Sql как найти четные числа

Bit rusty on the old sql.

Can you help?

Given a number eg 1 or 2 or 4 I need to determine if it’s even or odd number and do some calculation
depending if even or odd .

How do you detect that in sql (sql server 2000)
thanks a lot

asked Nov 26, 2009 at 19:55

Use the modulus operator n % 2. It returns 0 if the number is even, and 1 if the number is odd.

declare @t table(num int)
insert into @t select 1 union all select 2 union all select 3 union all select 4

select 
    num
    ,case when num % 2 = 0 then 'Even' else 'Odd' end as Status
from @t 

Output:

num Status

1   Odd
2   Even
3   Odd
4   Even

e.g. If the number is even(multiply by 1) or odd (multiply by 2) then divide by 10 and get the remainder

declare @myNumber int ,@result int
set @myNumber = 16 
select  
    Result = 
    (case when @myNumber % 2 = 0 then @myNumber * 1 else @myNumber * 2 end) %10     

Result

6

when @myNumber = 11 then

Result

2

Hope this helps

answered Nov 27, 2009 at 3:10

I am using the same thing in MS SQL SP as follows:

IF @current_number % 2 = 0 SET @something = 1 

— or —

IF @current_number % 2 = 0 exec sp_whatever

answered Nov 4, 2013 at 19:29

Use the modulus operator (%).

x % 2

will tell you if x is even or odd.

answered Nov 26, 2009 at 19:58

NOT INTENDED AS A SERIOUS ANSWER…BUT IT WORKS.

Please reserve the following for giving to people you don’t like who simply want a quick answer without wanting to understand the solution (i.e people wanting to cheat).

--Enter the whole integer you want to check
declare @number int = 115

--The check is independent of sign
set @number = abs(@number)
declare @OriginalNumber int = @number

--Firstly, we need to peform the Wilhelm Leibniz conversion, 64 length to allow for very big numbers
declare @WilhelmLeibnizConversion varchar(64) = ''

declare @currentBit int = power(8,exp(log(1))+1)

while @currentBit > 0
begin

    set @WilhelmLeibnizConversion=convert(char(1), @number % 2) + @WilhelmLeibnizConversion

    set @number = convert(int, (@number / 2)) 

    set @currentBit-=1

end

--Although checking the 1 bit value of the Wilhelm Leibniz conversion is usually enough, for robust code you should also include the Kimmo Eriksson Factors one and two.
declare @KimmoErikssonFactor1 int = (@OriginalNumber + 1) % 2
declare @KimmoErikssonFactor2 int = (@OriginalNumber - 1) & 1

--Now check all 3 for 100% confirmation on the parity of your original number.
select case when right(@WilhelmLeibnizConversion,1) = 0 and (@KimmoErikssonFactor1 + @KimmoErikssonFactor2 <> 0) then 'Even' else 'Odd' end

answered Jul 4, 2018 at 9:15

Let’s say for a table STATION.

Schema:

ID NUMBER

CITY VARCHAR

STATE VARCHAR

-- You can use any of the mentioned criteria to fetch the even ID.

-- 1. MOD() Fucntion

select distinct CITY from STATION as st where MOD(st.id,  2) = 0

-- 2. % function

select distinct CITY from STATION as st where st.id % 2 = 0

You could check the 1-bit of the hex value of the number. If that bit is on, then it is odd.

DECLARE @Int INT

SELECT CASE WHEN @Int&0x0001<>0 THEN 'ODD' ELSE 'EVEN' END

USE AdventureWorks;
GO

SELECT BusinessEntityID,
    CASE BusinessEntityID % 2
    WHEN 0 THEN 'Even' ELSE 'Odd' END AS "Status"
FROM HumanResources.Employee;
GO

задан 12 июл 2012 в 11:16

MOD(N,M)
%
Значение по модулю (подобно оператору % в C). Возвращает остаток от деления N на M

MySQL четное/нечетное.

Для MSSQL & PostgreSQL так (Х – целое, проверяемое значение):

select case X % 2 when 0 then 'четное' else 'нечетное' end

ответ дан 28 июл 2012 в 6:09