I have a dataset, df, that has a column of values that are in MB. I would like to transform into TB.
MB
10000000
20000000
Desired
TB
9.09
18.18
Doing
select MB AS 'TB', (CONVERT([int],round([MB]/((1024)*(1024)),(0)))) AS TB from df
However, the result I get is
MB
0
0
I am still researching. Any suggestion is appreciated
/ is integer division in SQL Server.
It means, that for example
SELECT 4 / 5
will return 0.
But, if you write
SELECT 4 / 5.0
you'll get 0.8
5.0 is treated as decimal type and all values in the expression are converted to decimal and division is no longer integer.
So, you can use 1024.0 constant in the expression, and all the values in it will be converted to decimal type and division will not be integer.
In the question you say that you want to show results with two decimal places, so you should not convert result to int.
select
[MB]
,round([MB]/(1024.0*1024.0), 2) AS TB
from df
I am exactly doing this Sum(2322933.99/1161800199.8)*
100
I should get
1.9 something but I am getting 64. Something
can anyone guide my y this division in snowflake giving wrong results
I tried them converting into decimal values and tried with Formula div0()
Nothing worked
I guess that your database table has 33 rows. So you get 33 * 1.9 (because of SUM), which is about 64.
My guess, with the few details that you gave us:
sum(x)/sum(y) is different than sum(x/y)
1/2 + 2/4 + 4/8 = 1.5
(1+2+4)/(2+4+8) = 0.5
Try writing sum(total gross weight)/sum(total cases filled) instead of sum(total gross weight /total cases filled).
I am currently outputting values out to 6 decimal places, and would like to round up the 6th place regardless of the integer value.
I have been using a CEILING() function so far which has worked great for values 1-9 on rounding up; however, in situations where I have the 7th decimal as 0 (ex: 2705.1520270), the function does not round up to 2705.152028.
select CEILING(price*1000000)/1000000 as PriceRound
from tc_alcf a (nolock)
Here is one approach:
SELECT ROUND(2705.1520270 + 0.0000005, 6);
2705.1520280
Demo
We can add 0.0000005 to the input and then just use SQL Server's ROUND function to 6 decimal places. This works because values with a sixth decimal place between 0 and 0.4999 (repeating) would become 5 to 0.9999 (repeating), meaning they would round up to the next digit. And values with already have 5 or greater in the sixth decimal place would not be bumped up to the next digit.
This problem should be familiar to many developers as the rounding half up problem.
Add 1 and use FLOOR():
select floor(price*1000000 + 1)/1000000 as PriceRound
from tc_alcf a
Or you can also shift the decimal by multiplying with the power function
CEILING(2705.1520275 * POWER(10,6)) / POWER(10,6)
How to handle decimal numbers in solidity?
If you want to find the percentage of some amount and do some calculation on that number, how to do that?
Suppose I perform : 15 % of 45 and need to divide that value with 7 how to get the answer.
Please help. I have done research, but getting answer like it is not possible to do that calculation. Please help.
You have a few options. To just multiply by a percentage (but truncate to an integer result), 45 * 15 / 100 = 6 works well. (45 * 15%)
If you want to keep some more digits around, you can just scale everything up by, e.g., some exponent of 10. 4500 * 15 / 100 = 675 (i.e. 6.75 * 100).
I'm trying to figure out decimal data type of a column in the SQL Server. I need to be able to store values like 15.5, 26.9, 24.7, 9.8, etc
I assigned decimal(18, 0) to the column data type but this not allowing me to store these values.
What is the right way to do this?
DECIMAL(18,0) will allow 0 digits after the decimal point.
Use something like DECIMAL(18,4) instead that should do just fine!
That gives you a total of 18 digits, 4 of which after the decimal point (and 14 before the decimal point).
You should use is as follows:
DECIMAL(m,a)
m is the number of total digits your decimal can have.
a is the max number of digits you can have after the decimal point.
http://www.tsqltutorials.com/datatypes.php has descriptions for all the datatypes.
The settings for Decimal are its precision and scale or in normal language, how many digits can a number have and how many digits do you want to have to the right of the decimal point.
So if you put PI into a Decimal(18,0) it will be recorded as 3?
If you put PI into a Decimal(18,2) it will be recorded as 3.14?
If you put PI into Decimal(18,10) be recorded as 3.1415926535.
For most of the time, I use decimal(9,2) which takes the least storage (5 bytes) in sql decimal type.
Precision => Storage bytes
1 - 9 => 5
10-19 => 9
20-28 => 13
29-38 => 17
It can store from 0 up to 9 999 999.99 (7 digit infront + 2 digit behind decimal point = total 9 digit), which is big enough for most of the values.
You can try this
decimal(18,1)
The length of numbers should be totally 18. The length of numbers after the decimal point should be 1 only and not more than that.
In MySQL DB decimal(4,2) allows entering only a total of 4 digits. As you see in decimal(4,2), it means you can enter a total of 4 digits out of which two digits are meant for keeping after the decimal point.
So, if you enter 100.0 in MySQL database, it will show an error like "Out of Range Value for column".
So, you can enter in this range only: from 00.00 to 99.99.
The other answers are right. Assuming your examples reflect the full range of possibilities what you want is DECIMAL(3, 1). Or, DECIMAL(14, 1) will allow a total of 14 digits. It's your job to think about what's enough.
request.input("name", sql.Decimal, 155.33) // decimal(18, 0)
request.input("name", sql.Decimal(10), 155.33) // decimal(10, 0)
request.input("name", sql.Decimal(10, 2), 155.33) // decimal(10, 2)