Problem about [\\w#@'] +

Quote from teacher's notes:

[^\w#@']+ (Matches one or more character that is not in word based characters, #, @ and apostrophe)

Please check resources on the web... And play around... It comes with experience... I will try my best, but you have to dig on your own, trying to apply everywhere:

OK. Here is my best explanation

I. Brackets [] mean symbol. Take example from https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html

When you write [a] it means only one symbol 'a' will work, more symbols or other symbols will not work, please try to play on the website that Craig suggests online Regex Tester.

When you write [ab] - it means it can be 'a' or 'b' -only one symbol

So coming back to Regex you want: Take a look at [] without PLUS symbol. With brackets we define a symbol

Lets look inside brackets:

^ character means except. Examples:

[^a] means any character but not 'a', so 'b', 'c', 'd' and all symbols that are not 'a' will work

Lets get back to [^\w]

\w is the same as [a-zA-Z_0-9] which means any character from 'a' to 'z' or 'A' to 'Z' and from 0 to 9.

So [^\w] means not 'a', 'z', not 'A-Z', not 0-9. It could be '+', '-', '=', and others ...

When we write:

[^\w#@] we write characters that want to exclude, which means not 'a-z', not 'A-Z', not 0-9, not '#' and not '@'

Now we come to plus at the end, plus is nothing else but one or more times, which means that can be combinations of the symbols that we don't want.

Craig wants to exclude punctuation signs, which means that he wants to exclude '?', '=', ',' and others... and with PLUS multiple combinations of them, like many trailing spaces...

I still strongly suggest to watch workshop:

https://teamtreehouse.com/library/regular-expressions-in-java

And read documentation:

https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html

It comes with experience. Try to play with online regex tester