Bad delimiter in Bash shell

Question

I'm trying to iterate over few lines of bash script and echo all the non comment lines.

Within my loop I have the following command:

echo $line | cut -d" #" -f1

However the interpreter throws at me the following error:

cut: bad delimiter

What I did wrong?

possible duplicate of [Is it possible to use a string as a delimiter in unix cut command?](http://stackoverflow.com/questions/857286/is-it-possible-to-use-a-string-as-a-delimiter-in-unix-cut-command) — , Aug 15 '14 at 00:14
The `-d` option appears to only allow a single character as the delimiter. — , Aug 15 '14 at 00:15
It's not clear what you intend `-d" #"` to do. `#` introduces a comment in bash, but it needn't be preceded by a space. — Keith Thompson, Aug 15 '14 at 00:38
@KeithThompson: Actually it does need to be, mostly. "a word beginning with # causes that word and all remaining characters on that line to be ignored." So `(#This is a comment` -- because `(` is self-delimiting -- but `[#This is not a comment`. (Of course, the fact that a # is preceded by a space doesn't prove it to be a comment. It could be in a quoted word.) — rici, Aug 15 '14 at 03:28

Warwick · Accepted Answer · 2014-08-15T01:04:02.437

4

As mentioned in the comments, the -d parameter to cut expects only one character. try using nawk instead for your example:

echo $line | nawk 'BEGIN {FS=" #" } ; { print $1 }'

Or to just print lines that don't begin with a " #", use grep:

grep -v " #" <file>

Or to print only lines that begin with a hash, or just white space and a hash, I'd use Perl:

perl -n -e 'unless (/(\S+)*#/) {print}' <file>

edited Aug 15 '14 at 01:04

answered Aug 15 '14 at 00:22

Warwick

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2
11

Any particular reason to use `nawk` rather than `awk`? – Keith Thompson Aug 15 '14 at 00:37
What if the line starts with `#` or tab character? – John B Aug 15 '14 at 00:38
@Keith On my Solaris 10 system, awk did not produce the desired output. nawk did. The OP does not specify the system they are using. – Warwick Aug 15 '14 at 00:51
@John - Agreed, but from the example the OP gave, their lines appear to start with a space followed by a hash. But I'll change my answer to give that option too. – Warwick Aug 15 '14 at 01:02
@Warwick: Good point -- but I think Solaris is unusual in still having a very old version of `awk` as the default. – Keith Thompson Aug 15 '14 at 01:29
@KeithThompson Solaris `/usr/bin/awk` is broken and `/usr/xpg4/bin/awk` should be used instead if GNU `awk` or `nawk` isn't available. – jaypal singh Aug 15 '14 at 03:11

John B · Answer 2 · 2014-08-15T00:36:42.483

You could use awk:

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

Only prints lines that do not start with a # in the first field.

If you want to test if a string is a comment:

if [[ ! $(awk '$1 !~ /^#/' <<<"$string") ]]; then
    echo "'$string' is a comment"
else
    echo "'$string' is not a comment"
fi

Or another way using bash only:

if [[ ! ${string%\#*} || ${string%\#*} == +([[:space:]]) ]]; then
    echo "'$string' is a comment"
else
    echo "'$string' is not a comment"
fi

score 0 · Answer 3 · answered Aug 15 '14 at 20:51

Just for completeness, using sed:

sed '/^\S*#/d' file

This essentially says output all lines that do not start with any whitespace followed by a #. This is a modern version of sed, so older ones you might have to use something else for the whitespace match.

Bad delimiter in Bash shell

3 Answers3