Sed replace only lines with matching groups

sed

Currently, I have this command:

sed 's/\([^C]*\)/\1AB/'

It inserts "AB" before the first instance of "C" on a line. But it is also adding "AB" at the end of every line without a "C".

How can I have it only edit when it finds "C"?

Best Answer

What's wrong with simply replacing the C?

sed 's/C/ABC/'

This will only replace the first instance (if you wanted to replace them all, you would add g at the end).

OUTPUT

NAME_A  12,1
NAME_B  21,2

That addresses lines beginning with a letter, pulls in the next if there is one, and substitutes a tab character for the newline.

note that the s/\n/<tab>/ bit contains a literal tab character here, though some seds might also support the \t escape in its place

To handle a recursive situation you need to make it a little more robust, like this:

sed '$!N;/^[[:alpha:]].*\n[^[:alpha:]]/s/\n/    /;P;D' <<\DATA
NAME_A
NAME_B
12,1  
NAME_C
21,2
DATA

OUTPUT

NAME_A
NAME_B  12,1
NAME_C  21,2

That slides through a data set always one line ahead. If two ^[[:alpha:]] lines occur one after the other, it does not mistakenly replace the newline, as you can see.

Replace only on the first matching line with sed

{ sed -Ee'/[0-9]+/{s///;q;}'; cat; } <in >out

^should work w/ a BSD sed. but apparently it doesn't.

and so:

sed -e'/[0-9][0-9]*/{s///;:b' -e'n;bb' -e\} <in >out

...should work with any of them.

Best Answer

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Sed – Replacing Lines Containing a Pattern

OUTPUT

OUTPUT

Replace only on the first matching line with sed

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