Extracting part of a string bounded by special symbols - bash

Question

Hello I am passing strings for example /bin/bash/Xorg.tar.gz to my script which is
for i in $*; do
echo "$(expr match "$i" '\.*\.')"
done
I expect to return Xorg only but it returns 0,any ideas why?

Answer 1

It seems weird that your string would be /bin/bash/Xorg.tar.gz (kinda looks like /bin/bash is a directory or something) but either way, you can use standard parameter expansion to get the part you want:
i=${i##*/}
i=${i%%.*}
First remove everything up to the last /, then remove everything from the first ..

Answer 2

expr match directive attempts to match complete input not partial.
However, you can use builtin BASH regex for this:
[[ "$i" =~ .*/([^./]+)\. ]] && echo "${BASH_REMATCH[1]}"
This will print Xorg for your example argument.

Answer 3

The immediate fix (leaving the loop aside):
$ expr '/path/to/Xorg.tar.gz' : '.*/\([^.]*\)'
Xorg
Note:
: is needed after the input string to signal a regex-matching operation.
Note: expr <string> : <regex> is the POSIX-compliant syntax; GNU expr also accepts expr match <string> <regex>, as in your attempt.
expr implicitly matches from the start of the string, so .*/ must be used to match everything up to the last /
\([^.]*\) is used to match everything up to, but not including, the first . of the filename component; note the \-escaping of the ( and ) (the capture group delimiters), which is needed, because expr only supports (the obsolescent and limited) BREs.
Using a capture group ensures that the matched string is output, whereas by default the count of matching chars. is output.
As for the regex you used:
'\.*\.': \.* matches any (possibly empty) sequence (*) of literal . chars. (\.), implicitly at the start of the string, followed by exactly 1 literal . (\.).
In other words: you tried to match 2 or more consecutive . chars. at the start of the string, which is obviously not what you intended.
Because your regex doesn't contain a capture group, expr outputs the count of matching characters, which in this case is 0, since nothing matches.
That said, calling an external utility in every iteration of a shell loop is inefficient, so consider:
Tom Fenech's helpful answer, which only uses shell parameter expansions.
anubhava's helpful answer, which only uses Bash's built-in regex-matching operator, =~
If you don't actually need a shell loop and are fine with processing all paths with a single command using external utilities, consider this:
basename -a "$#" | cut -d'.' -f1
Note: basename -a, for processing multiple filename operands, is nonstandard, but both GNU and BSD/macOS basename support it.
To demonstrate it in action:
# Set positional parameters with `set`.
$ set -- '/path/to/Xorg.tar.gz' '/path/to/another/File.with.multiple.suffixes'
$ basename -a "$#" | cut -d'.' -f1
Xorg
File