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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!

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3 Answers 3

Reset to default
101

You just need to negate the match using !:

sed -i '/^##Input/! s/foo/bar/g' myfile
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  • 4
    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
    Commented Jun 7, 2018 at 21:56
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    @bballdave025: See the addresses section of the GNU sed manual to learn more.
    – Dennis Williamson
    Commented Jun 8, 2018 at 4:07
-5

You got to escape # as in \#.

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    There's no need to escape the #. It has no special meaning to sed.
    – Dennis Williamson
    Commented Feb 10, 2011 at 5:49
-10

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 \{
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  • I fail to see how it's an 'ugly request' but your comment might be called that. Also it doesn't really answer the question. I mean okay it might be understandable what that means but to someone asking the question do you really think they'll understand that sed invocation? And who said anything about a json file anyway ?
    – Pryftan
    Commented Oct 18, 2022 at 9:25