CODEWITHSUNDEEP
bashtextescapingecho
phil
phil

Reputation: 3

in bash linux how to echo escaped text as output of a command

I want to print a text example 

 '(]\{&}$"\n

the escaped version I have is this:

"'(]\\{&}$\"\\n"

I tried the following:

cat $CLIPBOARD_HISTORY_FILE | sed "$2!d" |  sed 's/^.\(.*\).$/\1/'
cat $CLIPBOARD_HISTORY_FILE | sed "$2!d" |  sed 's/^.\(.*\).$/\1/'  | eval 'stdin=$(cat); echo  "$stdin"'
VAR1=$(cat $CLIPBOARD_HISTORY_FILE | sed "$2!d" |  sed 's/^.\(.*\).$/\1/')
VAR2="'(]\\{&}\$\"\\n"
VAR3=$VAR1
echo "1  '(]\\{&}\$\"\\n"
echo "2  $VAR1"
echo "3  $VAR2"
echo "4  $VAR3"
echo -e "5  $VAR1"
echo -e "6  $VAR2"
echo -e "7  $VAR3"


$ 
'(]\\{&}$\"\\n
'(]\\{&}$\"\\n
1  '(]\{&}$"\n
2  '(]\\{&}$\"\\n
3  '(]\{&}$"\n
4  '(]\\{&}$\"\\n
5  '(]\{&}$\"\n
6  '(]\{&}$"

7  '(]\{&}$\"\n

echoing the text directly works, but not if it comes from a command.... what am I not seeing or understanding? Thanks for the help!

Upvotes: 0

Views: 108

Answers (1)

Jonathan Leffler
Jonathan Leffler

Reputation: 753870

In general, it's best to enclose material in single quotes rather than double quotes; then you only have to worry about single quotes. Thus:

$ x="'"'(]\{&}$"\n'
$ printf "%s\n" "$x"
'(]\{&}$"\n
$ printf "%s\n" "$x" | sed -e "s/'/'\\\\''/g" -e "s/^/'/" -e "s/$/'/"
''\''(]\{&}$"\n'
$

The use of printf is important; it doesn't futz with its data, unlike echo. The '\'' sequence is crucial; it stops the current single quoted string, outputs a single quote and then restarts the single quoted string. That output is 'sub-optimal'; the initial '' could be left out (and similarly the final '' could be left out if the data ends with a single quote):

$ printf "%s\n" "$x" | sed -e "s/'/'\\\\''/g" -e "s/^/'/" -e "s/$/'/" -e "s/^''//" -e "s/''$//"
\''(]\{&}$"\n'
$

If you really must have double quotes around the data, rather than single quotes, then you have to escape more ($`\" need protection), but the concept is similar.

Upvotes: 1

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