将一个命令的输出作为参数传递给另一个 [英] Passing output from one command as argument to another

问题描述

我有这个用于:

for i in `ls -1 access.log*`; do tail $i |awk {'print $4'} |cut -d: -f 1 |grep - $i > $i.output; done

ls 将提供access.log，access.log.1，access.log.2等.
tail (尾部)将给我每个文件的最后一行，如下所示:192.168.1.23 - - [08/Oct/2010:14:05:04 +0300]等.等等.
awk + ​​cut 将提取日期(2010年10月8日-但在每个access.log中不同)，这将允许我为其 grep 并将输出重定向到一个单独的文件.

ls will give access.log, access.log.1, access.log.2 etc.
tail will give me the last line of each file, which looks like: 192.168.1.23 - - [08/Oct/2010:14:05:04 +0300] etc. etc. etc
awk+cut will extract the date (08/Oct/2010 - but different in each access.log), which will allow me to grep for it and redirect the output to a separate file.

但是我似乎无法将awk + ​​cut的输出传递给grep.

But I cannot seem to pass the output of awk+cut to grep.

所有这些的原因是那些访问日志包含具有多个日期(06/Oct，07/Oct，08/Oct)的行，而我只需要具有最新日期的行.

The reason for all this is that those access logs include lines with more than one date (06/Oct, 07/Oct, 08/Oct) and I just need the lines with the most recent date.

我该如何实现?

谢谢.

推荐答案

作为注释，tail显示最后10行.

As a sidenote, tail displays the last 10 lines.

可能的解决方案是这样grep:

A possible solution would be to grepthis way:

for i in `ls -lf access.log*`; do grep $(tail $i |awk {'print $4'} |cut -d: -f 1| sed 's/\[/\\[/') $i > $i.output; done

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