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;
Using SQL in PostgreSQL I need to select all the rows from my table called "crop" when the first digit of the integer numbers in column "field_id" is 7.
select *
from crop
where (left (field_id,1) = 7)
First, you know that the column is a number, so I would be inclined to explicitly convert it, no matter what you do:
where left(crop::text, 1) = '7'
where crop::text like '7%'
The conversion to text is simply to be explicit about what is happening and it makes it easier for Postgres to parse the query.
More importantly, if the value has a fixed number of digits, then I would suggest using a numeric range; something like this:
where crop >= 700000 and crop < 800000
This makes it easier for Postgres to use an index on the column.
Try with cast, like this:
select *
from crop
where cast(substring(cast(field_id as varchar(5)),1,1) as int) = 7
where 5 in varchar(5) you should put number how long is your integer.
I have this example:
CREATE TABLE test
(
VALUE NUMBER (5, 2)
);
INSERT INTO test
VALUES (6.3);
SELECT * FROM test;
In table test, I have a value of 6.3.
I have an application in .NET who queries this table and returns a single value of 6.3.
Assuming the value is stored in the s variable like this:
Dim s As Single = 6.3
Dim d As Double = CDbl(s)
.NET converts the single into a double, and the variable d has this value: 6.3000001907348633. I understand that is a different representation of the 6.3 value, but when I'm using this d value in a operation like this:
(795407.2 * d) / 100 = 50110,6551171188
in PL\SQL when I'm querying the table, I want to get the 6.3000001907348633 value instead of 6.3.
How can I convert this, or there is a datatype that does that? I've tried the BINARY_FLOAT and DOUBLE PRECISION types, but they don't convert this.
EDIT: Using 6.3 value I get: (795407.2 * d) / 100 = 50110,6536. if I round the value to get two decimal places (and assuming that this is a monetary value, I get 50110,66€ and 50110,65€ respectively).
Try using the BINARY_FLOAT Data Type in your SQL.
It represents a 32-bit floating point number in Oracle:
SQL> select cast(6.3 as BINARY_FLOAT) num,
2 to_char(cast(6.3 as BINARY_FLOAT),'0.00000000') num_to_char
3 from dual;
NUM NUM_TO_CHAR
---------- -----------
6,3E+000 6.30000019
Edit:
Sorry, I was too fast and didn't read all your trials
For compatibility purposes, try convert your number / literal to BINARY_FLOAT and then to BINARY_DOUBLE:
SQL> select TO_CHAR(cast(cast(6.3 as BINARY_FLOAT) as BINARY_DOUBLE),'0.000000000000000000') MyNum from dual;
MYNUM
---------------------
6.300000190734863300
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
I've got an Sqlite database where one of the columns is defined as "TEXT NOT NULL". Some of the values are strings and some can be cast to a DOUBLE and some can be case to INTEGER. Once I've narrowed it down to DOUBLE values, I want to do a query that gets a range of data. Suppose my column is named "Value". Can I do this?
SELECT * FROM Tbl WHERE ... AND Value >= 23 AND Value < 42
Is that going to do some kind of ASCII comparison or a numeric comparison? INTEGER or REAL? Does the BETWEEN operator work the same way?
And what happens if I do this?
SELECT MAX(Value) FROM Tbl WHERE ...
Will it do string or integer or floating-point comparisons?
It is all explained in the Datatypes In SQLite Version 3 article. For example, the answer to the first portion of questions is
An INTEGER or REAL value is less than any TEXT or BLOB value. When an INTEGER or REAL is compared to another INTEGER or REAL, a numerical comparison is performed.
This is why SELECT 9 < '1' and SELECT 9 < '11' both give 1 (true).
The expression "a BETWEEN b AND c" is treated as two separate binary comparisons "a >= b AND a <= c"
The most important point to know is that column type is merely an annotation; SQLite is dynamically typed so each value can have any type.
you cant convert text to integer or double so you wont be able to do what you want.
If the column were varchar you could have a chance by doing:
select *
from Tbl
WHERE ISNUMERIC(Value ) = 1 --condition to avoid a conversion from string to int for example
and cast(value as integer) > 1 --rest of your conditions