regExp - use of \D* before \d{2}?

[MSPa1nT]

Hej!

I’m struggling a bit to understand the regExp used in the lookahead challange when we need to match a password with greater than 5 digits and at least 2 numbers.

/(?=\w{6,})(?=\D*\d{2})/

I dont understand why we use the \D* before \d{2}.
as per my understanding we use the \D* to match anything that is not a digit, 0 or more times. Why is this needed before the \d{2} ?

[gaac510]

let regex1 = /\w{6,}\D*\d{2}/;
let regex2 = /(?=\w{6,})(?=\D*\d{2})/;

I’ll try to explain this by comparing how regex1 and regex2 work differently.

regex1

• regex1 doesn’t start with /^, so it could start matching at any string position (think zero-based indices).
• Say regex1 started matching at string index 2 (i.e. 3rd character of the string), if successful with the \w{6,} part, it would have matched at least 6 alphanumeric characters (including underscore). At this point, the smallest index we could be at is 8, and regex1 would start look for zero or more non-digits followed by exactly two digits.
• All in all, regex1 will match a string that’s at least 8 characters long. The string would have 6 alphanumeric characters appearing before 2 digits, and the two groups may be separated by zero or more alphanumeric characters followed by zero or more non-digits.

regex2

• Similar to regex1, regex2 doesn’t start with /^, so it could start matching at any string index.
• Again, say regex2 started matching at string index 2, if successful with the (?=\w{6,}) part, it would have matched at least 6 alphanumeric characters, although this time, we wouldn’t be at or after index 8. We’d be back at index 2 because assertions aren’t matched. Starting again at index 2, regex2 would proceed to check the (?=\D*\d{2}) part, which denotes 2 digits preceded by zero or more non-digits.
• It may already become obvious to you that, if \D* was omitted, regex2 would be checking whether the character at precisely index 2 and index 3 were digits.
• All in all, regex2 will match a string that’s at least 6 characters long. The string would have 6 alphanumeric characters, themselves including at least one of two consecutive digits or appearing before such digits.

[azrul.naut]

So regex2 could start matching at any string position as it does not start with /^.

But without the \D*, the second lookahead would only be checking at precisely index 2 instead?

Is this because it is the second lookahead? I understand why it works with the \D*, but I don’t see why it doesn’t work without.

[gaac510]

@azrul.naut

So regex2 could start matching at any string position as it does not start with /^ .

yes

@azrul.naut

But without the \D* , the second lookahead would only be checking at precisely index 2 instead?

yes. in the example index 2 and 3, because we assumed regex2 started matching at index 2, as well as the {2} part.

@azrul.naut

Is this because it is the second lookahead?

Assertions assert but don’t match, and remember, once a regex starts checking at a certain index, it only moves on to the next character if the current character is matched.

[azrul.naut]

@gaac510

once a regex starts checking at a certain index, it only moves on to the next character if the current character is matched

Ah this explains it very clearly to me.

Thank you so much for taking the time to explain in detail. I really appreciate it.

[gaac510]

No worries. Regex can be really confusing when you first learn about it. When I was learning it a lot of my brain cells suffered while I was trying to wrap my head around some of the rules, lookahead/lookbehind being the worst of the bunch.