[英]How can I extract a file name from a grep result?
I am trying to get a list of all programs executed on system startup. 我正在尝试获取系统启动时执行的所有程序的列表。
My game plan is as follows: 我的游戏计划如下:
grep -r
the /etc/init.d
and /etc/rc.d/*
directories grep -r
/etc/init.d
和/etc/rc.d/*
目录 ./...
) ./...
) To that end, I used the following: 为此,我使用了以下内容:
egrep -r '^\s*/|\$\(\s*/|\`\s*/' /etc/rc.d/* /etc/init.d
Since it's searching files in the directories, the results list the file it was found in and the full line. 由于它是在目录中搜索文件,因此结果将列出在其中找到的文件以及整行。 I would like to now pipe the results into something to get just the file name and the pipe that into
sort|uniq
to get a simplified list. 现在,我想将结果通过管道传递给某种文件,以获取文件名,通过管道将管道传递给
sort|uniq
以获取简化列表。 I think I can use awk somehow, but I am not so familiar with it. 我想我可以以某种方式使用awk,但是我对此并不熟悉。
Example Result: 结果示例:
/etc/init.d/foo: foo=$(/bin/echo hello)
/etc/init.d/bar: bar=$(/bin/echo world)
/etc/rc.d/init.d/foobar: /bin/false
Desired Output: 所需输出:
/bin/echo
/bin/false
If you add -h option to egrep, filename will not be shown. 如果在egrep中添加-h选项,则不会显示文件名。
egrep -hr '^\s*/|\$\(\s*/|\`\s*/' /etc/rc.d/* /etc/init.d | sed -e 's/\($(\|)\)//g'
That sed regex will delete all the "$(" and ")" sed regex将删除所有的“ $(”和“)”
OK, thanks to the help from alb3rtobr, I was able to get it: 好的,感谢alb3rtobr的帮助,我能够得到它:
egrep -hor '(^\s*/|\$\(\s*/|\`\s*/)[^ ]*' /etc/rc.d/* /etc/init.d | sed -e 's/\($(\|)\|^\s*\)//g' | sort | uniq
I modified the egrep to continue matching until encountering a space and then added the -o
option to return only the matched pattern. 我修改了egrep以继续匹配,直到遇到空格为止,然后添加了
-o
选项以仅返回匹配的模式。 I also modified the sed to trim leading whitespace. 我还修改了sed以修剪前导空白。
EDIT: Changed to match pattern while not a space character 编辑:更改为匹配模式,而不是空格字符
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