I have a couple of strings (nvarchar data type), one is a whole number and one has decimal points trailing. My goal is to remove decimals and have all values as a whole number.
I tried the code below but it gives me an error for the value with no decimals. Is there a way to accomplish this without a case expression. I'll be using this new column in a join.
SELECT [SOW]
--,LEFT([SOW], CHARINDEX('.', [SOW])-1) as 'TestColumn'
FROM [dbo].[t_Schedule_kdm]
WHERE sow in ('15229.11','11092')
Output:
11092
15229.11
My desired Output:
11092
15229
Just append a dot character so that you'll always find an index:
LEFT(SOW, CHARINDEX('.', SOW + '.') - 1)
It's not clear whether you need to cast the result of that expression to an integer value.
Convert first to the most precision number you could ever have e.g. decimal(9,2) then convert to an int. You can't convert directly from a decimal string to an int.
SELECT [Value]
, CONVERT(int,CONVERT(decimal(9,2),[Value]))
FROM (
VALUES ('15229.11'),('11092')
) x ([Value]);
Related
I have a column (RCV1.ECCValue) in a table which 99% of the time has a constant string format- example being:
T0-11.86-273
the middle part of the two hyphens is a percentage. I'm using the below sql to obtain this figure which is working fine and returns 11.86 on the above example. when the data in that table is in above format
'Percentage' = round(SUBSTRING(RCV1.ECCValue,CHARINDEX('-',RCV1.ECCValue)+1, CHARINDEX('-',RCV1.ECCValue,CHARINDEX('-',RCV1.ECCValue)+1) -CHARINDEX('-',RCV1.ECCValue)-1),2) ,
However...this table is updated from an external source and very occasionally the separators differ, for example:
T0-11.86_273
when this occurs I get the error:
Invalid length parameter passed to the LEFT or SUBSTRING function.
I'm very new to SQL and have got myself out of many challenges but this one has got me stuck. Any help would be mostly appreciated. Is there a better way to extract this percentage value?
Replace '_' with '-' to string in CHARINDEX while specifying length to the substring
'Percentage' = round(SUBSTRING(RCV1.ECCValue,CHARINDEX('-',RCV1.ECCValue)+1, CHARINDEX('-',replace(RCV1.ECCValue,'_','-'),CHARINDEX('-',RCV1.ECCValue)+1) -CHARINDEX('-',RCV1.ECCValue)-1),2) ,
If you can guarantee the structure of these strings, you can try parsename
select round(parsename(translate(replace('T0-11.86_273','.',''),'-_','..'),2), 2)/100
Breakdown of steps
Replace . character in the percentage value with empty string using replace.
Replace - or _, whichever is present, with . using translate.
Parse the second element using parsename.
Round it up to 2 digits, which will also
automatically cast it to the desired numeric type.
Divide by 100
to restore the number as percentage.
Documentation & Gotchas
Use NULLIF to null out such values
round(
SUBSTRING(
RCV1.ECCValue,
NULLIF(CHARINDEX('-', RCV1.ECCValue), 0) + 1,
NULLIF(CHARINDEX('-',
RCV1.ECCValue,
NULLIF(CHARINDEX('-', RCV1.ECCValue), 0) + 1
), 0)
- NULLIF(CHARINDEX('-', RCV1.ECCValue), 0) - 1
),
2)
I strongly recommend that you place the repeated values in CROSS APPLY (VALUES to avoid having to repeat yourself. And do use whitespace, it's free.
I am trying to truncate a decimal which has 2 numbers after decimal point in presto that should display on truncating the number without floating values and display the full decimal with floating values when there are numbers from 1 to 9 after decimal point. I have used the following query but it does not do the job and still I am ending up with numbers having zeroes after decimal point.
select column1,case when right(cast(column1 as varchar),7)='.000000' then truncate(column1) else column1 end from table1;
Using varchar pads extra zeroes to the right and hence are the extra zeroes I have used in the above expression after the decimal point
Please let me know what has to done to truncate the decimal only when it has zeroes as the floating values
The thing is truncate(x) → double Returns x rounded to integer by dropping digits after decimal point, but it is double, not integer. And displaying double without non-significant zeroes is a GUI job, it displays all of them or not displays non-significant zeroes. For example when I am using Presto Qubole, it does not displays .000000 if nothing else except 0s after dot. So the problem is in tool you are using probably.
For example this works fime in Presto on Qubole:
with mydata as (
select 123.00000 as figure union all
select 123.0123 )
select case when regexp_like(cast(figure as varchar),'\d+\.0+$') then truncate(figure) else figure end
from mydata
Result:
123.0123
123
But in your GUI it may not work the same because in second line is not integer, it is decimal(8,5), wrap in the typeof() function and you will see, and GUI decides how to display decimal(8,5).
You said:
Using varchar pads extra zeroes to the right and hence are the extra
zeroes I have used in the above expression after the decimal point
No, the result of your expression is not varchar, varchar is being implicitly converted to decimal or double, check using typeof().
If you want it to work not depending on tool you are using, convert to varchar and transform explicitly:
select case when regexp_like(cast(figure as varchar),'\d\.0+$') --all zeroes, change according to your requirements
then regexp_replace(cast(figure as varchar),'\.0+$','') --remove fractional part
else cast(figure as varchar) --we need same type in case
end as result
from mydata
This will work guaranteed because result is varchar and displayed as is.
All that expression can be simplified:
--remove .0+ if no 1-9 after dot:
select regexp_replace(cast(figure as varchar),'\.0+$','')
from mydata
I wanted to double check my logic for a query in SQL Server.
The idea is that I am able to feed the following values and it will make sure the result is a decimal with four trailing digits.
Possible values for #LABORQTY:
1,200
1,200.42
1200 (Integer)
1200.42
1200 (As a String)
1200.42 (As a String)
When the value is a string, it will give the error:
Error converting data type nvarchar to numeric.
Here is my code:
CONVERT(DECIMAL(12, 4), REPLACE(#LABORQTY, ',', ''))
The output each time though should be decimal:
1200.4200
Your question is really confused, but I'll answer according to the following parameters:
#laborqty is a VARCHAR
#laborqty may somehow come to contain any of the following values:
'1200'
'1200.42'
'1,200'
'1,200.42'
In which case CONVERT(DECIMAL(12, 4), REPLACE(#LABORQTY, ',', '')) will indeed produce a decimal with up to 4 digits of fractional precision. Whether your query tool/programming language will output it as 1200.4200 or not is another matter entirely; it might well just output 1200.42 and drop the trailing zeroes
If you're getting Error converting data type varchar to numeric. still, there is some other character data (not comma) in your numeric string
If you definitely want the trailing zeroes, format it into a string before you output
FORMAT(CONVERT(decimal(12,4), '1200.42'), '0.0000')
This will generate a string with 4 trailing zeroes
you can use :
select CAST ( REPLACE( '1,200.4' , ',','') AS decimal(17,4))
I have some sales numbers in a string column that I need to convert to some format so that i can calculate them with each other but I get this error while trying to convert them.
Conversion failed when converting the varchar value '-6.353,35' to data type int.
I'm not allowed to lose any money by rounding it up. It doesnt mather but in what type i convert as long as im not rounding them up. What's your thoughts?
For example i have -6.353,35 and 300,30 and i want to sum them too -6.053,05
try this...
select convert(int,convert(float,replace('-6.353,35',',','')))
as there are (,) Commas it cannot be converted to float,so remove the (,)commas after converting to float we can convert to int
If you want decimal values, then you should use float
select convert(float,replace('-6.353,35',',',''))
Edit
Like #marc_s suggested, it is preferred to use decimal rather than float
select convert(decimal,replace('-6.353,35',',',''))
Firstly, you want to convert this to a decimal value, not an int, as you will lose anything after decimal point otherwise.
The problem here is the separators that are being used when storing your values.
With your numbers, you have points . to represent thousand separators and commas , to represent decimal points.
To enable you to convert the value to a valid decimal you need to have it in a format that SQL Server can process.
A simple way to do this would be to remove/replace the problem characters before trying to convert the value:
DECLARE #val AS VARCHAR(10) = '-6.353,35'
-- step 1: remove thousand separator
SET #val = REPLACE(#val, '.', '')
SELECT #val AS RemoveThousandSeparator
-- step 2: replace decimal separator with decimal point
SET #val = REPLACE(#val, ',', '.')
SELECT CONVERT(DECIMAL(18,2), #val) AS DecimalPointAdded
-- to do it in one statement:
SET #val = '-6.353,35'
SELECT CONVERT(DECIMAL(18,2),
REPLACE(REPLACE(#val, '.', ''), ',', '.')) AS NumericeRepresentation
Aside from this, you would be much better off storing numeric values in the correct format in the first place to avoid this kind of workaround.
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