我想出了下面的脚本来列出从服务器日志文件中存储所有呼叫到服务的服务器日志文件,这些调用花费的时间超过10000000微秒(我们日志中的ServiceDuration的值以微秒记录)。
servicedurationlimit.awk
#!/bin/sh
dir1=$1/*.log
# for each log file in the input dir
for file1 in $dir1
do
echo $file1":"
awk '/#BeginLogEntry|ServiceDuration/ {
#get slow running service's information
if ($1 == "#BeginLogEntry")
{
split($0, a, "\t");
servinfo=a[3]" ServiceAt:"a[2];
} else {
getline;
if ($0 > 10000000)
{
print servinfo", ServDur:"$0
}
}
}' $file1
done
在运行脚本时,我得到以下错误:
./servicedurationlimit.awk /path/to/server/logs/
./servicedurationlimit.awk: line 12: syntax error near unexpected token `$0,'
./servicedurationlimit.awk: line 12: ` split($0, a, "\t"); '
你能帮我理解可能导致这种情况的原因吗?
以下是示例日志文件(有2个日志条目):
#BeginLogEntry 04.13 20:11:11.671 BROWSE_ALL
@Properties LocalData
IsJson=1
UserTimeZone=utc
@end
@IdcRSet AuditProps
2
auditPropertyKey
auditPropertyValue
ServiceDuration
62818
ServiceStartTime
{ts '2015-04-13 20:11:11.671'}
@end
#EndLogEntry 04.13 20:11:11.671 BROWSE_ALL
#BeginLogEntry 04.13 21:12:11.671 BROWSE_SOME
@Properties LocalData
IsJson=1
UserTimeZone=utc
@end
@IdcRSet AuditProps
2
auditPropertyKey
auditPropertyValue
ServiceDuration
162818123
ServiceStartTime
{ts '2015-04-13 21:12:11.671'}
@end
#EndLogEntry 04.13 21:12:11.671 BROWSE_SOME
以下是在包含上述日志条目的日志文件上运行脚本后我期望的输出。
BROWSE_SOME ServiceAt:04.13 21:12:11.671, ServDur: 162818123
awk
版本信息
$ awk --version
GNU Awk 3.1.5
答案 0 :(得分:5)
我正在运行GNU Awk 4.0.2并且您的代码在我的框中出现此错误:
awk: cmd. line:2: #get slow running services
awk: cmd. line:2: ^ syntax error
./test.sh: line 11: syntax error near unexpected token `$0,'
./test.sh: line 11: ` split($0, a, "\t");'
但我相信您只需要从评论中删除'单引号:
#get slow running service's