I have a float variable-field1 in 2 tables-table1 & table2. When I query the table and check the values of the field both look identical but when I find their difference it gives a difference instead of zero.
Field1(Table1) value---84.4660194174757
Field2(Table2) value---84.4660194174757
Differnce---1.4210854715202E-14
Why would I get this problem?
Use ROUND to limit the decimal places
Use ABS(value1-value2) < 0.00001 with some suitable value
Don't use float
Related
I found weird or strange behavior of Round function in MSSQL for real column type. I have tested this issue in Azure SQL DB and SQL Server 2012
Why #number=201604.125 Return 201604.1 ?
Why round(1.12345,10) Return 1.1234500408 ?
-- For Float column it working as expected
-- Declare #number as float,#number1 as float;
Declare #number as real,#number1 as real;
set #number=201604.125;
set #number1=1.12345;
select #number as Realcolumn_Original
,round(#number,2) as Realcolumn_ROUND_2
,round(#number,3) as Realcolumn_ROUND_3
, #number1 as Realcolumn1_Original
,round(#number1,6) as Realcolumn1_ROUND_6
,round(#number1,7) as Realcolumn1_ROUND_7
,round(#number1,8) as Realcolumn1_ROUND_8
,round(#number1,9) as Realcolumn1_ROUND_9
,round(#number1,10) as Realcolumn1_ROUND_10
Output for real column type
I suspect what you are asking here is why does:
DECLARE #n real = 201604.125;
SELECT #n;
Return 201604.1?
First point of call for things like this should be the documentation: Let's start with float and real (Transact-SQL). Firstly we note that:
The ISO synonym for real is float(24).
If we then look further down:
float [ (n) ] Where n is the number of bits that are used to store the
mantissa of the float number in scientific notation and, therefore,
dictates the precision and storage size. If n is specified, it must be
a value between 1 and 53. The default value of n is 53. n value
Precision Storage size
1-24 7 digits 4 bytes
So, now we know that a real (aka a float(24)) has precision of 7. 201604.125 has a precision of 9, that's 2 too many; so off come that 2 and 5 in the return value.
Now, ROUND (Transact-SQL). That states:
Returns a numeric value, rounded to the specified length or precision.
When using real/float those digits aren't actually lost, as such, due to the floating point. When you use ROUND, you are specifically stating "I want this many decimal places". This is why you can then see the .13 and the .125, as you have specifically asked for those. When you just returned the value of #number it had a precision of 7, due to being a real, so 201604.1 was the value returned.
My query is relatively simple but would like to know the root cause
Sql query to convert Float to Decimal gives incorrect results.
Example:
Select CAST(CONVERT(DECIMAL(25,8), CONVERT(FLOAT,338193293.16))AS VARCHAR(255))
Expected result: 338193293.16000000
Actual Result: 338193293.16000003
Where did the extra 3come from?
The number of digits are too much - no matter that after decimal point they are only 2. Use shorter number e.g. 38193293.16 or use FLOAT(24). For normal FLOAT there is "Precision = 7 digits".
Using T-SQL and Microsoft SQL Server I would like to specify the number of decimal digits when I do a division between 2 integer numbers like:
select 1/3
That currently returns 0. I would like it to return 0,33.
Something like:
select round(1/3, -2)
But that doesn't work. How can I achieve the desired result?
The suggestions from stb and xiowl are fine if you're looking for a constant. If you need to use existing fields or parameters which are integers, you can cast them to be floats first:
SELECT CAST(1 AS float) / CAST(3 AS float)
or
SELECT CAST(MyIntField1 AS float) / CAST(MyIntField2 AS float)
Because SQL Server performs integer division. Try this:
select 1 * 1.0 / 3
This is helpful when you pass integers as params.
select x * 1.0 / y
It's not necessary to cast both of them. Result datatype for a division is always the one with the higher data type precedence. Thus the solution must be:
SELECT CAST(1 AS float) / 3
or
SELECT 1 / CAST(3 AS float)
use
select 1/3.0
This will do the job.
I understand that CASTing to FLOAT is not allowed in MySQL and will raise an error when you attempt to CAST(1 AS float) as stated at MySQL dev.
The workaround to this is a simple one. Just do
(1 + 0.0)
Then use ROUND to achieve a specific number of decimal places like
ROUND((1+0.0)/(2+0.0), 3)
The above SQL divides 1 by 2 and returns a float to 3 decimal places, as in it would be 0.500.
One can CAST to the following types: binary, char, date, datetime, decimal, json, nchar, signed, time, and unsigned.
Looks like this trick works in SQL Server and is shorter (based in previous answers)
SELECT 1.0*MyInt1/MyInt2
Or:
SELECT (1.0*MyInt1)/MyInt2
Use this
select cast((1*1.00)/3 AS DECIMAL(16,2)) as Result
Here in this sql first convert to float or multiply by 1.00 .Which output will be a float number.Here i consider 2 decimal places. You can choose what you need.
If you came here (just like me) to find the solution for integer value, here is the answer:
CAST(9/2 AS UNSIGNED)
returns 5
I was surprised to see select 0.7/0.9 returning 0.8 in Teradata given they're already as floats/decimal numbers! I had to do cast(0.7 as float) to get the output that I was after.
When using literals, the best way is to "tell" SQL
which type you mean.
if you want a decimal result, add decimal point ".0" to your numbers:
SELECT 1.0 / 3.0
Result
0.333333
if you want a float (real) result, add "e0" to your numbers:
SELECT 1e0 / 3e0
Result
0.333333333333333
I use SQL Server 2005 and need to test whether values in a column that's metadata has been specified as DECIMAL(18.3) actually contains data that has values to the right of the Decimal point, and if so, what these values are.
I've read a few articles that only discuss how to drop off the decimal places or how to round the values, but not how to ONLY display what is stored to the right of the decimal point.
Your help would be greatly appreciated.
Kind Regards,
Ignacio.
Try:
SELECT a - FLOOR(a)
FROM ...
SELECT decimalnumber - FLOOR(decimalnumber) AS decimalpart
FROM mytable
WHERE decimalnumber - FLOOR(decimalnumber) > 0
This may not always work the way you expect it to. The problem occurs when you have negative numbers. You can think of FLOOR as a type of rounding, where it always rounds down to the next whole number. Floor(3.14) = 3, and Floor(-3.14) = -4.
To get the value of a number after the decimal point, you can use the ParseName function, which will work for positive and negative numbers.
Select ParseName(-3.9876, 1)
Select ParseName(-3.1234, 1)
Select ParseName(3.9876, 1)
Select ParseName(3.1234, 1)
I have a list of values such as "12000","12345","123456" that need to be converted to currency ("120.00", "123.45", "1234.56"). The only way I know is to convert the value to a string, copy the first strlen()-2 characters to one string (dollars) and the remainging two digits to another string(cents) and then write them as the following:
printf("%s.%s", dollars, cents);
printf("$%.2f", value/100);
Don't use floats for storing or representing monetary amounts. Use longs (if you need more than 4 billion cent use llongs). Its usually a good idea to represent currency in its minimum usable unit, example use 10000 to represent 100Euro). Then the correct way to format these values (assuming 100 cent to the euro or dollar) is:
printf( "%d.%02d", value/100, value%100);
Hope that makes sense...
Calculations with currency values is a complex subject but you cant go far wrong is you always aim to have a rounded answer to the nearest currency unit (cent for example) and always make sure that rounding errors are calculated for (example, to divide 1 dollar three ways you should end up with 33+33+34 or 33+33+33+1).
to prefix values less than $1.00 with 0, use:
printf( "$%0.2f", value / 100.0 );
This will result in $0.25 if value = 25