Bash读取文件名并使用awk返回版本号

Question

I am trying to use one or two lines of Bash (that can be run in a command line) to read a folder-name and return the version inside of the name.

So if I have myfolder_v1.0.13 I know that I can use echo "myfolder_v1.0.13" | awk -F"v" '{ print $2 }' echo "myfolder_v1.0.13" | awk -F"v" '{ print $2 }' and it will return with 1.0.13 .

But how do I get the shell to read the folder name and pipe with the awk command to give me the same result without using echo ? I suppose I could always navigate to the directory and translate the output of pwd into a variable somehow?

Thanks in advance.

Edit: As soon as I asked I figured it out. I can use

result=${PWD##*/}; echo $result | awk -F"v" '{ print $2 }'

and it gives me what I want. I will leave this question up for others to reference unless someone wants me to take it down.

Answer 1

But you don't need an Awk at all, here just use bash parameter expansion.

string="myfolder_v1.0.13"
printf "%s\n" "${string##*v}"
1.0.13

Answer 2

You can use

basename "$(cd "foldername" ; pwd )" | awk -Fv '{print $2}'

to get the shell to give you the directory name, but if you really want to use the shell, you could also avoid the use of awk completetly: Assuming you have the path to the folder with the version number in the parameter "FOLDERNAME":

echo "${FOLDERNAME##*v}"

This removes the longest prefix matching the glob expression "*v" in the value of the parameter FOLDERNAME.