I'm using the below script to take numbers from my table and round them based on the value. If the number is above 15, I want the value to round off to one decimal place. If the number is below 15, I want the value to round off to three decimal places. The formula is working on numbers above 15, but will not place a ".0" next to whole values. I suspect that Access will not do this. Am I right? If there is a better way to do this please let me know.
SELECT Tbl_Formulas.QtyPer
, Round([QtyPer],IIf([QtyPer]<=15,3,1)) AS Expr1
FROM Tbl_Formulas;
You could use Format to display .0 or .000 even for integer values:
SELECT
Tbl_Formulas.QtyPer,
Format([QtyPer], IIf([QtyPer]<=15, "0.000", "0.0")) AS Expr1
FROM Tbl_Formulas;
Related
I am trying to round my value to one decimal, below is sample which I tried.
INPUT DESIRED OUTPUT
129.700 129.7
129.769 129.8
I have tried
SELECT CAST(ROUND(('129.768'),0) AS DECIMAL(10,1))
FROM Table1
is not giving correct value.
Please help.
You want one digit after the decimal place, so that would be an argument of 1, not 0:
SELECT CAST(ROUND('129.768', 1) AS DECIMAL(10, 1))
I'm not sure why you feel the need to convert back to a decimal -- unless you are saving the value to a table and want to control the type. This also does what you want:
SELECT ROUND('129.768', 1)
I was wondering if there was any way to condense this select statement?
as well as if it is possible to round each result to a certain amount of decimals?
It is set up correctly and giving me the right result but I was just wondering if there was any way to tighten it up? I am using SQLite
SELECT AVG(eFG),
AVG(OPP_eFG),
AVG(TOV_PCT),
AVG(OPP_TOV_PCT),
AVG(ORB_PCT),
AVG(DRB_PCT),
AVG(FTA_RATE),
AVG(OPP_FTA_RATE)
You can simply use the ROUND() function
SQLite round() function rounds a floating-point value t up to a number of digits to the right of the decimal point. If the 2nd argument (rounded digits) is omitted, it is assumed to be 0.
SELECT ROUND(AVG(eFG),2),
ROUND(AVG(OPP_eFG),2),
ROUND(AVG(TOV_PCT),2),
ROUND(AVG(OPP_TOV_PCT),2),
ROUND(AVG(ORB_PCT),2),
ROUND(AVG(DRB_PCT),2),
ROUND(AVG(FTA_RATE),2),
ROUND(AVG(OPP_FTA_RATE),2)
I have some value with data type Numeric(28,10) (e.g. 128000,0000000000). I want to round it up to 2 significances and convert it into string. What is wrong with this?
convert(varchar,round(isnull(td2.Qty,0),2))
where td2.Qty is that value. It coverts it to string, but doesn't round it. Thanks in advance
It does round, but it keeps displaying the zeros because this is how numerics are always displayed.
If you need to stop displaying zeros, convert the value to a different type after the rounding, e.g. float or numeric(28,2):
convert(varchar, cast(round(isnull(td2.Qty,0),2) as numeric(28,2)))
SELECT CAST(round(isnull(128000,0000000000),2)AS FLOAT)
SELECT CAST(round(isnull(128000,0000000000),2)AS NUMERIC(28,2))
Let me explain the use of ROUND() function in SQL. It will not return the decimal value for the given length. E.g. round(isnull(td2.Qty,0),2) will not return friction value length 2, but rather
(e.g. 128000.0000000000) to (e.g. 128000.00)
instead of
(e.g. 128000.12340000000) to (e.g. 128000.120000000000).
It returns only after decimal 2 value and remain all '0'. In your case you want ROUND to truncate. so you can use
Cast(round(isnull(td2.Qty,0),2)as decimal(18,2));
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
When rounding up to 2 decimal places, the value 4.01132141 would be rounded to 4.02 because it exceeds 4.01.
How can you do this in PL/SQL?
One way would be to do ceil(value*100)/100, but that seems inelegant. Not sure there's any way to make round behave the way you want.
The function to 'round up' is CEIL, but it generates an integer.
The function to 'round down' is FLOOR, but it too generates an integer.
The function to 'round nearest' is ROUND, and it allows you to specify a number of decimal places (dp).
Note that CEIL rounds to an integer; to round to 2 dp, you'd have to multiply by 100, use CEIL, and divide by 100.
To get the answer reasonably directly, use:
ROUND(value+0.005, 2)
This works because, for the example data of 4.01132141, the value passed to ROUND is 4.01632, and when rounded to 2 dp, that becomes 4.02. If the value started as 4.0593, say, then the value passed to ROUND would be 4.0643, which when rounded to 2 dp becomes 4.06, as required.
There are a couple of tricky bits there:
If the number of dp varies, the value to be added (0.005 in the example) varies. You could create a table to hold the number of decimal places in one column and the rounding value to add in the other. Alternatively, you could use an expression with powers of 10, etc.
Deciding on the correct behaviour for negative numbers. Does -4.01132141 become -4.02 or -4.01? You might need to play with SIGN and ABS functions to get that to work as you want.
I faced the same issue and came up with the following statement, it has worked fine so far.
select 4.01132141+(mod((ceil(4.01132141)-4.01132141)*1000,10)/1000) from dual
select 4.01132141, CEIL(4.01132141*100)/100 from dual
You can use this for Rounding up in PLSQL:
ROUND( UrNo+ (5 / POWER(10, DecimalPlaces + 1)) , DecimalPlaces)