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Size: 1625
Comment: ${x##*([$' \t\n'])} is wrong -- you don't want to match zero or more of those characters. You want +, for one or more.
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Size: 1623
Comment: Remove GNUism (sed s/x\+//). Ordinary Kleene closure (*) is perfectly adequate here. (Likewise in the extglob part, but I left that alone.)
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| x=$(echo "$x" | sed -e 's/^[[:space:]]\+//' -e 's/[[:space:]]\+$//') | x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//') |
How can I trim leading/trailing white space from one of my variables?
There are a few ways to do this -- none of them elegant.
First, the most portable way would be to use sed:
x=$(echo "$x" | sed -e 's/^ *//' -e 's/ *$//') # Note: this only removes spaces. For tabs too: x=$(echo "$x" | sed -e $'s/^[ \t]*//' -e $'s/[ \t]*$//') # Or possibly, with some systems: x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
One can achieve the goal using builtins, although at the moment I'm not sure which shells the following syntax supports:
# Remove leading whitespace:
while [[ $x = [$' \t\n']* ]]; do x=${x#[$' \t\n']}; done
# And now trailing:
while [[ $x = *[$' \t\n'] ]]; do x=${x%[$' \t\n']}; doneOf course, the preceding example is pretty slow, because it removes one character at a time, in a loop (although it's good enough in practice for most purposes). If you want something a bit fancier, there's a bash-only solution using extglob:
shopt -s extglob
x=${x##+([$' \t\n'])}; x=${x%%+([$' \t\n'])}
shopt -u extglobRather than specify each type of space character yourself, you can use character classes. Two character classes that are useful for matching whitespace are space and blank.
More info: ctype/wctype(3), re_format/regex(7), isspace(3).
shopt -s extglob
x=${x##+([[:space:]])}; x=${x%%+([[:space:]])}
shopt -u extglobThere are many, many other ways to do this. These are not necessarily the most efficient, but they're known to work.