With sed, how can I replace word within a matching line

regular expressionsed

I am trying to write a sed expression to detect lines similar to this:

s0:12345:respawn:/sbin/agetty -8 -s 115200 ttyS0 linux

and replace the ttys0 with something_else

I do not want to detect commented out lines (beginning with #)

I can replace the whole line, but how to I just replace the substring, leaving the rest of the line intact ?

Best Answer

One way to do this is as follows:

echo 's0:12345:respawn:/sbin/agetty -8 -s 115200 ttyS0 linux' | 
    sed -e '/^[^#]/ s/ttyS0/foo/'

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.

Decreasing a number with sed

For that purpose is better use awk so as it support arithmetic operations

cat $1 | awk -v d=$2 '/BookmarkPageNumber:/{$2-=d}1'

Best Answer

Related Solutions

Sed – Replacing Lines Containing a Pattern

OUTPUT

OUTPUT

Decreasing a number with sed

Related Question