How to echo a variable containing an unescaped dollar sign in bash

问题

If I have a variable containing an unescaped dollar sign, is there any way I can echo the entire contents of the variable?

For example something calls a script:

./script.sh "test1$test2"

and then if I want to use the parameter it gets "truncated" like so:

echo ${1}
test1

Of course single-quoting the varaible name doesn't help. I can't figure out how to quote it so that I can at least escape the dollar sign myself once the script recieves the parameter.

回答1:

The variable is replaced before the script is run.

./script.sh 'test1$test2'

回答2:

The problem is that script receives "test1" in the first place and it cannot possibly know that there was a reference to an empty (undeclared) variable. You have to escape the $ before passing it to the script, like this:

./script.sh "test1\$test2"

Or use single quotes ' like this:

./script.sh 'test1$test2'

In which case bash will not expand variables from that parameter string.

回答3:

by using single quotes , meta characters like $ will retain its literal value. If double quotes are used, variable names will get interpolated.

回答4:

As Ignacio told you, the variable is replaced, so your scripts gets ./script.sh test1 as values for $0 and $1.

But even in the case you had used literal quotes to pass the argument, you shoudl always quote "$1" in your echo "${1}". This is a good practice.

来源：https://stackoverflow.com/questions/3923623/how-to-echo-a-variable-containing-an-unescaped-dollar-sign-in-bash