p (precision) :
The maximum total number of decimal digits that can be stored, both to the left and to the right of the decimal point.
s (scale) :
The maximum number of decimal digits that can be stored to the right of the decimal point.

SIR HERE THE DECIMAL VALUE ACCEPT

DECIMAL(TOTAL DIGIT,DECIMAL POINT AFTER DIGIT)

SO USE

SELECT CAST(@mymoney AS DECIMAL(7,2)) MoneyInt;

IT WILL RUN

Precision includes Scale.

So Decimal (5,2) actually means, we can have 3 digits before decimal and 2 digits after decimal.

From above example, (3+2) digits is called as precision and 2 digits after decimal is callled as scale.

In your example, you have 5 digits before decimal and 2 digits after decimal, therefore the correct decimal should be Decimal (7,2).

~ IM.

DECLARE @mymoney MONEY;
SET @mymoney = 12345.67;
SELECT CAST(@mymoney AS DECIMAL(7,2)) MoneyInt;

Just adding description for whay it’s filing when we rty to convert it to Decimal(5,2)

In decimal decalaration, DECIMAL(P,S), P (precision) specifies the number of digits on both side of decimal point. 12345.67 has 7 digits ( of 2 digits are of scale), so we need decimal(5,2) for storing this number.

Even if we tries to convert the number to DECIMAL(7,3), we get same error as we specify scale as 3 and so, we can have only 4 digits on left side of decimal point.

-Chintak

The overflow problem is because the length of DECIMAL(5,2) is small to support this value.

In this case, the value 12345.67 has length (precision) of 7 digits, when 2 digits are dedicated to “.67″ (scale), then you can convert this number to a DECIMAL(7,2) and not to a DECIMAL(5,2).

DECLARE @mymoney MONEY;
SET @mymoney = 12345.67;
SELECT CAST(@mymoney AS DECIMAL(7,2)) MoneyInt;
GO

But you can lost the 2 digits dedicated to “.67″, if you use a small decimal with a compatible precision without scale:

DECLARE @mymoney MONEY;
SET @mymoney = 12345.67;
SELECT CAST(@mymoney AS DECIMAL(5,0)) MoneyInt;
GO

Now it will not through any error.
Since @mymoney variable has 5 digits before decimal and 2 digits after decimal. So we have to put 7 insead of 5 in precision of DECIMAl Datatype, because 5 + 2 = 7 means total no of digits that DECIMAL datatype may contain before decimal and after decimal place.
For understanding, you can also see it like this DECIMAL((7-2),2).

Have a look the detail given below for more clarity-
DECIMAL(p,s)
p (precision)
The maximum total number of decimal digits that can be stored, both to the left and to the right of the decimal point.
s (scale)
The maximum number of decimal digits that can be stored to the right of the decimal point.

as per the bol.

In Transact-SQL statements, a constant with a decimal point is automatically converted into a numeric data value, using the minimum precision and scale necessary. For example, the constant 12345.67 is converted into a numeric value with a precision of 7 and a scale of 2.

bol link : http://msdn.microsoft.com/en-us/library/ms191530.aspx

section : Converting decimal and numeric Data

Precision is the total number of digits in a number.

Scale is the number of digits to the right of the decimal point in a number.

For Example the number 8714.235 has a

precision – 7

scale – 3

DECLARE @mymoney MONEY;
SET @mymoney = -12345.67;
SELECT CAST(@mymoney AS DECIMAL(7,2)) MoneyInt;

Because…

Here first part specifies total number of decimals. i.e. number of characters before “Period” (.) + number of characters after “Period” (.)

5+2 = 7

Second part sepcifies how many characters should be after “Period” (.)

DECLARE @mymoney MONEY;
SET @mymoney = 12345.67;
SELECT CAST(@mymoney AS DECIMAL(abc,2)) MoneyInt;

you have 7 digits in @mymoney (including decimals) so you put abc=7 or greater than 7

DECLARE @mymoney MONEY;
SET @mymoney = 12345.67;
SELECT CAST(@mymoney AS DECIMAL(,2)) MoneyInt;

you have 7 digits in @mymoney (including decimals)