Sort text in shell using last word before specific delimiter

Question

How do I sort a text file alphabetically in shell using the last word before a certain delimiter as key?

Eg the following file:

Manfred1 Mustermann1 (Berlin)
Manfred Siegfried Müller (Aachen)
Gerd A. Meier (Stuttgart)

In this case the delimiter would be " (" and I would like to sort using the last name of the persons in the list. The number of words before the last name is varying.

Given a line of text I know how to extract the key:

s="Manfred1 Mustermann1 (Berlin)"
w=${s%% (*}
l=${w##* }
echo $l

This displays "Mustermann1", but I did not get any further.

EDIT Using l0b0's answer, this is the complete code that did it:

while read s; do
w=${s%% (*}
l=${w##* }
printf "%s\t%s\n" "$l" "$s"
done <names.txt | sort -k 1 | cut  -f 2 > names_sorted.txt

where names.txt contains the text and the sorted list will be printed to names_sorted.txt

Answer 1

Once you've got the sort key ( $l ) you can proceed like this:

1. Print the sort key followed by a tab character followed by the original line, as in printf %s\\t%s\\n .
2. Use sort 's --key option to sort by the first column (tab should be the default key separator).
3. Remove the sort key and tab character by using awk or grep .