How to use bash variables in sed command?

It is important to understand that bash and sed are two completely independent things. When you give bash a command, it first processes it according to its rules, in order to come up with a utility name and a set of arguments for that utility (in this case sed), and then calls the utility with the arguments.

Probably $DIR contains a slash character. Perhaps it looks something like /usr/home/codyline/src.

So when bash substitutes that into the argument to the sed command:

"s/${DIR}//g"

the result is

s//usr/home/codyline/src//g

which is what is then passed to sed. But sed can't understand that commabnd: it has (many) too many / characters.

If you really want to use sed for this purpose, you need to use a delimiter other than /, and it needs to be a character you are confident will never appear in $DIR. Fortunately, the sed s command allows you to use any character as a delimiter: whatever character follows the s is used as the delimiter. But there always must be exactly three of them in the command.

For example, you might believe that no directory path contains a colon (:), in which case you could use:

sed "s:${DIR}::g"

Of course, someday that will fail precisely because you have a directory with a colon in its name. So you could make things more general by using bash's substitute-and-replace feature to backslash-escape all the colons:

sed "s:${DIR//:/\:}::g"

But you could have used this bash feature in order to avoid the use of sed altogether:

NEWBASE=${NAME//$DIR}

Unfortunately, you can't nest bash substitute-and-replaces, so you need to do them sequentially:

NEWBASE=${NEWBASE//.\/}

Note: I used ${var//...}, which is the equivalent of specifying the g flag in a sed s command, but I really don't know if it is appropriate. Do you really expect multiple instances of $DIR in a single path? If there are multiple instances, do you really want to remove all of them? You'll have to decide.