In the example below the result of the query instead of returning 7 is returned 5 .
DECLARE @valor money = 10.6879
SELECT LEN(@Valor)
The impression I get is that when the number is of type money , only 2 decimal places are considered.
What function would I use to return the exact number of characters of a money number?
First create the function
CREATE FUNCTION dbo.countDigits(@A float) RETURNS tinyint AS
BEGIN
declare @R tinyint
IF @A IS NULL
RETURN NULL
set @R = 0
while @A - str(@A, 18 + @R, @r) <> 0
begin
SET @R = @R + 1
end
RETURN @R
END
GO
Then just use
SELECT LEN(cast(10.6879 as decimal)) + dbo.countDigits(10.6879) + + 1
//número antes da virgula + Número de casas depois da vírgular + 1 = 7
Convert using CONVERT to type DECIMAL ( reference here ):
DECLARE @valor money = 10.6879
SELECT LEN(CONVERT(decimal(6,4),@Valor)); //7
The solution to your problem is simple:
DECLARE @Valor MONEY = 10.6879
SELECT LEN(CONVERT(FLOAT, @Valor)) -- 7
The difference for the @LeonardoBonetti example is the data type of the conversion ( FLOAT instead of DECIMAL(6, 4) ).