How to extract part of string in bash? Suppose I have a string product_name_v1.2. This name is not fixed as it can be changed. However the one thing that is certain is that there will be a version umber in the string in a format like v#.#.
What I want to do is extract the v1.2 and 1.2 from the string. Any suggestions?
var=product_name_v1.2
#from the beginning strip (#) longest (##) glob ending in v (*v)
echo "${var##*v}
Bash has regexp matching too (with [[ and =~) but basic string manipulation should be enough for your use case.
(You can also check out bash(1) for more information).
In any POSIX shell, this is straightforward:
str=product_name_v1.2
version=${str##*_v}
echo $version
#=> 1.2
The construction ${name##pattern} removes the longest prefix of the contents of the parameter $name that matches the glob pattern, leaving only what comes after it.
Assuming you want to extract digits and dots from the string:
input="product_name_v1.2_more_stuff_here"
echo "${input//[^0-9.]/}"
1.2
There are several standard tools you may use, e.g.:
expr product_name_v1.2 : 'product_name_\(.*\)'expr product_name_v1.2 : '.*_\(v.*\)'echo product_name_v1.2 | sed -ne's/^.*_\(v[0-9.]\+\)$/\1/p'The choice mainly depends on how complex the match needs to be. expr only supports a limited anchored regular expression match, which suffices in the examples above.
