How to eliminate only continuous duplicates but not all duplicates in a select query (MySQL)?
I have a table like this:
01-Jul-17 100
02-Jul-17 100
03-Jul-17 300
04-Jul-17 300
05-Jul-17 500
06-Jul-17 500
07-Jul-17 300
08-Jul-17 400
09-Jul-17 100
10-Jul-17 100
What I want to output is (in this order) by eliminating the continuous duplicates but not all duplicates:
100
300
500
300
400
100
I cannot select Distinct, as it will eliminate the second instances of 300, 100. Is there a way to achieve this result in MySQL?
Thanks!
You want to get the previous value. If the dates really have no gaps or duplicates, just do:
select t.*
from t left join
t tprev
on t.col1 = date_add(tprev.col1, interval 1 day)
where tprev.col2 is null or tprev.col2 <> t.col2;
EDIT:
If the dates don't meet these conditions, then you can use variables:
select t.*
from (select t.*,
(@rn := if(@v = col2, @rn + 1,
if(@v := col2, 1, 1)
)
) as rn
from t cross join
(select @v := 0, @rn := 0) params
order by t.col1
) t
where rn = 1;
Note that MySQL does not guarantee the order of evaluation of expressions in the SELECT. So variables should not be assigned in one expression and then used in another -- they should be assigned in a single expression.