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- Users: Christian Feard
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bash script bad substitution error message
revised and simplified version using the new -n declaration : unset a b c x declare -A a b ; for i in k{1..2} ; do a[a$i]=a_val_$i b[b$i]=b_val_$i ; done ; echo -e "${a[@]}\n${b[@]}" declare -n x declare -a c=( a b ) for x in "${c[@]}" ; do for k in "${!x[@]}" ; do echo -n "$k = " ; echo...- Christian Feard
- Post #3
- Forum: UNIX Scripting
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bash script bad substitution error message
Well, I'm new to tek-tips (I've just registered) and I see your question has still no answer. I'm pretty sure you have the answer since, but anyway, here is mine: parameter expansion does not accept nesting (you can't do ${${var}}. You can have an indirection with '!' ( ${!var} ), but in the...- Christian Feard
- Post #2
- Forum: UNIX Scripting
