如何理解awr报告中的db time?

dhcwenOra · 发表于 2007-12-20 15:44

Snap Id Snap Time                Sessions Cursors/Session
Begin Snap: 9337    12-Dec-07 00:00:24 169          9.5
End Snap: 9338    12-Dec-07 01:00:26 183          9.0
Elapsed:    60.03 (mins)
DB Time:    85.87 (mins)

发现10g里的awr报告中多了一个db time，不知道如何理解

dhcwenOra · 发表于 2007-12-20 15:53

还有一个问题：
top 5事件里经常出现cpu time，看文档解释说是cpu等待的时间，就设任何事情不做，不明白!

dhcwenOra · 发表于 2007-12-20 16:05

ding yi xia !

randylin2006 · 发表于 2007-12-20 16:18

DB可以理解成处理说有所有事务的时间，包括CPU time，sql运行时间和其他等待事件等等

alantany · 发表于 2007-12-20 16:18

1. db time is the aggregation of all forground sessiones'time。
2. cpu time is not a wait event ,is the cost of cpu units.

dhcwenOra · 发表于 2007-12-20 16:37

感觉你的理解还是不对：下面有一个awr一个部分报告：
Time Model Statistics
Total time in database user-calls (DB Time): 6691.6s
Statistics including the word "background" measure background process time, and so do not contribute to the DB time statistic
Ordered by % or DB time desc, Statistic name

Statistic Name          Time (s)             % of DB Time
sql execute elapsed time  6,711.30             100.29
DB CPU                   4,225.97             63.15
parse time elapsed       269.75                4.03
hard parse elapsed time 178.18                2.66
failed parse elapsed time 155.11                2.32
RMAN cpu time (backup/restore) 119.46          1.79
PL/SQL execution elapsed time  44.38          0.66
sequence load elapsed time    7.72             0.12
PL/SQL compilation elapsed time 6.36          0.10
connection management call elapsed time 1.68    0.03
hard parse (sharing criteria) elapsed time 0.06  0.00
repeated bind elapsed time       0.02       0.00
inbound PL/SQL rpc elapsed time    0.01       0.00
hard parse (bind mismatch) elapsed time 0.01    0.00
DB time                         6,691.59
background elapsed time       8,912.91
background cpu time             222.48
background elapsed time 8,912.91
background cpu time 222.48

按照你的理解应该是db time=DB CPU + sql execute elapsed time 。。。。。
不对阿