Issue
foo=username
bar=foo
a=$(eval echo \$$bar) # same as ${!bar}
echo $(eval echo \$${a^^}) # expected val
How to get the value of foo through the variable bar， Same as the above output
I know it wants a variable name not a string here.
echo $(eval echo \$${${!bar}^^}) # error bad substitution
Solution
I believe this is what you want to start with:
foo=username
bar=$foo
... where username is the literal string, you want to put in the variable, named foo, and $foo is the value of the variable foo, which you want to put in the variable, named bar.
If you want to create another variable, let's say a, which contains the value of bar, you just do:
a=$bar
Answered By - Dominique Answer Checked By - Marie Seifert (WPSolving Admin)