bash 使用 awk 命令递归搜索
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Recursive search using awk command
提问by atul329
I have 100 log files and I want awk to search for a given pattern between a given timestamp recursively. Log files look something like this
我有 100 个日志文件，我希望 awk 递归搜索给定时间戳之间的给定模式。日志文件看起来像这样
2010-03-24 07:00:01 ZZZZC941948879 RUFFLES 222.222.222.222 GET / - 80 - 220.181.7.113 HTTP/1.1
2010-03-24 07:00:23 ZZZZC941948879 RUFFLES 222.222.222.222 GET
Code is
代码是
awk -v "b=$date1" -v "e=$date2" ' >= b && <= e' log.txt > output
grep -i "21things" output
I am able to search for the pattern but for single file only. Is it possible using awk command to search recursively?
我可以搜索模式，但只能搜索单个文件。是否可以使用 awk 命令进行递归搜索？
Thanks for the help..!!
谢谢您的帮助..！！
回答by glenn Hymanman
If your logs are all in the same directory, use a shell wildcard:
如果您的日志都在同一目录中，请使用 shell 通配符：
awk -v "b=$date1" -v "e=$date2" ' >= b && <= e' *.log
Note that awk can do what grep does, so you don't need the temp file:
请注意，awk 可以执行 grep 所做的操作，因此您不需要临时文件：
awk -v "b=$date1" -v "e=$date2" -v patt="21things" '
>= b && <= e && tolower(##代码##) ~ patt
' *.log
If you have GNU awk, use -v IGNORECASE=1and remove the tolowerfunction.
如果您有 GNU awk，请使用-v IGNORECASE=1并删除该tolower函数。
