I need Urgent help please

  • Thread starter Thread starter Dimitris
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Dimitris

In a table I have a column with numbers from 1-1000. I need to sort them by
odds and evens. For example 1,3,5,7,9,11.....999 and after 999 the next
number should be 2,4,6,8,10....1000. Is it possible? and how?
Thanks
Dimitris
 
Dimitris said:
In a table I have a column with numbers from 1-1000. I need to sort them by
odds and evens. For example 1,3,5,7,9,11.....999 and after 999 the next
number should be 2,4,6,8,10....1000. Is it possible? and how?
Thanks
Dimitris
Click to expand...

SELECT * FROM some_table ORDER BY some_column Mod 2 DESC, some_column
 
Baz said:
SELECT * FROM some_table ORDER BY some_column Mod 2 DESC, some_column
Click to expand...
Hi Baz,

Could you explain how that works for the benefit of my small brain?

Thanks.
Keith.
 
Keith said:
Hi Baz,

Could you explain how that works for the benefit of my small brain?

Thanks.
Keith.
Click to expand...

The Mod (modulus) operator performs a division, but it returns the remainder
instead of the result. So:

1 Mod 2 = 1
2 Mod 2 = 0
3 Mod 2 = 1
4 Mod 2 = 0

and so on.

So, "some_column Mod 2" returns 1 when some_column is odd, and returns 0
when some_column is even.

Therefore, a descending sort by "some_column Mod 2" puts all the odd values
of some_column before the even values.

Finally, the ascending sort of some_column puts the values in ascending
order within odd and even.
 
Baz said:
The Mod (modulus) operator performs a division, but it returns the
remainder
instead of the result. So:

1 Mod 2 = 1
2 Mod 2 = 0
3 Mod 2 = 1
4 Mod 2 = 0

and so on.

So, "some_column Mod 2" returns 1 when some_column is odd, and returns 0
when some_column is even.

Therefore, a descending sort by "some_column Mod 2" puts all the odd
values
of some_column before the even values.

Finally, the ascending sort of some_column puts the values in ascending
order within odd and even.
Click to expand...
Thanks Baz, I was not familiar with the Mod operator and couldn't find
anything about it in the help. That's quite a neat solution.

Regards,
Keith.
 
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