extract part of string using sed

Question

ls lib/oracle-11.2.0.3.0.txt | sed 's/lib.\([oracle.*]\)\.txt/\1/'

It is giving the whole string instead of just oracle part until .txt What am I doing wrong?

I can do it using awk as follows, but, not sure why sed is not giving the wanted result.

echo "lib/oracle-11.2.0.3.0.txt" | awk -F/ '{print substr($2,1,index($0,".txt")-1);}'

Solution 1

[oracle.*] means "one of the characters o, r, a, c, l, e, ., or *". Consequently, your regex will only match something like

lib+c.txt

and not the actual filename you're passing it. If you remove the [ and ] from the regex, then it will work fine:

ls lib/oracle-11.2.0.3.0.txt | sed 's/lib.\(oracle.*\)\.txt/\1/'

However, a much simpler way of doing that is

basename lib/oracle-11.2.0.3.0.txt .txt

or, if you really want the file to come from stdin:

ls lib/oracle-11.2.0.3.0.txt | xargs -I{} basename {} .txt

Solution 2

Here are a few more ways of doing this:

1. Perl

   echo "lib/oracle-11.2.0.3.0.txt" | perl -pe 's/.+(oracle.+)\.txt/$1/'
   

2. sed

   echo "lib/oracle-11.2.0.3.0.txt" | sed 's/.*\(oracle.*\)\.txt/\1/'
   

3. cut

   echo "lib/oracle-11.2.0.3.0.txt" | cut -d'/' -f 2 | cut -d '.' -f 1-5
   

4. basename and bash

   echo "lib/oracle-11.2.0.3.0.txt" | while read n; do 
     echo $(basename ${n/.txt//}); 
   done
   

Solution 3

How about using cut

echo "lib/oracle-11.2.0.3.0.txt" | cut -c5-19