It needs to by backwards because you want multiple results. Lets make an
example and use small numbers:
Your list - the values you are looking at 1,2,3,4,5,6,7,8,9,10.
Lets say we want to find the list 1, 6, 11, 16 or (5 * x + 1) where x is
0,1,2,3,4.
So (mod(1,5)) = 1, (mod(2,5)) = 2, (mod(3,5)) = 3 .... (mod(5,5))=0,
(mod(6,5))=1....
We want all of the results that have the result of 1.
Now after I made a long example and re-read your question, I don't think it
makes any difference which way you have it 1 = (mod(x,y)) or (mod(x,y) = 1,
but the trick is you need to col name in the mod function and not as the
result.
Troy Sosamon
===== Original Message from [EMAIL PROTECTED] at 11/21/02 1:45 pm
>that is interesting, why is it 1 = instead of = 1? Will it run faster that
>way?
>
>----- Original Message -----
>From: "Troy Sosamon" <[EMAIL PROTECTED]>
>To: "Multiple recipients of list witango-talk" <[EMAIL PROTECTED]>
>Sent: Thursday, November 21, 2002 12:39 PM
>Subject: RE: Witango-Talk: modulus
>
>
>> You should be able to get this to work, but you need to think backwards.
>> Try something like this:
>>
>> select ..... where 1 = (mod(colname,50))
>>
>> Remember you want to look at the mod of the col value and find the results
>> that equal 1.
>>
>> Troy Sosamon
>>
>> -------- Original Message --------------
>>
>> hello, i have a table, and for paging reasons, i want to get every entry
>> where:
>>
>> count modulus 50=1
>> and
>> count modulus 50=0
>>
>> this would make it so i could make a drop down list that had ranges like:
>>
>> 1..50
>> 51..100
>> 101..150
>>
>> etc.
>>
>> problem is i cant find how to do modulus in SQL. Is there a way? If not
>is
>> there another way to do this without a for loop?
>>
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>
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