CODEWITHSUNDEEP
bashsedenvironment-variables
RSFalcon7
RSFalcon7

Reputation: 2311

How to select the nth line of a file from a variable in bash?

Possible duplicate: Bash tool to get nth line from a file

I need to select the nth line a file, this line is defined be the variable PBS_ARRAYID

The accept solution in the another question (link above) is:

sed 'NUMq;d' job_params

I'm trying to adapt for the variable like (actually I try lots of stuff, but is the one that makes more sense):

sed "${PBS_ARRAYID}q;d" job_params

But I get the following error:

sed: -e expression #1, char 2: invalid usage of line address 0

What am I doing wrong?

Upvotes: 2

Views: 1975

Answers (2)

RSFalcon7
RSFalcon7

Reputation: 2311

Your solution is correct:

sed "${PBS_ARRAYID}q;d" job_params

The only problem is that sed considers the first line to be line 1 (thanks rici), so PBS_ARRAYID must be in range [1,X], where X is the number of lines on the input file, or:

wc -l job_params

Upvotes: 1

Jotne
Jotne

Reputation: 41446

Here is an awk example.

Lets say we have this file:

cat file
1 one
2 two
3 three
4 four
5 five
6 six
7 seven
8 eight
9 nine

Then we have theses variable:
var="four"
number=2

Then this awk gives:

awk '$0~v {f=NR} f && f+n==NR' v="$var" n="$number" file
6 six

Upvotes: 0

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