Extract range of lines using sed

Question

I have defined two variables as follows:

var1=$(unzip -c ./*.zip | grep -n "Channel8"| cut -f1 -d":")
var2=$(unzip -c ./*.zip | grep -n "Channel10"| cut -f1 -d":")

I have a very big file and I would like to extract the range of lines between $var1 and $var2 using sed. I am trying the following

sed -n '/"$var1","$var"2p' $(unzip -c ./*.zip)

But with no success. Could you give an explanation why and how to fix it? Thanks.

Within single quotes, nothing is expanded and you keep literal `"$var1"` in your command instead of what they expand to. — Benjamin W., Jul 20 '18 at 20:15
Possible duplicate of [shell variables in sed script](https://stackoverflow.com/questions/7006910/shell-variables-in-sed-script) — Benjamin W., Jul 20 '18 at 20:17
Or https://stackoverflow.com/questions/3306007/replace-a-string-in-shell-script-using-a-variable — Benjamin W., Jul 20 '18 at 20:19

Barmar · Answer 1 · 2018-07-20T21:03:18.480

3

Variables aren't expanded inside single quotes. Also, you need to pipe the output of unzip to sed, not use it as command-line arguments.

unzip -c ./*.zip | sed -n "${var1},${var2}p"

But it seems like you're doing this the hard way, reading the zip file 3 times. Just use the pattern you want to match as the range:

unzip -c ./*.zip | sed -n '/^extracting:.*Channel8/,/^extracting:.*Channel10/p'

edited Jul 20 '18 at 21:03

answered Jul 20 '18 at 20:20

Barmar

Why are you using :/,/ in between sed? – JPV Jul 20 '18 at 20:21
@JPV A sed address can be a line number or a pattern. If it's a pattern, it goes between two slashes. – Benjamin W. Jul 20 '18 at 20:24
1

This being said, I don't think the colons should be here, they come from `grep -n` in the original question. – Benjamin W. Jul 20 '18 at 20:25
I did it in the hard way because maybe I need to select 2 lined up from var1 and 4 lines up from var2. So I want to modify my variables – JPV Jul 20 '18 at 20:26
@BenjaminW. No, they come from `unzip -c`. It displays `filename:` – Barmar Jul 20 '18 at 20:39
My `unzip -c` doesn't display any filenames followed by a colon. – Benjamin W. Jul 20 '18 at 20:56
You're right, I misread the output. I saw `Archive:`, I thought that was a filename. – Barmar Jul 20 '18 at 21:03

score 2 · Accepted Answer · answered Jul 20 '18 at 20:17

2

You can use:

unzip -c ./*.zip | sed -n "$var1,$var2 p"

Fixes are:

answered Jul 20 '18 at 20:17

anubhava

Why the character p is with white space? – JPV Jul 20 '18 at 20:23
3

If there is no space before `p` then shell will attempt to expand `var2p` without any command and will generate syntax error – anubhava Jul 20 '18 at 20:25

Poshi · Answer 3 · 2018-07-20T20:29:42.447

0

Use double quotes to expand the vars:

sed -n "${var1},${var2}p" $(unzip -c ./*.zip)

edited Jul 20 '18 at 20:29

answered Jul 20 '18 at 20:17

Poshi

3 Answers3