How to globally replace strings in lines NOT starting with a certain pattern

Question

I want to globally replace the string foo with the string bar, using sed. This should only be done for lines which do NOT start with the string ##Input.

I can't get it to work. I tried things like this but reached a point where I'm not sure if I know what I'm doing:

sed -i '/^##Input/ s/foo/bar/g' myfile

Please help!

score 91 · Accepted Answer · answered Feb 10 '11 at 05:48

91

You just need to negate the match using !:

sed -i '/^##Input/! s/foo/bar/g' myfile

answered Feb 10 '11 at 05:48

Dennis Williamson

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This is so useful! I didn't realize you could give sed an input that matches the line before you run the search and replace on that line. This helps me immensely! – bballdave025 Jun 07 '18 at 21:56
1

@bballdave025: See the [addresses](https://www.gnu.org/software/sed/manual/sed.html#sed-addresses) section of the GNU `sed` manual to learn more. – Dennis Williamson Jun 08 '18 at 04:07

score -5 · Answer 2 · answered Feb 10 '11 at 05:46

-5

You got to escape # as in \#.

answered Feb 10 '11 at 05:46

vpit3833

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There's no need to escape the `#`. It has no special meaning to `sed`. – Dennis Williamson Feb 10 '11 at 05:49

score -9 · Answer 3 · answered Aug 01 '16 at 16:40

-9

An ugly answer for an ugly request (i.e. they get what they asked for):

echo \{
for file in *.json; do
    sed -n '/^[\{\}]/! s/\([^\,]\)$/\1,/; /^[\{\}]/!p' $file
done
echo \{

answered Aug 01 '16 at 16:40

Keith Tyler

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How to globally replace strings in lines NOT starting with a certain pattern

3 Answers3

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