CODEWITHSUNDEEP
bashvariablessedreplace
Michael Gruenstaeudl
Michael Gruenstaeudl

Reputation: 1783

Replacing string with variable in bash when variable starts with special characters

Replacing a string with the contents of a variable using sed by enclosing the search expression with double (") instead of single (') quotes is well documented.

$ astring="Liftoff in [sec]"
$ for s in 3 2 1; do echo $astring | sed -e "s/\[sec\]/$s/"; done
Liftoff in 3
Liftoff in 2
Liftoff in 1

However, how do I conduct the above replacement if the variable content starts with special characters? For example, the variable contents could start with a period forwardslash (./), which is often the case if local file paths are passed as variables?

for s in ./3 ./2 ./1; do echo $astring | sed -e "s/\[sec\]/$s/"; done
sed: -e expression #1, char 14: unknown option to `s'
sed: -e expression #1, char 14: unknown option to `s'
sed: -e expression #1, char 14: unknown option to `s'

Upvotes: 0

Views: 95

Answers (2)

rici
rici

Reputation: 241741

For this particular use case, there is no need to use sed, and the native bash solution is simpler, more efficient, and less fiddly (because you don't need to worry about delimiters or other special characters in the replacement string):

$ for s in ./3 ./2 ./1; do echo "${astring/\[sec\]/$s}"; done
Liftoff in ./3
Liftoff in ./2
Liftoff in ./1

See the list of bash parameter expansion syntaxes in the Bash manual.

Upvotes: 0

Gilles Quénot
Gilles Quénot

Reputation: 185171

You just have to use another delimiter : 

for s in ./3 ./2 ./1; do echo "$s" | sed -e "s|\[sec\]|$s|"; done

to avoid conflict with your input

Upvotes: 1

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