Need regular expression for sed

Question

I am trying to create a script that will amnog other things enable source repositories in sources.list (Ubuntu). As you may or may not know the format of the repository file is something like this:

deb http://archive.ubuntu.com/ubuntu/ xenial-updates main restricted
# deb-src http://archive.ubuntu.com/ubuntu/ xenial-updates main restricted
# etc.

So I would like to enable source repository (uncomment lines with deb-src) but only for those preceded by the uncommented deb line. GNU sed should be perfect for this but I'm not able to make this work. This is what I have right now:

sed -i 's/^# deb-src/deb-src/gm' /etc/apt/sources.list

What I would like to have is something like this but it should replace only the commented line not everything:

sed -i 's/^deb .*$^# deb-src/deb-src/gm' /etc/apt/sources.list

So is there a way to make regex for sed that will only replace a part of matched string?

In the default sources.list there are several entries where both deb and deb-src lines are commented. In those cases I want them to remain commented. — Darko Miletic, Feb 22 '18 at 11:11
Have you had a look at https://stackoverflow.com/questions/1251999/how-can-i-replace-a-newline-n-using-sed? — Biffen, Feb 22 '18 at 11:13
@DarkoMiletic You'll need to match the newline somehow. E.g. `s/^deb (.*)# deb-src \1/…/g` — Biffen, Feb 22 '18 at 11:17
In the replace mode sed replaces COMPLETE matched string with supplied one. So if I have 's/^deb .*$^# deb-src/deb-src/gm' it will replace the complete deb line and deb-src line with uncommented deb-src — Darko Miletic, Feb 22 '18 at 11:18

score 3 · Accepted Answer · answered Feb 22 '18 at 11:37

3

Use -z in addition to -r:

sed -rz -i 's/^(deb .*\s+)^# deb-src/\1deb-src/gm' /etc/apt/sources.list

Capturing groups are necessary in order to refer to them in replacement.

answered Feb 22 '18 at 11:37

revo

47,783
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That's it :) Thanks – Darko Miletic Feb 22 '18 at 11:40

Sundeep · Answer 2 · 2018-02-22T11:34:36.943

I personally prefer awk/perl for matching two consecutive lines

awk '/^# *deb-src/ && p ~ /^deb /{sub(/^# */, "")} {p=$0} 1'

p=$0 saves previous line
/^# *deb-src/ && p ~ /^deb / if current line starts with comment and has deb-src and previous line starts with deb
sub(/^# */, "") remove comment
1 idiomatic way to print current record, including any changes made
See also: awk save modifications in place

With sed, you might need something like

sed '/^deb /{n; /^# *deb-src/ s/# *//p; D}'

or

sed '$!N; /^deb .*\n# *deb-src/ s/\n# */\n/p; D'

but I'm not very sure if this would handle all cases. See https://www.gnu.org/software/sed/manual/sed.html#advanced-sed for details. The m modifier comes into play only if you have multiple lines in pattern space. By default, sed uses newline as line separator and only has single line in pattern space. The manual link has more details

Thanks for this detailed response but I prefer the answer provided by user @revo. — Darko Miletic, Feb 22 '18 at 11:45

Zumo de Vidrio · Answer 3 · 2018-02-22T11:33:26.070

If your sources.list is only composed of lines like:

deb http://................
deb-src http://............

# deb http://...........
# deb-src http://.........

deb http://................
# deb-src http://............

And you don't care of removing the commented lines, you can create dinamically a new sources.list file by modifying only the lines after the ones starting by deb that are not commented:

grep -A1 "^deb " sources.list | sed "s/# deb-src/deb-src/g"

With this you grep only the lines starting by ^deb and the following one, and then you uncomment the next line (the one starting by # deb-src)

You can then send the output to a new file which can be renamed by sources.list afterwards:

grep -A1 "^deb " sources.list | sed "s/# deb-src/deb-src/g" > sources_new && mv -f sources_new sources.list

But this will only work if you have this exact syntax in your sources.list file.

Need regular expression for sed

3 Answers3