bash - grep for expression containing variable

Question

I want to add this command grep -co '5' $INFILE in a bash script. The problem is that instead of 5 i want to use a variable in its place so i write:

V=`grep -co '$L' $INFILE`

but it does not work since the $ is used to describe internally the end of line in grep. How can i make it work? Is there an escape sequence for $ to make it use its bash meaning of the value of a variable?

Answer 1

Use " instead of ' to allow expansion of your $L variable.