Parsing 解析aws cloudfront日志

Parsing 解析aws cloudfront日志,parsing,amazon-cloudfront,logfiles,Parsing,Amazon Cloudfront,Logfiles,我有一个shell脚本,它解析aws cloudfront日志 awk '{print $1","$2","$4","$5","$8","$9","(substr($11,1,7))","$12 }' access_log | grep cid= | sed -e 's/\/data//g;s/\/videos//g;s/\/images//g;s/\/hls//g;s/\/rss//g;s/\/xml//g;s/cid=//g' > stats.txt 我正在尝试合并字段$1和$2,因

我有一个shell脚本,它解析aws cloudfront日志

awk '{print $1","$2","$4","$5","$8","$9","(substr($11,1,7))","$12 }' access_log | grep cid= | sed -e 's/\/data//g;s/\/videos//g;s/\/images//g;s/\/hls//g;s/\/rss//g;s/\/xml//g;s/cid=//g' > stats.txt
我正在尝试合并字段$1和$2,因此日期时间戳类似于以下格式: 日期和时间 0000-00-00:00:00

它当前在两个字段中:日期、时间 2012-12-23 20:59:47


感谢您的帮助

脚本的awk部分在您希望格式化为datetime的日期和时间部分之间添加了一个逗号。这导致“YYYY-MM-DD HH:MM:SS”显示为“YYYY-MM-DD,HH:MM:SS”

要获得所需的结果,请将脚本更改为:

awk '{print $1" "$2","$4","$5","$8","$9","(substr($11,1,7))","$12 }' access_log | grep cid= | sed -e 's/\/data//g;s/\/videos//g;s/\/images//g;s/\/hls//g;s/\/rss//g;s/\/xml//g;s/cid=//g' > stats.txt
这应该正确插入到datetime字段中