How to check if string is a comment?

Question

How do I check if string starts with '#' ?

#path /var/dumpfolder

I tried awk for

awk -F '' '{print $1}'

Is there any better way ?

How do you want to check it? Just view? Or do you want messages? — konsolebox, Sep 05 '13 at 22:06
I basically want to check if string is commented if so, I want to remove that comment '#' from string and keep it. Yes a message or storing string without the comment. @Pills : string is in a variable in bash ? — run_time_rookie, Sep 05 '13 at 22:09

Aleks-Daniel Jakimenko-A. · Accepted Answer · 2013-09-05T23:13:30.797

I still cannot understand what are you trying to do, but there is bash in your tags, so here is a bash solution to check if string starts with '#'

string='#path /var/dumpfolder'

if [[ $string == \#* ]]; then
    echo 'yes'
else
    echo 'no'
fi

or:

if [[ ${string:0:1} == '#' ]]; then
    echo 'yes'
else
    echo 'no'
fi

If you want to remove # character then use:

echo "${string#\#}" # prints 'path /var/dumpfolder'

This will remove # character if it exists. If it doesn't exist then the string will be left intact.

Edit:

If sometimes there are leading spaces, then use regexp:

if [[ $string =~ ^[[:space:]]*\#.*$ ]]; then
    echo "${string#*\#}"
else
    echo "$string"
fi

what if there is a space in front of the comment? – progrenhard Sep 05 '13 at 22:48 — progrenhard, Sep 05 '13 at 22:48

score 2 · Answer 2 · edited Feb 08 '17 at 14:46

2

^\s*[#](.*)$

Regular expression visualization

Edit live on Debuggex

here is a regular expression ..... if that's what you are looking for.

^ denotes the start of the string
\s* means a one more spaces to take care of the case where there could be a space in front of the #
[#] or just # works this checks that the first letter is a #
(.*) grabs all(most) of the characters and throws them into a capture group
$ stops at the end of the string

This should work to remove the #'s in your code and spaces in front of the comment

sed 's/\^\s*[#](.*)$/\1/'

edited Feb 08 '17 at 14:46

Community

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answered Sep 05 '13 at 22:08

progrenhard

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I just need to check if commented if so remove comment and extract rest of string. The regex gives me complete string with even '#' included and I want only rest of string – run_time_rookie Sep 05 '13 at 22:26
1

Why ``#`` in ``[]`` ? Seems like there's no need. Also why ``.*`` is in ``( )`` ? ``^\s#.*$`` should work as well. – Aleks-Daniel Jakimenko-A. Sep 05 '13 at 22:36
@Aleks-DanielJakimenko For clarification you're right! I could leave it out. Also about the .*, I like putting stuff I want to match into capture groups just In case later i want to use them for something other than matching.(putting the values into an array) – progrenhard Sep 05 '13 at 22:38

score 1 · Answer 3 · answered Sep 05 '13 at 22:09

If you just want to find #path or path (without #) pattern in file, do:

grep -E '^#?path' file

To just find #path exactly, don't add ?:

grep -E '^#path' file

More exlicit is to add a space, but your separator could be a tab as well:

grep -E '^#path ' file

score 1 · Answer 4 · answered Sep 05 '13 at 22:15

I basically want to check if string is commented if so, I want to remove that comment '#' from string and keep it. Yes a message or storing string without the comment.

so

sed 's/^\s*#*//'

example (with file):

kent$  cat f
#c1
 #c2
foo
bar

kent$  sed 's/^\s*#*//' f
c1
c2
foo
bar

score 0 · Answer 5 · answered Sep 06 '13 at 05:14

0

Using awk

awk '{sub(/^#/,"")}1' file

answered Sep 06 '13 at 05:14

Jotne

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How to check if string is a comment?

5 Answers5

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