I'm trying to uncomment file content using sed but with regex (for example: [0-9]{1,5})
# one two 12
# three four 34
# five six 56
The following is working:
sed -e 's/# one two 12/one two 12/g' /file
However, what I would like is to use regex pattern to replace all matches without entering numbers but keep the numbers in the result.
For complying sample question, simply
sed 's/^# //' file
will suffice, but if there is a need to remove the comment only on some lines containing a particular regex, then you could use conditionnal address:
sed '/regex/s/^# //' file
So every lines containing regex will be uncomented (if line begin with a #)
... where regex could be [0-9] as:
sed '/[0-9]/s/^# //' file
will remove # at begin of every lines containing a number, or
sed '/[0-9]/s/^# \?//' file
to make first space not needed: #one two 12, or even
sed '/[0-9]$/s/^# //' file
will remove # at begin of lines containing a number as last character. Then
sed '/12$/s/^# //' file
will remove # at begin of lines ended by 12. Or
sed '/\b\(two\|three\)\b/s/^# //' file
will remove # at begin of lines containing word two or three.
#"; you want to "remove the # on lines that contain digits".) For regex, you could say [0-9] to have it only affect lines with digits in them, or [0-9]$ if the number has to be at the end.
sed -e 's/^#\s*\(.*[0-9].*\)$/\1/g' filename
should do it.
Is the -i option for replacement in the respective file not necessary? I get to remove leading # by using the following:
sed -i "s/^# \(.*\)/\1/g" file
In order to uncomment only those commented lines that end on a sequence of at least one digit, I'd use it like this:
sed -i "s/^# \(.*[[:digit:]]\+$\)/\1/g" file
This solution requires commented lines to begin with one space character (right behind the #), but that should be easy to adjust if not applicable.
If you only want those lines uncommented which contain numbers, you can use this:
sed -e 's/^#\s*\(.*[0-9]+.*\)/\1/g' file
The following sed command will uncomment lines containing numbers:
sed 's/^#\s*\(.*[0-9]\+.*$\)/\1/g' file
I find it. thanks to all of you
echo "# one two 12" | grep "[0-9]" | sed 's/# //g'
or
cat file | grep "[0-9]" | sed 's/# //g'
sed. I have made a significant change tot he Highest score answer (edits in review) and ask if @starryknight64 please review and choose an answer.