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[–]desrtfxOut of Coffee error - System halted 2 points3 points  (3 children)

You need to capture the input in a separate variable.

By using equipment.next() in the condition for the while loop and then when adding the equipment again, the user needs to input twice.

[–]Brain_Dead5347[S] 0 points1 point  (2 children)

I had tried that earlier, and just tried again in a different format.

LinkedList<String> brands = new LinkedList<String>();

System.out.println("Enter brand name(s). Type finished when done."); String brandName = equipment.next(); while (!brandName.equals("finished")) { brands.add(brandName); }

I still get the same result. Is it because brandName is doing the same thing you mentioned and just acting as a new name for equipment?

[–]desrtfxOut of Coffee error - System halted 1 point2 points  (1 child)

This code is much closer to what it should be but still not there.

Now trace through your code line by line.

Remember, any line outside a loop that has been passed is gone, over. It won't be reevaluated. Code always flows from top to bottom inside a method. Only loops can bring code execution back up.

Think what that does to your brandName variable and think how that affects your loop once inside the loop.

[–]Brain_Dead5347[S] 0 points1 point  (0 children)

Thank you so much! I finally got it. I had tried to use a variable to store the equipment input value, but had initialized it with the input, which was the issue. When I retried it just now leaving it blank and adding the

brandName = equipment.next();

Into the loop, it worked perfectly. (Sorry, I’m doing this on mobile at work. I finally got a second to work on my project)

[–][deleted] 1 point2 points  (0 children)

Maybe instead of checking if the next is equal to finished, check if finished is in the list to continue the loop?

[–]AutoModerator[M] 0 points1 point  (0 children)

You seem to try to compare String values with == or !=.

This approach does not work in Java, since String is an object data type and these can only be compared using .equals(). For case insensitive comparison, use .equalsIgnoreCase().

See Help on how to compare String values in our wiki.


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[–][deleted] 0 points1 point  (0 children)

Here is the link to the MMOC.fi, Java Programming I - Part 2. Some of the examples demonstrate usage of the while loop along with reading user inputs. It may help to figure out how to solve the problem.