how can i get the string between two specific characters, only in bash [with out using grep or sed]
e.g
input=hostname~web:sit
I want to extract web from the above input
${hostname#*~} gives me output as web:sit, but i need only web in the output
From the other post i can see
% strips from end of $var, up to pattern. But not sure how to apply it.
Any help please
Do it in two steps:
input=hostname~web:sit
rightpart=${input#*~} # remove prefix up to "~" (included)
output=${rightpart%:*} # remove suffix from ":" (included)
echo $output
Using extglob in bash, you can do this in single step:
shopt -s extglob
input='hostname~web:sit'
echo "${input//@(*~|:*)/}"
web
Here @(*~|:*) matches a substring from start to ~ character OR a substring from : to end. Using // we replace all such instances with an empty string.
There is a sed solution as well:
sed -E 's/.*~([^:]+):.*/\1/' <<< "$input"
web
Another option, using cut wich is available from the GNU Core Utilities:
~:input='hostname~web:sit'
echo "$input" | cut -d '~' -f2 | cut -d ':' -f1
# web