Bash command variable assignment [duplicate]

Question

If I assign a variable:

testThis='echo "This is a test"'

If I use $testThis, it works in a script.

But, what if I want to skip a line ? So I try:

testThis='echo; echo "This is a test"'

and this fails!

Can't figure it out after much effort trying $() command substitution and all sorts of quoting.

Don't use variables to store commands. Use functions. See: BashFAQ/050: I'm trying to put a command in a variable, but the complex cases always fail! — John Kugelman
– John Kugelman, Commented Jan 18, 2022 at 14:58

chepner · Accepted Answer · 2022-01-18 15:01:39Z

2

A parameter expansion is not reparsed. The following

testThis='echo; echo "This is a test"'
$testThis

is equivalent to

echo ';' echo \"This is a test\"

The semicolon and the double quotes are both literal parts off the string, not shell syntax. After word-splitting and quote removal, the shell identifies the command echo with six arguments:

;
echo
"This
is
a
test"

Use a function instead:

testThis () {
    echo; echo "This is a test"
}

testThis

answered Jan 18, 2022 at 15:01

chepner

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