Commands by houghi001 (2)

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find an unused unprivileged TCP port
Not really better - just different ;) There's probably a really simple solution out there somewhere...

Lists all listening ports together with the PID of the associated process
The PID will only be printed if you're holding a root equivalent ID.

Introduction to user commands
Tested on debian and ubuntu. Translations could be useless, so "LANG=C man intro" is a better alternative.

Printing multiple years with Unix cal command
src: http://tinyapps.org/weblog/nix/200907090700_linux_cal_print_multiple_years.html

GZip all files in a directory separately

Which processes are listening on a specific port (e.g. port 80)
swap out "80" for your port of interest. Can use port number or named ports e.g. "http"

Create a single-use TCP (or UDP) proxy
Redirect the local port 2000 to the remote port 3000. The same but UDP: $ nc -u -l -p 2000 -c "nc -u example.org 3000" It may be used to "convert" TCP client to UDP server (or viceversa): $ nc -l -p 2000 -c "nc -u example.org 3000"

Which processes are listening on a specific port (e.g. port 80)
swap out "80" for your port of interest. Can use port number or named ports e.g. "http"

Print a row of 50 hyphens
essentially the ruby one, but perhaps has a larger installed base

a find and replace within text-based files, to locate and rewrite text en mass.
syntax follows regular command line expression. example: let's say you have a directory (with subdirs) that has say 4000 .php files. All of these files were made via script, but uh-oh, there was a typo! if the typo is "let's go jome!" but you meant it to say "let's go home!" find . -name "*.php" | xargs perl -pi -e "s/let\'s\ go\ jome\!/let\'s\ go\ home\!/g" all better :) multiline: find . -name "*.php" | xargs perl -p0777i -e 's/knownline1\nknownline2/replaced/m' indescriminate line replace: find ./ -name '*.php' | xargs perl -pi -e 's/\".*$\"/\new\ line\ content/g'


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