RE: Calculating archivedlog file size per day
- From: "Post, Ethan" <Ethan.Post@xxxxxx>
- To: <oracle-l@xxxxxxxxxxxxx>
- Date: Mon, 14 Mar 2005 09:59:07 -0600
Here is a quick report I run to see when archive logs are being
generated. Doesn't sort to pretty but tells you what hours of the day
are busiest.
set lines 140
col dt format a6 heading "Date"
col "00" format 99
col "01" like "00"
col "02" like "00"
col "03" like "00"
col "04" like "00"
col "05" like "00"
col "06" like "00"
col "07" like "00"
col "08" like "00"
col "09" like "00"
col "10" like "00"
col "11" like "00"
col "12" like "00"
col "13" like "00"
col "14" like "00"
col "15" like "00"
col "16" like "00"
col "17" like "00"
col "18" like "00"
col "19" like "00"
col "20" like "00"
col "21" like "00"
col "22" like "00"
col "23" like "00"
col "24" like "00"
select substr(dt,1,6) dt,
sum(decode(hh,1,1,0)) "01",
sum(decode(hh,2,1,0)) "02",
sum(decode(hh,3,1,0)) "03",
sum(decode(hh,4,1,0)) "04",
sum(decode(hh,5,1,0)) "05",
sum(decode(hh,6,1,0)) "06",
sum(decode(hh,7,1,0)) "07",
sum(decode(hh,8,1,0)) "08",
sum(decode(hh,9,1,0)) "09",
sum(decode(hh,10,1,0)) "10",
sum(decode(hh,11,1,0)) "11",
sum(decode(hh,12,1,0)) "12",
sum(decode(hh,13,1,0)) "13",
sum(decode(hh,14,1,0)) "14",
sum(decode(hh,15,1,0)) "15",
sum(decode(hh,16,1,0)) "16",
sum(decode(hh,17,1,0)) "17",
sum(decode(hh,18,1,0)) "18",
sum(decode(hh,19,1,0)) "19",
sum(decode(hh,20,1,0)) "20",
sum(decode(hh,21,1,0)) "21",
sum(decode(hh,22,1,0)) "22",
sum(decode(hh,23,1,0)) "23",
sum(decode(hh,24,1,0)) "24"
from
(
select trunc(first_time) dt,
to_number(to_char(first_time,'HH24')) hh
from v$archived_log
)
group
by substr(dt,1,6)
/
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