This regex passes in fiddle but doesn't pass here

JavaScript
1

This post may…be in the wrong place but this code also works, I’m showing it so maybe admin can pass it in future? I don’t see why it SHOULDN’T pass :wink:

Your code so far


let username = "JackOfAllTrades";
let userCheck =/^[a-z]\w+[0-9]*$|^[a-z]\w+\[0-9][0-9]+$/i // Change this line
//(\w+)
///^\w\w+[0-9]*$|^\w\w+\[0-9][0-9]+$/i (this line also passes in fiddle)
let result = userCheck.test(username);

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/93.0.4577.82 Safari/537.36

Challenge: Restrict Possible Usernames

Link to the challenge:

2

The \w+ in your regex is causing issues with the matching. Remember that \w matches [a-zA-Z0-9_] so you may end up matching more than you expected. Let’s break it down using examples that shouldn’t pass:

D9 - invalid because two-character usernames can’t end with a digit.
Notv4lid0 - invalid because digits can only be at the end.

And we’ll only need to look at the first part of your regex:

^[a-z]\w+[0-9]*$

In the case of username D9, it is considered valid by the first part of your regex, which checks that:

  • The first character is [a-z] (D)
  • There are one or more characters that match [a-zA-Z0-9_] (9)
  • And there are zero or more characters that match [0-9] at the end (true, there’s nothing)

In the case of the username Notv4lid0, the first part of the regex is also the problem:

  • The first character is [a-z] (N)
  • There are one or more characters that match [a-zA-Z0-9_] (otv4lid0, greedy matching)
  • And there are zero or more characters that match [0-9] at the end (true, there’s nothing left).

If you replace the \w+ in the first part of your regex, you should be able to remove the part after the | completely and pass the exercise.

3

You may have seen the solution in the guide, but here’s what I came up with that works in the problem:

Answer 
/^[a-z]([a-z]+|[0-9][0-9]+)[0-9]*$/i

or a shorter equivalent

/^[a-z]([a-z]+|[0-9]{2,})[0-9]*$/i
Explanation 
^[a-z]

Match a single a-z character at the beginning of the string, as we know there will be one.

([a-z]+|[0-9][0-9]+)
([a-z]+|[0-9]{2,})

Next comes either one or more a-z characters, with greedy matching, or two or more numerals (which comes up in the exercise).

[0-9]*$

Then, there are zero or more numerals at the end.

4

Your solution shouldn’t work though…because the question involved has an “option” so you’d need an “or” operator here

5

Yeah, I see what you mean here…one would NEED the [a-z] with the i, although…one could also use the \D (for non-digits, in place of [a-z])

6

That’s what this is for:

([a-z]+|[0-9]{2,})

The “or” is in the middle.

7

You can also always check your regex on a site like regexr or regex101. Use the strings from the tests. They also give a rundown of what is being matched.

1 Like
8

My apologies, I saw it (or) when I looked again, didn’t see before, my bad…

9

Yeah, I know about these, I just used fiddle because it works with js that’s all :wink:

10

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