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 occurrence of the second number within the first and returns what is left to complete the first number i.e. 2(5) gives 10 adding 3 gives 13 therefore MOD(13,5)=3

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 the records are selected. These types of queries will never work:

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 @m = MOD(321,11) SELECT @m Error: Msg 195, Level 15, State 10, Line 2 'MOD' is not a recognized built-in function name. Why I can't use this function from the link above?

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 = (200/10) ----> 20 ln_months_mod = (130/10) ----> 13 ln_months_mod = (123/10) ----> 10 Last one need to return as 12 So I need, ln_months_mod = MOD (123/10) like that

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. What you need is not a loop ending when MOD(v_counter_1, 2) = 0, but a loop that scans all the rows, simply printing the values for the only rows that match your criteria:

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 details you can use this code. select * from (select ename,job, ROWNUM AS rn from emp) where mod(rn, 2) <> 0;

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 rank from tenter t) tt where c_date between :p_from_date and :p_to_date order by l_uid, c_time asc;