Echo a string with a variable in it without expanding/evaluating it [closed]

Question

I've searched everywhere. Tried echo and print. Tried single and double quotes. But I have parsed data and assigned it to a variable and would like to then evaluate it for if there is a variable within it. I will then replace the variable with a wildcard and search for the file.

Example:

var="file.$DATE.txt"

### Where it goes wrong-  Needs to identify that $DATE is within the $var varaible.
test=$(echo "$var"|grep '\$')
if [[ $test ]]
then
    ### I would use whatever fix is discovered here as well
    test=$(echo $test|sed 's/\$[a-zA-Z]*/\*/')
fi

### (Actually pulling from remote machine to local)
cat $test > /tmp/temporary.file

Here is at least one of my many failures:

PROMPT> file=blah.$DATE
PROMPT> test=$(echo "$file"|grep '\$')
PROMPT> echo $test
PROMPT>
PROMPT>

I know it has something to do with expansion, but have no idea how to work it out. Any help would be appreciated. Thanks!

Answer 1

If you need $date inside the variable var:

var='file.$date.txt'

That will keep the $ inside the variable:

$ echo "$var" | grep '\$'
file.$date.txt

Answer 2

Use single quotes around variables to prevent shell expansion. Example echo '$file' will not expand $file.

Edit after comment below:

You can escape the $ sign int the var variable with var="file.\$DATE.txt".