Please explain the following syntax:
i##*.
i%.*
I understand what it\'s doing, but I want to know the general pattern (why/how is i
Look for the section "Substring removal" in this manual:
${string##substring}
Deletes longest match of
$substring
from front of$string
.${string%substring}
Deletes shortest match of
$substring
from back of$string
.
See Parameter Expansion in man bash
:
${parameter#word} ${parameter##word}
Remove matching prefix pattern. The word is expanded to produce a pattern just as in pathname expansion. If the pattern matches the beginning of the value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the
#
case) or the longest matching pattern (the##
case) deleted. If parameter is @ or *, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.${parameter%word} ${parameter%%word}
Remove matching suffix pattern. The word is expanded to produce a pattern just as in pathname expansion. If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the
%
case) or the longest matching pattern (the%%
case) deleted. If parameter is @ or *, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with @ or *, the pattern removal operation is applied to each mem- ber of the array in turn, and the expansion is the resultant list.
In short, #
removes the pattern from the left, %
from the right, doubling the symbol makes the matching greedy. (Mnemonic: #
is to the left of %
on most keyboards).