使用 Sed 和 Regex 替换字符串

Question

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.

Answer 1

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 .

Answer 2

sed -e 's/^#\s*\(.*[0-9].*\)$/\1/g' filename

应该这样做。

Answer 3

如果您只想取消注释包含数字的那些行，您可以使用以下命令：

sed -e 's/^#\s*\(.*[0-9]+.*\)/\1/g' file

Answer 4

以下sed命令将取消注释包含数字的行：

sed 's/^#\s*\(.*[0-9]\+.*$\)/\1/g' file

Answer 5

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.

Answer 6

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'