.b = (1/4) * 100
> Something similar is coded somewhere in one of our apps and is causing
> incorrect reporting.
> Thanks!|||It gives you control of the input datatype so you can determine over which datatypes the operation
will occur.
--
Tibor Karaszi, SQL Server MVP
http://www.karaszi.com/sqlserver/default.asp
http://www.solidqualitylearning.com/
Blog: http://solidqualitylearning.com/blogs/tibor/
"gracie" <gracie@.discussions.microsoft.com> wrote in message
news:D037A0AB-7A95-4285-9471-3695FAB55CFC@.microsoft.com...
> Perfect. Thanks.
> Since they both return the correct result, why is specifying the datatype
> for numeric literals better?
>
> "gracie" wrote:
>> Hi,
>> What's wrong with this:
>> DECLARE @.a as float
>> DECLARE @.b as float
>> SET @.a = (0.25) * 100
>> SET @.b = (1/4) * 100
>> print @.a
>> print @.b
>> Returns:
>> 25
>> 0
>> Why is the following returning 0 and not 25?
>> SET @.b = (1/4) * 100
>> Something similar is coded somewhere in one of our apps and is causing
>> incorrect reporting.
>> Thanks!|||gracie wrote:
> Perfect. Thanks.
> Since they both return the correct result, why is specifying the datatype
> for numeric literals better?
>
Otherwise, how can you guarantee which numeric datatype, scale and
precision the server will pick? Not specifying the datatype may work
the way you expect it today, but the next version of SQL Server may
change things. In fact, the rules for implict casting of datatypes have
changed several times in different releases of SQL Server.
--
David Portas, SQL Server MVP
Whenever possible please post enough code to reproduce your problem.
Including CREATE TABLE and INSERT statements usually helps.
State what version of SQL Server you are using and specify the content
of any error messages.
SQL Server Books Online:
http://msdn2.microsoft.com/library/ms130214(en-US,SQL.90).aspx
--

multiply by 1/4

Hi,
What's wrong with this:
DECLARE @.a as float
DECLARE @.b as float
SET @.a = (0.25) * 100
SET @.b = (1/4) * 100
print @.a
print @.b
Returns:
25
0
Why is the following returning 0 and not 25?
SET @.b = (1/4) * 100
Something similar is coded somewhere in one of our apps and is causing
incorrect reporting.
Thanks!
gracie wrote:
> Hi,
> What's wrong with this:
> DECLARE @.a as float
> DECLARE @.b as float
> SET @.a = (0.25) * 100
> SET @.b = (1/4) * 100
> print @.a
> print @.b
> Returns:
> 25
> 0
> Why is the following returning 0 and not 25?
> SET @.b = (1/4) * 100
> Something similar is coded somewhere in one of our apps and is causing
> incorrect reporting.
> Thanks!
Do:
SET @.b = (1/4.0) * 100 ;
Otherwise you get an integer division, which equals 0.
Better still, specify a datatype for numeric literals:
SET @.b = (CAST(1 AS FLOAT)/CAST(4 AS FLOAT)) * 100 ;
David Portas, SQL Server MVP
Whenever possible please post enough code to reproduce your problem.
Including CREATE TABLE and INSERT statements usually helps.
State what version of SQL Server you are using and specify the content
of any error messages.
SQL Server Books Online:
http://msdn2.microsoft.com/library/ms130214(en-US,SQL.90).aspx
|||Perfect. Thanks.
Since they both return the correct result, why is specifying the datatype
for numeric literals better?
"gracie" wrote:

> Hi,
> What's wrong with this:
> DECLARE @.a as float
> DECLARE @.b as float
> SET @.a = (0.25) * 100
> SET @.b = (1/4) * 100
> print @.a
> print @.b
> Returns:
> 25
> 0
> Why is the following returning 0 and not 25?
> SET @.b = (1/4) * 100
> Something similar is coded somewhere in one of our apps and is causing
> incorrect reporting.
> Thanks!
|||It gives you control of the input datatype so you can determine over which datatypes the operation
will occur.
Tibor Karaszi, SQL Server MVP
http://www.karaszi.com/sqlserver/default.asp
http://www.solidqualitylearning.com/
Blog: http://solidqualitylearning.com/blogs/tibor/
"gracie" <gracie@.discussions.microsoft.com> wrote in message
news:D037A0AB-7A95-4285-9471-3695FAB55CFC@.microsoft.com...[vbcol=seagreen]
> Perfect. Thanks.
> Since they both return the correct result, why is specifying the datatype
> for numeric literals better?
>
> "gracie" wrote:
|||gracie wrote:
> Perfect. Thanks.
> Since they both return the correct result, why is specifying the datatype
> for numeric literals better?
>
Otherwise, how can you guarantee which numeric datatype, scale and
precision the server will pick? Not specifying the datatype may work
the way you expect it today, but the next version of SQL Server may
change things. In fact, the rules for implict casting of datatypes have
changed several times in different releases of SQL Server.
David Portas, SQL Server MVP
Whenever possible please post enough code to reproduce your problem.
Including CREATE TABLE and INSERT statements usually helps.
State what version of SQL Server you are using and specify the content
of any error messages.
SQL Server Books Online:
http://msdn2.microsoft.com/library/ms130214(en-US,SQL.90).aspx

multiply by 1/4

Hi,
What's wrong with this:
DECLARE @.a as float
DECLARE @.b as float
SET @.a = (0.25) * 100
SET @.b = (1/4) * 100
print @.a
print @.b
Returns:
25
0
Why is the following returning 0 and not 25?
SET @.b = (1/4) * 100
Something similar is coded somewhere in one of our apps and is causing
incorrect reporting.
Thanks!gracie wrote:
> Hi,
> What's wrong with this:
> DECLARE @.a as float
> DECLARE @.b as float
> SET @.a = (0.25) * 100
> SET @.b = (1/4) * 100
> print @.a
> print @.b
> Returns:
> 25
> 0
> Why is the following returning 0 and not 25?
> SET @.b = (1/4) * 100
> Something similar is coded somewhere in one of our apps and is causing
> incorrect reporting.
> Thanks!
Do:
SET @.b = (1/4.0) * 100 ;
Otherwise you get an integer division, which equals 0.
Better still, specify a datatype for numeric literals:
SET @.b = (CAST(1 AS FLOAT)/CAST(4 AS FLOAT)) * 100 ;
David Portas, SQL Server MVP
Whenever possible please post enough code to reproduce your problem.
Including CREATE TABLE and INSERT statements usually helps.
State what version of SQL Server you are using and specify the content
of any error messages.
SQL Server Books Online:
http://msdn2.microsoft.com/library/ms130214(en-US,SQL.90).aspx
--|||Perfect. Thanks.
Since they both return the correct result, why is specifying the datatype
for numeric literals better?
"gracie" wrote:

> Hi,
> What's wrong with this:
> DECLARE @.a as float
> DECLARE @.b as float
> SET @.a = (0.25) * 100
> SET @.b = (1/4) * 100
> print @.a
> print @.b
> Returns:
> 25
> 0
> Why is the following returning 0 and not 25?
> SET @.b = (1/4) * 100
> Something similar is coded somewhere in one of our apps and is causing
> incorrect reporting.
> Thanks!|||It gives you control of the input datatype so you can determine over which d
atatypes the operation
will occur.
Tibor Karaszi, SQL Server MVP
http://www.karaszi.com/sqlserver/default.asp
http://www.solidqualitylearning.com/
Blog: http://solidqualitylearning.com/blogs/tibor/
"gracie" <gracie@.discussions.microsoft.com> wrote in message
news:D037A0AB-7A95-4285-9471-3695FAB55CFC@.microsoft.com...[vbcol=seagreen]
> Perfect. Thanks.
> Since they both return the correct result, why is specifying the datatype
> for numeric literals better?
>
> "gracie" wrote:
>|||gracie wrote:
> Perfect. Thanks.
> Since they both return the correct result, why is specifying the datatype
> for numeric literals better?
>
Otherwise, how can you guarantee which numeric datatype, scale and
precision the server will pick? Not specifying the datatype may work
the way you expect it today, but the next version of SQL Server may
change things. In fact, the rules for implict casting of datatypes have
changed several times in different releases of SQL Server.
David Portas, SQL Server MVP
Whenever possible please post enough code to reproduce your problem.
Including CREATE TABLE and INSERT statements usually helps.
State what version of SQL Server you are using and specify the content
of any error messages.
SQL Server Books Online:
http://msdn2.microsoft.com/library/ms130214(en-US,SQL.90).aspx
--

Friday, March 23, 2012

Multiply Rows with T - SQL

Multiply record fields with same mainID.

Multiply record fields with same mainID.

Multiply record fields with same mainID.

multiply by 1/4

multiply by 1/4

multiply by 1/4

multiple resultsets

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