Re: [MonetDB-users] Mathematical functions

AND for some reason when I apply the cast like this: (CAST(field1 AS NUMERIC(12,4))/CAST(field2 AS NUMERIC(12,4))) AS ratio the result is 0.2 when field1=3 and field2=5 - shouldn't the result be 0.6? On Thu, Nov 5, 2009 at 10:53 AM, Guillaume Theoret wrote:

MonetDB's "numeric" basically a decimal type.

Mysql has a good explanation:

http://dev.mysql.com/doc/refman/5.0/en/precision-math-decimal-changes.html

So basically 12,4 means 12 digits, 4 of them to the right of the decimal.

So 12345678.1234 is a representable number but if you try more digits on either side they will either be rounded in the case of the decimal side or you will get an overflow error in the case of the left side:

sql>select cast(12345678.1234 as decimal(12,4)); select cast(12345678.1234 as decimal(12,4)); +----------------+ | single_value | +================+ | 12345678.1234 | +----------------+ 1 tuple Timer 0.400 msec 1 rows sql>select cast(12345678.12349 as decimal(12,4)); select cast(12345678.12349 as decimal(12,4)); +----------------+ | single_value | +================+ | 12345678.1235 | +----------------+ 1 tuple Timer 0.422 msec 1 rows sql>select cast(123456789.1234 as decimal(12,4)); !SQLException:convert:too many digits (13 > 12) 0 tuples Timer 0.426 msec 0 rows

On Thu, Nov 5, 2009 at 10:02 AM, Roman Sokolyuk wrote:

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Thanks a lot - this works - could you tell me please what the 12,4 stand for?

On Wed, Nov 4, 2009 at 10:40 PM, Guillaume Theoret wrote:

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I had this problem today as well. What I did to solve it was this:

select cast(field2 as numeric(12,4)) / field1 * 100

This will force monetdb into not simply doing integer division.

On Wed, Nov 4, 2009 at 6:09 PM, Roman Sokolyuk wrote:

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Hi,

I am encountering the following issue:

SELECT field1, //returns 4 field2, //returns 1 ( ( field2 / field1) * 100) //this returns zero - not what I want - I

am

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looking for 25

Am I doing something wrong with type conversion (integer/floating point)?

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