Grep表达式排除了我想要的一些匹配项

Question

I'm working on a program that extracts the comments from a bash file and outputs them to a new file.

I need to ignore #'s encased in ' ' or " " quotations which I think I have done correctly.

grep -oe "[^\'\"\\]#[^\'\"].*" somefile >> somecomments

This extracts comments that are preceded by some text fine, eg

echo Sum: $Sum    # Displays the sum

will be converted to "# Displays the sum" in the output file. The problem is that lines beginning with # are now excluded for some reason eg

# Name
# Date

will not show up in the output file at all.

How do I fix my expression so that I can still exclude quotations in front of the # but have it extract lines beginning with #?

Answer 1

Check if following works for you:

grep -oe "[^\'\"\\]#[^\'\"].*" -e "^#.*" somefile >> somecomments

But as one of the comment say, you will have to take care of many exceptions.

What if there is a statement like

echo '<space> ########### Following code does this #########<space>'

You shall have to take all those chances in mind.