grep for lines not containing #

UNIX for Advanced & Expert Users
1

I'm trying to grep for lines not containing a #. I'm not getting the syntax right. This is what I have tried so far.

grep -v \\# /home/bob/.config/vlc/vlcrc
grep -v '\\#' /home/bob/.config/vlc/vlcrc
grep -v "\\\\#" /home/bob/.config/vlc/vlcrc

I tried to look at this thread of mine for ideas with no luck.

2

If your just ignoring the lines with # .. Then try the below..

$ grep -v "#" infile
3

That didn't work :(.

4

Jajan's code worked flawless for me. Do you mind posting a helpful output of the error you got? Having 316 posts in this forum, you should be familiar giving detailed answers to solve your problems, thanks.

5

can you post the output of

 
cat -v /home/bob/.config/vlc/vlcrc
6

My problem is I didn't get an error. I got a ton of blank lines. The file has 4.5k of lines. I'm not sure what the preferred method is to deal with huge files.

Here is 100 of the lines.

# Video filter module (string)
#video-filter=

# Video output filter module (string)
#vout-filter=

# Enable sub-pictures (boolean)
#spu=1

# On Screen Display (boolean)
#osd=1

# Text rendering module (string)
#text-renderer=

# Use subtitle file (string)
#sub-file=

# Autodetect subtitle files (boolean)
#sub-autodetect-file=1

# Subtitle autodetection fuzziness (integer)
#sub-autodetect-fuzzy=3

# Subtitle autodetection paths (string)
#sub-autodetect-path=./Subtitles, ./subtitles

# Force subtitle position (integer)
#sub-margin=0

# Subpictures filter module (string)
#sub-filter=

# Program (integer)
#program=0

# Programs (string)
#programs=

# Audio track (integer)
#audio-track=-1

# Subtitles track (integer)
#sub-track=-1

# Audio language (string)
#audio-language=

# Subtitle language (string)
#sub-language=

# Audio track ID (integer)
#audio-track-id=-1

# Subtitles track ID (integer)
#sub-track-id=-1

# Input repetitions (integer)
#input-repeat=0

# Start time (float)
#start-time=0.000000

# Stop time (float)
#stop-time=0.000000

# Run time (float)
#run-time=0.000000

# Fast seek (boolean)
#input-fast-seek=0

# Playback speed (float)
#rate=1.000000

# Input list (string)
#input-list=

# Input slave (experimental) (string)
#input-slave=

# Bookmarks list for a stream (string)
#bookmarks=

# DVD device (string)
#dvd=/dev/dvd

# VCD device (string)
#vcd=/dev/cdrom

# Audio CD device (string)
#cd-audio=/dev/cdrom

# UDP port (integer)
#server-port=1234

# MTU of the network interface (integer)
#mtu=1400

# Force IPv6 (boolean)
#ipv6=0
7

Try:

awk 'NF && $1!~/^#/' infile

or

grep -vE '^[[:space:]]*($|#)' infile

--
To only exclude comment lines that have # as the first character (lines that have whitespace before the # are printed), as well as empty lines

awk 'NF && !/^#/' infile
8

That's better than

Thanks.

9

I'm no expert with grep or regular expressions in general (started working with them a couple days ago), but wouldn't using this pattern in grep work?

^[^\r\n#]+[\r\n]

I haven't used grep from the terminal, but my text editor can use grep for search, and this matches all lines without # and ignores those with a #.

EDIT:

To match blank lines, change the + to a *

^[^\r\n#]*[\r\n]
10

Carriage returns (\r) are not part of a regular Unix file and a linefeed (\n) cannot be matched by regular grep, since grep is line based. + is part of an extended regular expression (ERE) so you would need to use grep -E

11

Ok. That's good to know. I suppose the end of line anchor ($) will work, then.

This seems to work on a tiny test file.

grep -e '^[^#]*$' testhash.txt

EDIT:

This is my file contents:

line1 has some text
line2 is a second line
li#e3 has a hash!
line4 the above line had a hash!
line# has a hash too
line6 - ^ had a hash.. bad line?
#ine7 hashline
line8 lastly the last line
#

line11 single hash and a blank line above

And this is the output:

line1 has some text
line2 is a second line
line4 the above line had a hash!
line6 - ^ had a hash.. bad line?
line8 lastly the last line

line11 single hash and a blank line above
12

But that would also filter out lines that use the # symbol somewhere on the line either as a partial comment or not as a comment. It also does not filter out empty lines..

To compare:

$ cat infile
# This is a test
echo hello # partial test

echo the line above contains tabs and spaces

echo the line above is empty
  echo "just want to print an #i character" ffsf
$ grep -e '^[^#]*$' infile

echo the line above contains tabs and spaces

echo the line above is empty
$ awk 'NF && $1!~/^#/' infile
echo hello # partial test
echo the line above contains tabs and spaces
echo the line above is empty
  echo "just want to print an #i character" ffsf
$
13

Ok, as that was my intention, I suppose I was misunderstanding what was desired.