I want to use $var in a shell brace expansion with a range, in bash.
Simply putting {$var1..$var2} doesn't work, so I went "lateral"...
The following works, but it's a bit kludgey.
# remove the split files
echo rm foo.{$ext0..$extN} rm-segments > rm-segments
source rm-segments
Is there a more "normal" way?
You may want to try :
eval rm foo.{$ext0..$extN}
Not sure whether this is the best answer, but it certainly is one.
{$one..$three}, because it is not a valid form of brace-expansion, which in this case expects integers... It only becomes a valid form after the $var expansion, which eval then passes through the same process to generate a normal brace sequence expansion... 1 2 3 QED ;) ... Summary: the simple presence of a brace pair does't trigger brace expansion... only a valid form triggers it.
– Peter.O
Feb 21 '11 at 6:49
a={0..9}; echo $a; there is no brace extension. Using eval , it works. So I think @mattdm's explanation is good.
– dr0i
May 22 '17 at 8:32
As you already realized, {1..3} expands to 1 2 3 but {foo..bar} or {$foo..$bar} don't trigger brace expansion, and the latter is subsequently expanded to replace $foo and $bar by their values.
A fallback on GNU (e.g. non-embedded Linux) is the seq command.
for x in `seq $ext0 $extN`; do rm foo.$x; done
Another possibility. if foo. contains no shell special character, is
rm `seq $ext0 $extN | sed 's/^/foo./'`
The simplest solution is to use zsh, where rm foo.{$ext0..$extN} does what you want.
function f_over_range {
for i in $(eval echo {$1..$2}); do
f $i
done
}
function f {
echo $1
}
f_over_range 0 5
f_over_range 00 05
Notes:
