I am just stuck in this situation. Is there any way in SQL
to get rid off from all numbers except two numbers after decimal point?
for example, lets say 125.36987 so as the output i want only 125.36.
Ive tried to multiply it to 100 then use cast it to int and divide by 100.0
but still getting not what i want.
YES,ROUND CAN SOLVE THAT.
SELECT ROUND(125.36987,2) FROM DUAL; YOU CAN GET 125.37
Or if YOU DON'T NEED round-off ， you can use 'trunc'
like this TRUNC(125.36987,2) 125.36
use ROUND()
select round(125.36987, 2, 1)
SQLFiddle Demo
ROUND
Here's the detail. the syntax,
ROUND ( numeric_expression , length [ ,function ] )
function
Is the type of operation to perform. function must be tinyint,
smallint, or int. When function is omitted or has a value of 0
(default), numeric_expression is rounded. When a value other than 0 is
specified, numeric_expression is truncated.
Related
SELECT
ROUND(152.7300, 2, 1), --returns 152.7300
ROUND('152.7300', 2, 1) --returns 152.72
Why does the second rounding expression result in downward rounding to .72 ?
I imagine it has something to do with the VARCHAR input being converted to a numeric, but I can't imagine how that would cause it to lose a decimal value like that.
When you pass a string literal to ROUND, SQL Server converts it to float*. In contrast, 152.7300 is treated as numeric:
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.
Exact float representation of 152.73 is 152.72999572753906, hence truncation** to the second decimal digit yields 152.72.
* You can check this by passing a non-numeric string, e.g. ROUND('foo', 2, 1). This will produce "Error converting data type varchar to float."
** Passing a value other than zero for the third parameter of ROUND indicates truncation of the number, instead of rounding.
https://learn.microsoft.com/en-us/sql/t-sql/functions/round-transact-sql
ROUND ( numeric_expression , length [ ,function ] )
function
Is the type of operation to perform. function must be tinyint, smallint, or int. When function is omitted or has a value of 0 (default), numeric_expression is rounded. When a value other than 0 is specified, numeric_expression is truncated.
SELECT CAST ((SUM(r.SalesVolume)/1000) AS decimal(3,3)) FROM RawData r
The above is a part of a query that I am trying to run but returns an error:
Lookup Error - SQL Server Database Error: Arithmetic overflow error converting int to data type numeric.
Not sure what this means.
The result column looks like(Without dividing by 1000 and casting):
Total_Sales_Volume
64146
69814
68259
56318
66585
51158
44365
49855
49553
88998
102739
55713
Tried casting as float but doesnt help.
The Problem is decimal(3,3) --> this means a number with 3 digit, 3 of them behind the decimal point. If you want a number like this 1234567.123 you would have do declare it as decimal(10,3)
Try this:
SELECT CAST ((SUM(r.SalesVolume)/1000.0) AS decimal(6,3)) FROM RawData r
decimal(3,3) means that you allow numbers with 3 digits in total, and 3 of these are behind the comma ... I think you meant decimal(6,3)
EDIT: In addition, you need to to divide by 1000.0, not by 1000.
If you divide by 1000, it is an integer division.
If you divide by 1000.0, then it becomes a decimal division, with commas.
Try following:
SELECT CAST ((SUM(r.SalesVolume)/1000) AS numeric(6,3)) FROM RawData r
I am using LIKE to return matching numeric results against a float field. It seems that once there are more than 4 digits to the left of the decimal, values that match my search item on the right side of the decimal are not returned. Here's an example illustrating the situation:
CREATE TABLE number_like_test (
num [FLOAT] NULL
)
INSERT INTO number_like_test (num) VALUES (1234.56)
INSERT INTO number_like_test (num) VALUES (3457.68)
INSERT INTO number_like_test (num) VALUES (13457.68)
INSERT INTO number_like_test (num) VALUES (1234.76)
INSERT INTO number_like_test (num) VALUES (23456.78)
SELECT num FROM number_like_test
WHERE num LIKE '%68%'
That query does not return the record with the value of 12357.68, but it does return the record with the value of 3457.68. Also running the query with 78 instead of 68 does not return the 23456.78 record, but using 76 returns the 1234.76 record.
So to get to the question: why having a larger number causes these results to change? How can I change my query to get the expected results?
The like operator requires a string as a left-hand value. According to the documentation, a conversion from float to varchar can use several styles:
Value Output
0 (default) A maximum of 6 digits. Use in scientific notation, when appropriate.
1 Always 8 digits. Always use in scientific notation.
2 Always 16 digits. Always use in scientific notation.
The default style will work fine for the six digits in 3457.68, but not for the seven digits in 13457.68. To use 16 digits instead of 6, you could use convert and specify style 2. Style 2 represents a number like 3.457680000000000e+003. But that wouldn't work for the first two digits, and you get an unexpected +003 exponent for free.
The best approach is probably a conversion from float to decimal. That conversion allows you to specify the scale and precision. Using scale 20 and precision 10, the float is represented as 3457.6800000000:
where convert(decimal(20,10), num) like '%68%'
When you are comparing number with LIKE it is implicitly converted to string and then matched
The problem here is that float number is not precise and when it is converted you can get
13457.679999999999999 instead of 13457.68
So to avid this explicitly format number in appropriate format(not sure how to do this in sql server, but it will be something like)
SELECT num FROM number_like_test
WHERE Format("0.##",num) LIKE '%68%'
The conversion to string is rounding your values. Both CONVERT and CAST have the same behavior.
SELECT cast(num as nvarchar(50)) as s
FROM number_like_test
Or
SELECT convert(nvarchar(50), num) as s
FROM number_like_test
provide the results:
1234.56
3457.68
13457.7
1234.76
23456.8
You'll have to use the STR function and correct format parameters to try to get your results. For example,
SELECT STR(num, 10, 2) as s
FROM number_like_test
gives:
1234.56
3457.68
13457.68
1234.76
23456.78
Pretty well solved already, but you only need to CAST once, not twice like the other answer suggests, LIKE takes care of the string conversion:
SELECT *
FROM number_like_test
WHERE CAST(num AS DECIMAL(12,6)) LIKE '%68%'
And here's a SQL Fiddle showing the rounding behavior: SQL Fiddle
It's probably because a FLOAT data type represents a floating point number which is an approximation of the number and should not be relied on for exact comparisons.
If you need to do a search that includes the float value you would need to either store it in a decimal data type (which will hold the exact number) or convert it to a varchar using something like the STR() function
I have a column X which is full of floats with decimals places ranging from 0 (no decimals) to 6 (maximum). I can count on the fact that there are no floats with greater than 6 decimal places. Given that, how do I make a new column such that it tells me how many digits come after the decimal?
I have seen some threads suggesting that I use CAST to convert the float to a string, then parse the string to count the length of the string that comes after the decimal. Is this the best way to go?
You can use something like this:
declare #v sql_variant
set #v=0.1242311
select SQL_VARIANT_PROPERTY(#v, 'Scale') as Scale
This will return 7.
I tried to make the above query work with a float column but couldn't get it working as expected. It only works with a sql_variant column as you can see here: http://sqlfiddle.com/#!6/5c62c/2
So, I proceeded to find another way and building upon this answer, I got this:
SELECT value,
LEN(
CAST(
CAST(
REVERSE(
CONVERT(VARCHAR(50), value, 128)
) AS float
) AS bigint
)
) as Decimals
FROM Numbers
Here's a SQL Fiddle to test this out: http://sqlfiddle.com/#!6/23d4f/29
To account for that little quirk, here's a modified version that will handle the case when the float value has no decimal part:
SELECT value,
Decimals = CASE Charindex('.', value)
WHEN 0 THEN 0
ELSE
Len (
Cast(
Cast(
Reverse(CONVERT(VARCHAR(50), value, 128)) AS FLOAT
) AS BIGINT
)
)
END
FROM numbers
Here's the accompanying SQL Fiddle: http://sqlfiddle.com/#!6/10d54/11
This thread is also using CAST, but I found the answer interesting:
http://www.sqlservercentral.com/Forums/Topic314390-8-1.aspx
DECLARE #Places INT
SELECT TOP 1000000 #Places = FLOOR(LOG10(REVERSE(ABS(SomeNumber)+1)))+1
FROM dbo.BigTest
and in ORACLE:
SELECT FLOOR(LOG(10,REVERSE(CAST(ABS(.56544)+1 as varchar(50))))) + 1 from DUAL
A float is just representing a real number. There is no meaning to the number of decimal places of a real number. In particular the real number 3 can have six decimal places, 3.000000, it's just that all the decimal places are zero.
You may have a display conversion which is not showing the right most zero values in the decimal.
Note also that the reason there is a maximum of 6 decimal places is that the seventh is imprecise, so the display conversion will not commit to a seventh decimal place value.
Also note that floats are stored in binary, and they actually have binary places to the right of a binary point. The decimal display is an approximation of the binary rational in the float storage which is in turn an approximation of a real number.
So the point is, there really is no sense of how many decimal places a float value has. If you do the conversion to a string (say using the CAST) you could count the decimal places. That really would be the best approach for what you are trying to do.
I answered this before, but I can tell from the comments that it's a little unclear. Over time I found a better way to express this.
Consider pi as
(a) 3.141592653590
This shows pi as 11 decimal places. However this was rounded to 12 decimal places, as pi, to 14 digits is
(b) 3.1415926535897932
A computer or database stores values in binary. For a single precision float, pi would be stored as
(c) 3.141592739105224609375
This is actually rounded up to the closest value that a single precision can store, just as we rounded in (a). The next lowest number a single precision can store is
(d) 3.141592502593994140625
So, when you are trying to count the number of decimal places, you are trying to find how many decimal places, after which all remaining decimals would be zero. However, since the number may need to be rounded to store it, it does not represent the correct value.
Numbers also introduce rounding error as mathematical operations are done, including converting from decimal to binary when inputting the number, and converting from binary to decimal when displaying the value.
You cannot reliably find the number of decimal places a number in a database has, because it is approximated to round it to store in a limited amount of storage. The difference between the real value, or even the exact binary value in the database will be rounded to represent it in decimal. There could always be more decimal digits which are missing from rounding, so you don't know when the zeros would have no more non-zero digits following it.
Solution for Oracle but you got the idea. trunc() removes decimal part in Oracle.
select *
from your_table
where (your_field*1000000 - trunc(your_field*1000000)) <> 0;
The idea of the query: Will there be any decimals left after you multiply by 1 000 000.
Another way I found is
SELECT 1.110000 , LEN(PARSENAME(Cast(1.110000 as float),1)) AS Count_AFTER_DECIMAL
I've noticed that Kshitij Manvelikar's answer has a bug. If there are no decimal places, instead of returning 0, it returns the total number of characters in the number.
So improving upon it:
Case When (SomeNumber = Cast(SomeNumber As Integer)) Then 0 Else LEN(PARSENAME(Cast(SomeNumber as float),1)) End
Here's another Oracle example. As I always warn non-Oracle users before they start screaming at me and downvoting etc... the SUBSTRING and INSTRING are ANSI SQL standard functions and can be used in any SQL. The Dual table can be replaced with any other table or created. Here's the link to SQL SERVER blog whre i copied dual table code from: http://blog.sqlauthority.com/2010/07/20/sql-server-select-from-dual-dual-equivalent/
CREATE TABLE DUAL
(
DUMMY VARCHAR(1)
)
GO
INSERT INTO DUAL (DUMMY)
VALUES ('X')
GO
The length after dot or decimal place is returned by this query.
The str can be converted to_number(str) if required. You can also get the length of the string before dot-decimal place - change code to LENGTH(SUBSTR(str, 1, dot_pos))-1 and remove +1 in INSTR part:
SELECT str, LENGTH(SUBSTR(str, dot_pos)) str_length_after_dot FROM
(
SELECT '000.000789' as str
, INSTR('000.000789', '.')+1 dot_pos
FROM dual
)
/
SQL>
STR STR_LENGTH_AFTER_DOT
----------------------------------
000.000789 6
You already have answers and examples about casting etc...
This question asks of regular SQL, but I needed a solution for SQLite. SQLite has neither a log10 function, nor a reverse string function builtin, so most of the answers here don't work. My solution is similar to Art's answer, and as a matter of fact, similar to what phan describes in the question body. It works by converting the floating point value (in SQLite, a "REAL" value) to text, and then counting the caracters after a decimal point.
For a column named "Column" from a table named "Table", the following query will produce a the count of each row's decimal places:
select
length(
substr(
cast(Column as text),
instr(cast(Column as text), '.')+1
)
) as "Column-precision" from "Table";
The code will cast the column as text, then get the index of a period (.) in the text, and fetch the substring from that point on to the end of the text. Then, it calculates the length of the result.
Remember to limit 100 if you don't want it to run for the entire table!
It's not a perfect solution; for example, it considers "10.0" as having 1 decimal place, even if it's only a 0. However, this is actually what I needed, so it wasn't a concern to me.
Hopefully this is useful to someone :)
Probably doesn't work well for floats, but I used this approach as a quick and dirty way to find number of significant decimal places in a decimal type in SQL Server. Last parameter of round function if not 0 indicates to truncate rather than round.
CASE
WHEN col = round(col, 1, 1) THEN 1
WHEN col = round(col, 2, 1) THEN 2
WHEN col = round(col, 3, 1) THEN 3
...
ELSE null END
Using SQL Server 2000
I want to reduce the decimal length
Query
Select 23/12 as total
Output is showing as 1.99999999999
I don't want to round the value, I want to diplay like this 1.99
Tried Query
Select LEFT(23/12, LEN(23/12) - 3) as total
The above query is working only if there is decimal value like 12.444444, but if the total is single digit means like 12 or 4 or 11...., i am getting error at run time.
How to do this.
Need Query Help
There is a very simple solution. You can find it in BOL. Round takes an optional 3rd argument, which is round type. The values are round or truncate.
ROUND numeric_expression , length [ ,function ] )
...
function Is the type of operation to perform. function must be
tinyint, smallint, or int. When function is omitted or has a value of
0 (default), numeric_expression is rounded. When a value other than 0
is specified, numeric_expression is truncated.
So just do
Select ROUND(cast(23 as float)/12, 2, 1) as total
That gives 1.91. Note, if you were really seeing 1.999 - something is really wrong with your computer. 23/12 = 1.916666666(ad infinitum). You need to cast one of the numbers as float since sql is assuming they're integers and doing integer division otherwise. You can of course cast them both as float, but as long as one is float the other will be converted too.
Not terribly elegant, but works for all cases:
Select CONVERT(float,LEFT(CONVERT(nvarchar, 23.0/12.0),CHARINDEX('.',CONVERT(nvarchar, 23.0/12.0)) + 2)) as total
Scalar Function
-- Description: Truncate instead of rounding a float
-- SELECT dbo.TruncateNumber(23.0/12.0,2)
-- =============================================
CREATE FUNCTION TruncateNumber
(
-- Add the parameters for the function here
#inFloat float,
#numDecimals smallint
)
RETURNS float
AS
BEGIN
IF (#numDecimals < 0)
BEGIN
SET #numDecimals = 0
END
-- Declare the return variable here
RETURN CONVERT(float,LEFT(CONVERT(nvarchar, #inFloat),CHARINDEX('.',CONVERT(nvarchar, #inFloat)) + #numDecimals))
END
GO