Bdtjtk

Can someone explain, how this sed command works here?

pkg info | sed -e 's/([^.]*).*/1/' -e 's/(.*)-.*/1/'

This command removes version numbers from packages and prints into stdout like this

yajl-2.1.0                     Portable JSON parsing and serialization library in ANSI C

youtube_dl-2018.12.03          Program for downloading videos from YouTube.com

zathura-0.4.1                  Customizable lightweight pdf viewer

zathura-pdf-poppler-0.2.9_1    Poppler render PDF plugin for Zathura PDF viewer

zip-3.0_1                      Create/update ZIP files compatible with PKZIP

zsh-5.6.2                      The Z shell

and turns into this

yajl

youtube_dl

zathura

zathura-pdf-poppler

zip

zsh

But I am having a hard time understanding the parts ([^.]*).* (.*)-.*. I understand the case of , -e, s. But those wildcards seems very cryptic here.

In your regex ([^.]*).*, ( which actually is ( is the start of a capturing group and then [^.]* captures every character except a literal dot and * means zero or more, then ) is the mark of closing of group that we started, then .* captures whatever remains after capturing group1.

Similar will be the explanation for (.*)-.* regex, where (.*) will capture everything greedily in capturing group but will stop at last hyphen - and then will match hyphen and further .* will match remaining text.

To explain with an example, lets take youtube_dl-2018.12.03.

Here, ([^.]*) will capture everything until dot, hence it will capture youtube_dl-2018 and then remaining .* will capture .12.03. Then it will be replaced by 1 which means youtube_dl-2018 will be passed to the next regex -e 's/(.*)-.*/1/'.

Then in your second regex, (.*)-.*, (.*) will capture youtube_dl and put in group1 because after that there is a hyphen and .* will capture remaining text which is 2018. And as it is replaced by 1 hence final text will become youtube_dl.

Seeing your data, I believe, you can also simplify your command to this, as your first regex in sed command seems redundant. Try this following command and see if it outputs same result?

pkg info | sed -e 's/(.*)-.*/1/'

You can only use this simplified command, as none of your data contains a . before a -, otherwise you should use your own command which has two sed rules.

Also, on another note, if you use -r, (or -E for OS X), for extended regex, you don't need to escape the parentheses and you can write your regex as,

pkg info | sed -r 's/([^.]*).*/1/' -r 's/(.*)-.*/1/'

It is a difficult way for saying:

Remove all substrings starting with a dot or hypen.

The part before the delimiter is matched and remembered.

Alternatives:

# Incorrect: removes from first, not last hypen:

#    pkg info | sed 's/[-.].*//'

#    pkg info | cut -d "-" -f1 | cut -d"." -f1

#    pkg info | awk -F "-|[.]" '{print $1}'

# The dot is not needed when you remove the substring starting with the last hypen

pkg info | sed 's/-[^-]*$//'

pkg info | rev | cut -d"-" -f2- | rev

pkg info | awk -F "[.]" '{print $1}' | awk -F "[-]" -vOFS='-' 'NF>1 { NF--;print;}'

1. 
   

   Silly invisible-text GNU grep method that works on the console,
   but which would fail if sent to a file or piped to a filter:

   
   
   

   pkg info | GREP_COLORS='ms=30;30;30' grep '-[^-]*s.*$'
   
   

   
   
   

   ---

   
   

   How it works: grep is used to find the last hyphen before a
   space, and everything after that, (i.e. everything we don't want
   to see), which grep shows in highlighted colors as defined in the
   GREP_COLORS environmental variable. Since the highlight colors
   30;30;30 is a black font, (on a black background), the unwanted
   text is invisible.

   
   
   

   If the terminal background is already black, GREP_COLORS='ms=30
   would be sufficient.

   
   


2. 
   

   sed method based on not printing the grep regex:

   
   
   

   pkg info | sed 's#(^.*)(-[^-]*[[:space:]].*$)#1#'
   
   

   
   
   

   ...this method can be sent to pipes and filters. Shorter version using GNU sed:

   
   
   

   pkg info | sed 's#(^.*)(-.*s.*)#1#'