How to multiply string value to longint in SQL - sql

I have the below data which I want to multiply together, column A times column B to get column C.
A has datatype string and B has datatype long.
A B
16% 894
15% 200
I have tried this expression in query cast(A as int)*B but it is giving me an error.

You can try below way -
select cast(left(A, patindex('%[^0-9]%', A+'.') - 1) as int)*B
from tablename

You need to remove the '%' symbol before attempting your cast. And assuming you are actually wanting to calculate the percentage, then you also need to divide by 100.00.
cast(replace(A,'%','') as int)/100.00*B
Note: You need to use 100.00 rather than 100 to force decimal arithmetic instead of integer. Or you could cast as decimal(9,2) instead of int - either way ensures you get an accurate result.
You may well want to reduce the number of decimal points returned, in which case cast it back to your desired datatype e.g.
cast(cast(replace(A,'%','') as int)/100.00*# as decimal(9,2))
Note: decimal(9,2) is just an example - you would use whatever precision and scale you need.

The syntax of the cast in SQL Server is CAST(expression AS TYPE);
As you cannot convert '%' to an integer so you have to replace that with an empty character
as below:
SELECT cast(replace(A,'%','') AS int);
Finally you can write as below:
SELECT (cast(replace(A,'%','') AS int)/100.00)*B as C;

Related

Using decimal in a SQL query

I have a column Amount defined in my SQL Server database as varchar(20), null.
It has values like this:
1.56867
2.0
2.0000
2
If the user in the client app enters 2 for search, then I need to pull all records from the db except value 1.56867.
So search results should yield:
2.0
2.0000
2
I tried using this in my SQL query but it's still returning value 1.56867.
CONVERT(decimal, myTable.Amout) = CONVERT(decimal, 2)
Can you advise?
Thanks.
You should really consider fixing your data model, and store numbers as numbers rather than strings.
That said, you need to specify a scale and a precision for the decimal, otherwise i defaults to decimal(38, 0) , which results in the decimal part being truncated. Something like:
convert(decimal(20, 5), myTable.Amout) = 2
This gives you 20 digits max, including 5 decimal digits. There is no need to explictly convert the right operand here (that's a literal number already).
If your column may contain values that are not convertible to numbers, you can use try_convert() instead of convert() to avoid errors.
The problem is that when converting the string '1.56867' to decimal, SQL server assumes you want 0 decimal places (as pointed out by #GMB), and rounds the value to the nearest integer (2).
select convert(decimal,'1.56867');
-------
2
Also, since your column is defined to hold character data, you should use the try_cast() function to avoid Error converting data type varchar to numeric. errors that would otherwise occur if non-numeric data is present in the column.
select *
from (values
('1.56867')
,('2.0')
,('2.0000')
,('2')
,(null)
,('non-numeric value')
) myTable(Amount)
where try_cast(Amount as decimal(38,19)) = 2
;
Amount
-----------------
2.0
2.0000
2
(3 rows affected)

sqlserver sum with minus

I use sqlserver 2012.
I have a query like this
SELECT SUM(TH.CLEAVE_EARN_DAY), SUM(TH.CLEAVE_DAY),
SUM(TH.CLEAVE_EARN_DAY) - SUM(TH.CLEAVE_DAY)
FROM TH_LEAVE_CARD TH
The result is 0, 14.5, -15
so -15 is wrong. Must be -14.5
any suggestion ?
This is what you can try
SELECT SUM(TH.CLEAVE_EARN_DAY), SUM(TH.CLEAVE_DAY),
SUM(TH.CLEAVE_EARN_DAY)*1.0 - SUM(TH.CLEAVE_DAY)
FROM TH_LEAVE_CARD TH
Multiplying with 1.0 will just give you back decimal value and taking away will give you what you asked for
Try converting all arguments to the same datatype and then do calculation:
SELECT
SUM(CAST(TH.CLEAVE_EARN_DAY AS DECIMAL(18,2))),
SUM(CAST(TH.CLEAVE_DAY AS DECIMAL(18,2))),
SUM(CAST(TH.CLEAVE_EARN_DAY AS DECIMAL(18,2))
- CAST(TH.CLEAVE_DAY AS DECIMAL(18,2))) AS substraction
FROM TH_LEAVE_CARD TH
Also you can combine:
SUM(TH.CLEAVE_EARN_DAY) - SUM(TH.CLEAVE_DAY)
to (if both column are NOT NULL):
SUM(TH.CLEAVE_EARN_DAY - TH.CLEAVE_DAY)
or (thanks Arvo for pointing this):
SUM(ISNULL(TH.CLEAVE_EARN_DAY,0) - ISNULL(TH.CLEAVE_DAY,0))
To perform mathematical operations on columns:
Used columns should be converted into same numeric/decimal data type.
To handle null values you may use ISNULL function.
Ex:
SELECT SUM(TH.CLEAVE_EARN_DAY), SUM(TH.CLEAVE_DAY),
SUM(cast (TH.CLEAVE_EARN_DAY) as decimal(5,1)) - SUM(cast ( (TH.CLEAVE_DAY) as decimal(5,1))
FROM TH_LEAVE_CARD
There is few reason why the result is not as per what you are expecting. In Sql Server any math operation that contains a null would result to null. for example sum(1,2,3,null,4) is equal to null. 1 + null also equal to null.
therefore it would be safer to use isnull function to assign a default value in case the value is null.
for mathematical operation. sql server would do the calculation based on the specified data type. for example int / int = int. therefore the result would be missled. because most of the time int / int = float.
it would be better to change the value to double prior to do any arithmetic operation.
below is the example after include the isnull function as well as cast to float.
SELECT SUM(CAST(ISNULL(TH.CLEAVE_EARN_DAY,0) as double)), SUM(cast(ISNULL(TH.CLEAVE_DAY,0) as double)),
SUM(cast(ISNULL(TH.CLEAVE_EARN_DAY,0) as double)) - SUM(cast(ISNULL(TH.CLEAVE_DAY,0) as double))
FROM TH_LEAVE_CARD TH

SQL: Unable to CAST a query

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

SQL server 'like' against a float field produces inconsistent results

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

How do I count decimal places in SQL?

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