Bash Variable from Command Line

I am having trouble populating a variable with a command line. I am sure I am formatting it incorrectly, but I am not sure how. I want to take a list of URLs, something like:

http://blah.com/test
http://blah.com/test2

and print out:
test
test2

I am using the following, but getting an error "./test.sh: line 3: =: command not found"

#!/bin/bash
while read LINE; do
   $NAME = `echo $LINE | cut -d \/ -f4`
   echo $NAME
done < url-list.txt

Any ideas on how I formatted the $NAME variable wrong?
stakorAsked:
Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

simon3270Commented:
You mustn't put the $ at the start when setting the value of NAME, and you mustn't have spaces round the = sign, so

   NAME=`echo $LINE | cut -d \/ -f4`
0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
simon3270Commented:
There are a couple of other things to consider. Firstly you don't need the backslash in front of the forward slash - you only need to do that if / is a "special character" in the location you are using it, i.e. one which has a particular effect.  For the delimiter in "cut", "/" is just a normal character like others, so just use "-d / -f4".  It isn't actually an error here to use the backslash, but it makes the code harder to read, and it may cause problems if the command gets more complex.

The other is the format of your input lines. Will they always be like that, with just one name after the domain, or might they be longer, such as http://example.com/images/test3 - in that case, do you want to print "image", or "test3", or "image/test3" - all are possible!
0
stakorAuthor Commented:
Awesome, thank you.
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Shell Scripting

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.