Shell Programming and Scripting

Hi gurus

input file:

Code:

1
2
3
4

desired output

Code:

1
2\
	2a
3
4

I tried

Code:

sed '2 s/$/\\\n\t2a/' files

everything works OK

but if i want to use variable instead of fixed line (2nd in this case), i have to replace ' witch " because of replacing bash variables, so I tried.

Code:

var=2
sed "$var s/$/\\\n\t2a/" files

but returns me:

Code:

1
2\n	2a
3
4
5

Could someone please explain it ? and post correct solution ? Thanks

Try:

Code:

sed "$var"'s/$/\\\n\t2a/' infile

Try one of these commands:

Code:

sed "$var"'s/$/\\\n\t2a/'

or

Code:

sed "$var s/$/\\\\\n\t2a/"

The shell interpretes one of the backslash as an escape character if you use double quotes.
Try to echo the sed command to see what happens:

Code:

$ var=2
$ echo "$var s/$/\\\n\t2a/"
2 s/$/\\n\t2a/          <- missing one backslash

and now with the above commands:

Code:

$ echo "$var s/$/\\\\\n\t2a/"
2 s/$/\\\n\t2a/
$
$ echo $var's/$/\\\n\t2a/'
2s/$/\\\n\t2a/

Thanks both, now it is clear

---------- Post updated at 09:28 PM ---------- Previous update was at 06:59 PM ----------

Just one question

those 2 command give the same output (I added -e to echo for interpreting special characters)

Code:

echo -e "s/$/\\n/"
echo -e "s/$/\n/"

why in the first is not new line escaped ? If I want this I have to type like

Code:

echo -e "s/$/\\\n/"

Only thing which gives me sence is if first \ escapes second, so second \ is written and it escapes \n

But also this gives me the same result as previous and this gives me no sence

Code:

echo -e "s/$/\\\\n/"

I assumed first escapes second (so far second \ is written because is escaped by first \), written \ (the second) escapes third \ (so third is also written) which escapes \n.
so my expected result is: s/$/\\\n/ and NOT s/$/\n/
Can someone explain that ?

I am asking because I have traouble write something like this into the script

Code:

sed "$res""s/$/\\\\\n"$name"\\\\\n"$name"\\\/

if i type it directly to bash it runs with no problems

Thanks a lot

please how to write this sed to bash script ?

What are the contents of the variables $res and $name?
Give a sample of some lines of the input file and the desired output with the command above.

Hi $res is number of line e.g 4
and $name is username e.g tux123

Now it finnaly works I have mistake somewhere in quotes in script, also could you please explain those escaping sequences ?

I am a little confusing with escaping (maybe it is for another post, cause it is bash and not sed)

why following code gives the same result ? I assuming that in 1st case first \ escapes \n so output should be first\nsecond
but it is
first
second

Code:

1. echo -e "first\\nsecond"
2. echo -e "first\nsecond"

also folowing gives the same result

Code:

echo -e "s/$/\\\n/"
echo -e "s/$/\\\\n/"

Also I was trying eval, but did not work, why ? It is the same string echoed and then exected.

Code:

eval `echo -e "cat grp | sed "$res""s/$/\\\\\n\t\t(-,"$name",time2change.co.uk)\\\\\n\t\t(-,"$name",time2loose.co.uk)/""`

Could you please explain that or point me to some literature ? Thanks