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MS Access Forum / New Users / September 2005I need Urgent help please
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 > 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 SELECT * FROM some_table ORDER BY some_column Mod 2 DESC, some_column Thanks Baz. You saved my day! >> In a table I have a column with numbers from 1-1000. I need to sort them > by [quoted text clipped - 4 lines] > > SELECT * FROM some_table ORDER BY some_column Mod 2 DESC, some_column >> In a table I have a column with numbers from 1-1000. I need to sort them > by [quoted text clipped - 4 lines] > > SELECT * FROM some_table ORDER BY some_column Mod 2 DESC, some_column Hi Baz, Could you explain how that works for the benefit of my small brain? Thanks. Keith. > >> In a table I have a column with numbers from 1-1000. I need to sort them > > by [quoted text clipped - 11 lines] > Thanks. > Keith. 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. > The Mod (modulus) operator performs a division, but it returns the > remainder [quoted text clipped - 16 lines] > Finally, the ascending sort of some_column puts the values in ascending > order within odd and even. 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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