Extract substring from ls

Question

I'm creating a little script and I have to list all my Tomcat binaries.
So far I was able to do this:
ls -1 | grep '\-tomcat\-' | cut -f3 -d'-'

This basically lists all the versions, but it adds the .zip or .tar.gz

5.5.17.zip
5.5.26.tar.gz
5.5.27.tar.gz
5.5.28.tar.gz
5.5.31.tar.gz
5.5.32.tar.gz

I would like to know how to remove the .zip and .tar.gz from the extracted strings.

Answer 1

Or simplify the whole approach:

ls apache-tomcat*|sed -r 's/^.*-([0-9.]+)\..*/\1/'

Less tools, and it gives you the version numbers.

P.S.: Following up on @Nemo's suggestion: we let shell globbing and prior knowledge take care of half the job (just list things that actually look like apache-tomcat). When piping ls' output to another tool the -1 is moot, so we get rid of that. sed takes the values coming from ls, matches beginning of line to the first - followed by a digit, the parenthesis remember all digits & literal periods, and then we match the rest of the string till end of line (implicit). And then the whole match gets replaced with the remembered digits & periods.

Answer 2

Pipe it through another cut:

ls -1 | grep '-tomcat-' | cut -f3 -d'-' | cut -f1-3 -d'.'

This will work as long as the versions all have three components. If the version is just 5.5, it won't work.

Another option would be just to use sed:

ls -1 | grep '-tomcat-' | cut -f3 -d'-' | sed 's/.tar.gz\|.zip//'

This will remove .tar.gz or .zip from the strings.