grep

Grep lines that do not begin with ‘#’ or ‘;’

 in Bash / GNU/Linux  tagged bash / grep / squid / start of line by

Recently, we wanted to modify  the squid configuration file, which is really really big!

wc -l /etc/squid/squid.conf
7898 /etc/squid/squid.conf

We wanted to find all active rules that are enabled to modify our proxy server. Out of those ~8K lines less than 20 are actually active configuration, the rest is documentation.

To find all active configuration lines we needed to find all lines that:

  • are not empty
  • do not start with #
  • do not start with ;

To do this we used the following grep command

grep "^[^#;]" /etc/squid/squid.conf

The first ^ refers to the beginning of the line, this way if in a line there is some configuration and after that there is a comment it will not be excluded by mistake. The rest, [^#;] matches any character which is not # or ;.

This is what was actually in my configuration file (out of ~8K lines)

acl SSL_ports port 443
acl Safe_ports port 80        # http
acl Safe_ports port 21        # ftp
acl Safe_ports port 443        # https
acl Safe_ports port 70        # gopher
acl Safe_ports port 210        # wais
acl Safe_ports port 1025-65535    # unregistered ports
acl Safe_ports port 280        # http-mgmt
acl Safe_ports port 488        # gss-http
acl Safe_ports port 591        # filemaker
acl Safe_ports port 777        # multiling http
acl CONNECT method CONNECT
http_access deny !Safe_ports
http_access deny CONNECT !SSL_ports
http_access allow localhost manager
http_access deny manager
http_access allow localhost
http_access deny all
http_port 3128
coredump_dir /var/spool/squid
refresh_pattern ^ftp:        1440    20%    10080
refresh_pattern ^gopher:    1440    0%    1440
refresh_pattern -i (/cgi-bin/|\?) 0    0%    0
refresh_pattern (Release|Packages(.gz)*)$      0       20%     2880
refresh_pattern .        0    20%    4320

Grep: Print only the words of the line that matched the regular expression, one per line

 in Bash / GNU/Linux  tagged grep / word by 
grep -oh "\w*$YOUR_PATTERN\w*" *

We used the following parameters on our command:

-h, –no-filename : Suppress the prefixing of file names on output. This is the default when there is only  one  file  (or only standard input) to search.
-o, –only-matching : Print  only  the matched (non-empty) parts of a matching line,  with each such part on a separate output line.

Also, we wrapped out pattern with the \w* that matches  all word-constituent characters on either side. The * character states that it should find 0 or more of those characters in the pattern to match.

How to: Extract all usernames that are logged in from who

 in Bash / GNU/Linux / HowTos  tagged bash / cut / GNU/Linux / grep / howtos / sort / unique / who by

who | cut -d ' ' -f 1 | sort -u

who: will show who is logged on

cut  -d ‘ ‘ -f 1: will remove all sections from each line except for column 1. It will use the space character as the delimiter for the columns

sort -u: it will sort the usernames and remove duplicate lines. So if a user is logged in multiple times you will get that username only once.
In case you want to filter out root user from this list you can do it as follows:

who | cut -d ' ' -f 1 | sort -u | grep -v 'root'

Linux Bash: How to find a string inside C/C++ Source and Header Files

 in Bash / GNU/Linux / Programming  tagged bash / find / GNU/Linux / grep / regex by

Just issue the following in the folder of the source codes and replace ‘FIND ME’ with the string you want to query:

find -O3 . -regex '.*\.\(c\|cpp\|h\)





You can change the contents of the regular expression to search in different file extensions (file types). 


 -exec grep 'FIND ME!' -sl '{}' \;

You can change the contents of the regular expression to search in different file extensions (file types).