Here is my code
select len(cast(code as float)),code
from tbl1
where code is not null
and this is the output:
I want a count of digits in the code column.
I don't understand why the last one is counted as 12 and not 8?
Cast it as an int instead:
select len(cast(code as int)), code
from tbl1
where code is not null;
Presumably, some sort of decimal values are getting counted.
Get the number's power of 10 and add 1. This works either if ints or reals to count the number of digits of the whole number part (note using LOG10 only works on positive numbers so I have applied ABS to get around this issue, may not be required for your data):
SELECT code, CASE WHEN Number = 0 THEN 1
ELSE FLOOR(LOG10(ABS(code))) + 1 AS NDigits
FROM tbl1
Related
I created a table with a column containing numbers listed from 1 to 100. I want to delete numbers that divide by 3 without any remainder. Who can recommend me a way (script) to do that. Only logic I could make in this problem is that if the sum of digits of a number can divide by 3 that means any number which correspond to that case could be divisible.
If you are looking to delete the rows with number divisible completely by 3, you can use built-in modulus function
You could say something like this
delete
from myTable
where colNumber%3 = 0
This query should solve your problem
DELETE FROM table WHERE (id % 3) = 0;
I want to get sum of data but I am getting wrong result
Example 1
Example 2
Result when doing sum
AS you can see from Example 1 - where p_key is 11020145101617761 and LC_Amount is 8.4 , 168 , -176.4 the sum of this is 0
similarly in Example 2 - where p_key is 1102014510615767 and LC_amount is
-571067.53, 543873.84 , 27193.69 the sum of this is also 0
but in the result when I do group by with p_key , I am not getting 0
I don't understand what is the reason behind this.
It's an example of IEEE-754 rounding errors. Note the numbers are all very close to zero, but juuuuuust off, see the exponent.
Wrap your SUM in ROUND():
SELECT ROUND( SUM( LC_Amount ), 10 )
...should do it.
This occurs because float is not true exact numeric type.
Try to use DECIMAL(10,2) type to avoid round errors.
There is a lot info about float type in the web.
Table_A
ID Number
-- ------
1 0
2 00
3 0123
4 000000
5 01240
6 000
The 'Number' column is data type varchar.
EDIT for clarity.
My question is, can I easily pull back all rows of data which contain a variable length string of 0's?
I have tried:
SELECT *
FROM Table_A
WHERE LEFT(Number,1) = '0' AND RIGHT(Number,1) = '0'
Using the above, it would still return the below, using the example table provided.
ID Number
-- ------
1 0
2 00
4 000000
5 01240
6 000
I was looking for a function which I could pass the LEN(Number) int into, and then it generates a string of a specfic character (in my case a string of 0's). I wasn't able to find anything though.
Oh, and I also tried adding a SUBSTRING to the WHERE clause, but sometimes the Number column has a number which has a 0's in the middle, so it still returned strings with other numbers except only 0.
SUBSTRING(Number,ROUND(LEN(Number)/2,0),1) = '0'
Any help is appreciated.
So, you want a string that doesn't contain anything that isn't a 0? Sounds like it's time for a double-negative:
SELECT *
FROM Table_A
WHERE NOT Number like '%[^0]%'
AND number like '0%' --But we do want it to contain at least one zero
(The final check is so that we don't match the empty string)
Answer:
Where number like '%0%'
Your can use this query :
SELECT * FROM Table_A WHERE Number LIKE '%0%';
It'll solve your problem.
SELECT *
FROM yourtable
WHERE len(Number) - len(replace(number,'0','')) >= 0
One more approach
You can use this following one also,you will get your expected result.
SELECT *
FROM Table_A
WHERE Nunber not like '%[1-9]%'
Thanks.
I Have to find the percentage difference between two numbers.
i am using the formula A-B/(A+B/2)
I.e
select round( ( A-B)/(A+B/2)) from dual;
where a and b are two numbers. Now if a and b is 0 then the divisor comes as zero. In this case what can be done to avoid error ? what is the percentage difference between two equal numbers or 0 ?
I was working on a query today which required me to use the following to find all odd number ID values
(ID % 2) <> 0
Can anyone tell me what this is doing? It worked, which is great, but I'd like to know why.
ID % 2 is checking what the remainder is if you divide ID by 2. If you divide an even number by 2 it will always have a remainder of 0. Any other number (odd) will result in a non-zero value. Which is what is checking for.
For finding the even number we should use
select num from table where ( num % 2 ) = 0
As Below Doc specify
dividend % divisor
Returns the remainder of one number divided by another.
https://learn.microsoft.com/en-us/sql/t-sql/language-elements/modulo-transact-sql#syntax
For Example
13 % 2 return 1
Next part is <> which denotes Not equals.
Therefor what your statement mean is
Remainder of ID when it divided by 2 not equals to 0
Be careful because this is not going to work in Oracle database. Same Expression will be like below.
MOD(ID, 2) <> 0
ID % 2 reduces all integer (monetary and numeric are allowed, too) numbers to 0 and 1 effectively.
Read about the modulo operator in the manual.
In oracle,
select num from table where MOD (num, 2) = 0;
dividend % divisor
Dividend is the numeric expression to divide. Dividend must be any expression of integer data type in sql server.
Divisor is the numeric expression to divide the dividend. Divisor must be expression of integer data type except in sql server.
SELECT 15 % 2
Output
1
Dividend = 15
Divisor = 2
Let's say you wanted to query
Query a list of CITY names from STATION with even ID numbers only.
Schema structure for STATION:
ID Number
CITY varchar
STATE varchar
select CITY from STATION as st where st.id % 2 = 0
Will fetch the even set of records
In order to fetch the odd records with Id as odd number.
select CITY from STATION as st where st.id % 2 <> 0
% function reduces the value to either 0 or 1
It's taking the ID , dividing it by 2 and checking if the remainder is not zero; meaning, it's an odd ID.
<> means not equal. however, in some versions of SQL, you can write !=