Help with Regular Expression in grep

I'm trying to formulate a regular expression in grep to return expired account names from the shadow file.

If I have a line in /etc/shadow like the following:
I want grep to return everything to the left of :

netuser, that is


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awk -F: '{print $1}' /etc/shadow
mattinvtAuthor Commented:
Almost there - I don't want to see lines that don't contain ":!"

So if I have

the expression only returns netuser1
Here you go -
awk -F: ' /:!/ {print $1}' /etc/shadow

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mattinvtAuthor Commented:
My understanding of GREP is that it is used to print the entire line, not to extract parts of the line.

Why not try this Perl 1-liner?

perl -ne '/^([^:]+)/; print "$1\n"' /etc/shadow

perl -e says that the next argument is to be executed like a program
perl -n says to loop through each line of the file

so perl -ne says loop through each line of the file and execute the argument (inside '');

The perl part is two lines:

/^([^:]+)/;  # <-- this is a regular expression that starts at the beginning of the line and stops at the first ":"
print "$1\n"; # <--- this prints the result (captured inside the () in the previous line) and adds a newline

Hope this helps!
Seeing your previous response, my regexp needs to be modified to

But of course the awk solution is more elegant.
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