BASH scripting, proper quoting of variable inside sed?

Question

This variable, Btr="/home/BB/PL/DDr/"; is successfully echoed so I know it should be available, but none of these sed work properly.

Variants of syntax I have tried, none of which work.

1.

    Lg=$(echo "${Prg}" | sed 's/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/'$Btr'\2/');

2.

    Lg=$(echo "${Prg}" | sed 's/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/'${Btr}'\2/');

3.

    Lg=$(echo "${Prg}" | sed 's/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/"$Btr"\2/');

4.

    Lg=$(echo "${Prg}" | sed 's/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/"${Btr}"\2/');

5.

    Lg=$(echo "${Prg}" | sed 's/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/'"$Btr"'\2/');

6.

    Lg=$(echo "${Prg}" | sed 's/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/"'$Btr'"\2/');

7.

    Lg=$(echo "${Prg}" | sed -e "s/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/'${Btr}'\2/");

8.

    Lg=$(echo "${Prg}" | sed -e 's/\(\/home\/in\/PL\/\)\(.*_Data.txt\)$/'${Btr}'\2/');

Objective is to change a line like this one:

| tee -a /home/in/PL/SomeFile_Data.txt

to this:

| tee -a /home/BB/PL/DDr/SomeFile_Data.txt

Why use / as a delimiter when your pattern contains lots of slashes already? That's asking for trouble. I'd suggest you use something else, say, comma. It will make the sed invokations much more readable. — fstd
– fstd, Commented Apr 22, 2014 at 0:02
@fstd some things don't seem to work well when you don't use slashes, though. Such as @PATTERN@d doesn't seem to work. — Alexej Magura
– Alexej Magura, Commented Apr 22, 2014 at 0:05
@AlexejMagura that's be \@PATTERN@d (note the escaping of the first @). Slash seesm to be special in that the escape can be omitted, but generally sed expects it to be there. — fstd
– fstd, Commented Apr 22, 2014 at 0:12

Alexej Magura · Accepted Answer · 2014-04-22 00:07:05Z

1

Without double quotes:

a=1
echo 'b' | sed 's/b/'$a'/'

With double quotes:

a=1
echo 'b' | sed "s/b/$a/"

With single quoted pattern and double quoted variable:

a=1
echo 'b' | sed 's/b/'"$a"'/'

edited Apr 22, 2014 at 0:07

answered Apr 22, 2014 at 0:01

Alexej Magura

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3 Comments

MountainMan Over a year ago

Tried that also;sed: -e expression #1, char x: unknown option to `s'

Alexej Magura Over a year ago

works fine for me. What version of sed are you using? NOTE some things may need to be escaped if you're using double quotes!

MountainMan Over a year ago

Ah, yes. One must be wary of that. But you guys got me sorted on this one, thx again. GNU bash, version 4.2.45(1)-release (i686-pc-linux-gnu)

Brad Lanam · Accepted Answer · 2014-04-22 00:08:26Z

0

@fstd's answer. Change the delimiter.

  Lg=$(echo ${Prg} | sed 's,/home/in/PL/\(.*_Data\.txt\)$,'$Btr'\1,')

answered Apr 22, 2014 at 0:08

Brad Lanam

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2 Comments

MountainMan Over a year ago

Whoa, why would choice of separator style affect functionality? Hunh. Anyhow, with double quotes around the whole sed, no quotes around the variable and using comma as changed seperator, this does the job. Tried to show it in here but these comments mangle code ;(

Alexej Magura Over a year ago

@MountainMan it only really helps when working with a pattern that contains the delimiter you're using, such as: sed 's/\/var\/log//' as sed 's_/var/log__'

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BASH scripting, proper quoting of variable inside sed?

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Without double quotes:

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With single quoted pattern and double quoted variable:

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With double quotes:

With single quoted pattern and double quoted variable:

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