On 2020-07-08 04:47, Mark Hansen wrote:
You got us reading three posts, so we're fully invested in your issue,
and then
you don't share the answer? That's just cruel :)
I'm invested in the amusement of the whole procession, but all the while
I'm thinking that command substitution inside a for loop is basic
shell syntax from POSIX:
$ for x in $(echo a b c) ; do echo $x ; done
a
b
c
so it's either a PEBKAC or some obscure version-specific Zsh problem
nobody cares
about any more.
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