Linux, Print all lines in a file, NOT starting with

Question

I would like to print the contents of a file, but all lines starting with # I want to ignore. I was trying some stuff with grep and awk, but it kept printing the whole file, or just printed the lines starting with #. I you could give me a push in the right way, or a grep/awk command that would print anyline in the file that does not start with #.

Solution 1

Use the -v option of grep to negate the condition:

grep -v '^#' file

Solution 2

You can use the ! operator:

awk '!/^ *#/ { print; }'

This negates the result of the match. I also included lines that start with spaces and then #, but you can tailor the regex how you like.

Solution 3

You could use grep to exclude all lines that begin with # using the -v option

grep -v '^#' filename

If you're a fan of sed:

sed '/^#/d' filename

Solution 4

This would also leave out lines with whitespace before the # :

awk '$1!~/^#/' file

or

grep -v '^[[:blank:]]*#' file

Solution 5

Here is the grep PCRE way,

grep -P '^(?!#)' file

View more solutions

Comments

[cmevoli]

Two good examples. But, in the sed example, you don't need the .* part.

[Tuxdude]

@cmevoli - yes point taken, since the presence of the # at the beginning of the line is enough for the delete command :)

[Sigur]

What about if there are leading blank spaces at beginning of line, before the #?

[choroba]

@Sigur: '^[[:space:]]*#'