Jun 27, 2014 · MOD(13,5): returns 3 whereas, REMAINDER (13,5) returns -2. A simple way to understand the difference is that MOD uses the floor function therefore it counts the …

Aug 22, 2012 · SELECT COUNT(*) FROM ( select v1, v2, rownum rn from Foo ) f WHERE mod(rn,2) = 0; The ROWNUM is the row's position ins the result set, and it is evaluated after …

Oct 26, 2021 · This is the documentation of the MOD() function. – Uwe Keim. Commented Oct 7 ... Oracle query using mod ...

Apr 26, 2014 · I wanted to use MOD function in SQL Server 2008R2 and followed this link but still got the message: 'MOD' is not a recognized built-in function name. DECLARE @m INT SET …

May 1, 2020 · Mod function in fast formula Content. Hi all, I need to know how do we use MOD function in fast formula. Example: ln_months_mod = (100/10) ----> 10 ln_months_mod = …

Aug 13, 2017 · For Oracle, you have to use the MOD function: WHERE MOD(column, 2) = 1 Share. Improve this answer. Follow ...

Mod Function. Oct 7, 2016 3:40PM ... Is there a function in the reports that would allow me to find the ...

Jan 11, 2017 · The issue is not in the type of your variables, but in the fact that your loops end at the first row that does not match MOD(v_counter_1, 2) <> 0, thus not scanning all the rows. …

Nov 16, 2013 · This one is comfortable in oracle SQL as well as MySQL. select * from (select ename,job, ROWNUM AS rn from emp) where mod(rn, 2) = 0; To find the ODD number of row …

Sep 30, 2018 · you might use : select decode(mod(tt.rank,2),0,'even','odd') as rank, tt.* from ( select t.c_date,t.c_name,t.c_time,l_uid, row_number() over (order by l_uid, c_time asc) as …