Delete text after ==

The awk -F'==' '{print $1}' archivo solution uses a field separator ( FS ) with multi-characters. This is valid as long as you are using a version of awk compatible with POSIX. For example, on Solaris it will not work.

So the question is: how to make it work?

So let's simplify: the file consists of lines of the form módulo==versión . Therefore, what we can do is eliminate = and everything that follows:

$ cut -d'=' -f1 fichero
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This is saying: separate the line based on = as a separator ( -d= ) and print the first resulting field ( -f1 ).

It may be a bit fragile, so you can also choose to use sed :

sed 's/=.*//' fichero

This does the same thing: delete from the first symbol = . However, it allows you to extend the command to something more complex, such as:

$ sed '/==/s/=.*//' fichero
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That performs this substitution only on lines that contain == . And if you hurry me, you can say:

sed -n '/==/s/=.*//p' fichero

To print only these lines ( -n inhibits printing by default and p prints the current line).

If you really want to use% co_of% awk, use:

$ awk 'match($0, /^(.+)==(.+)/, res) {print res[1]}' fichero
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As you can see, the syntax is match() . Therefore, it is a question of capturing those that interest us: in this case only the first, so in fact we could limit ourselves to saying match(línea, patrón, matriz de resultados) , without needing to capture the rest.

In a nutshell: match($0, /^(.+)==/, res) does not seem like the best solution here because depending on which environments the field separator with multi-characters may give you problems. Make your life easy by using awk in this case: you do not need to use such complex regular expressions when a sed sencillito already gives you everything you need.

A. VALUES OF THE LEFT OF ==

Option 1.

Capture everything that is before ==

^.*?(?=\=\=)/gm

Option 2.

Make a match without capturing the group from ==

Thanks @fedorqui

^.*?(?:==)/gm

DEMO

Result

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B. VALUES OF THE RIGHT OF ==

=.*

DEMO

Result

==0.7.9
==0.8.0
==1.6.5
==2.3.4
==0.4.0
==1.6

Try this command in Bash:

cat requirements.txt | grep -oP "\w+[-_]{0,1}\w+"

requirements.txt would be the pip requirements file.

The important thing is the regular expression to use and the one that I include includes the requirement of the hyphen or dash separator; I updated the example of @A. Cedano so you can see it live here .

If you need to save the result to a file (surely yes), you can obviously use the output redirection; that is:

cat requirements.txt | grep -oP "\w+[-_]{0,1}\w+" > salida.txt

I hope you serve, greetings.

The alternative in Python is as follows:

import re


r = re.compile('(?P<paquete>.+)(==)(?P<version>.+)'
for l in open('base.txt').readlines():
    print (r.search(l).group('paquete'))

• The re module that handles regular expressions is imported into the first line.
• Since we are going to apply the same search to all the lines, we create an object pattern or search pattern with the expression we are looking for:
  • The first group has name paquete and is formed with any character and any amount.
  • The second group is only a separator, formed by the signs of equality.
  • The third group is named version and is formed by the rest of the characters after the second group.
• We go through the requirements file, line by line,
• And we pass it as an argument to the search (which uses the previously compiled object) and only the result of the group paquete (ie the match ) is printed.