How would i delete a line at specific line number

Shell Programming and Scripting
1

Hi guys ,

I m writing a script to delete a line at particular location.

But i m unable to use variable for specifying line number.
for example.

sed -n '7!p' filename

works fine and deletes 7th line from my file
but

sed -n '$variable!p' filename

gives following error.

sed: -e expression #1, char 3: extra characters after command
2

use double quotes instead when using shell variables.

1 Like
3

+
i tried

sed -n "$variable!p" vm.cfg

and

sed -n '"$variable"!p' vm.cfg

None of them are working.
Please help.

4 
[house@leonov] cat in.file
1st line
2nd line
3rd line
[house@leonov] var=2; sed "${var}d" in.file
1st line
3rd line
1 Like
5

Another way:

sed ''${variable}'d' vm.cfg
    ^^
This is two single quotes, not a double!
1 Like
6

You don't really need quotes at all.

$ cat file1
a
b
c
d

$ X=3
$ sed ${X}d file1
a
b
d
1 Like
7

Thanks for your input . But now i would like to enter a line at specific line number .

How would i establish this using sed.

8

Like this

# cat file
12
13
14
15

# sed '/12/d' file
13
14
15
9

Like this?

sed '4s/.*/this is the new line\n&/' infile
one
two
three
this is the new line
four
five
10

Does it have to be sed?

# cat in
1
2
3
4

# export n=3
# export s="aaa"
# awk -vn=$n -vs=$s 'NR==n{print s}1' in
1
2
aaa
3
4
11

What do you mean by "enter a line"? Insert a new line, or?

$ cat file1
a
b
c
d

$ sed '3 i\
new line goes here' file1
a
b
new line goes here
c
d

# or
$ sed '3 a\
new line goes here' file1
a
b
c
new line goes here
d
12

insert variable $line at line 3:

sed "3s/^/$line\n/" infile
13

Hi ,

#!/usr/bin/perl

while (<>) {
if ($. == $linenum ) { print "test line \n"; }
print $_;
}

OR

#!/bin/sh

linecount=`cat test1.txt | wc -l`
linetoinsert="test line to insert"
linenumber=4

linecount_head=`expr $linenumber - 1`
linecount_tail=`expr $linecount - $linenumber + 1`

head -$linecount_head test1.txt;echo $linetoinsert;tail -$linecount_tail test1.txt