The first thing I tried to do, is get the regex matching what I DON'T want. This way, I could just flip it to NOT accept that same input. This is where I came up with the first part of this regex.
Now what I want to do at this point, is combine them together to look for BOTH conditions. However when I do that it seems to break, and only match 9 digit numbers that are identical throughout WITH dashes: "^(\d)\1{2}-(\d)\1{1}-(\d)\1{3}$|^(\d)\1{8}$". This can be seen here: https://regex101.com/r/lPnksf/1.
I may be getting a little ahead of myself here, but in order to show my work as much as possible, I also tried flipping those regex separately, which also did not work as expected.
I would expect the two expressions (when flipped) to match any 9 digit number (with or without dashes) where all numbers are not identical. How ever this does not happen at all.
This is the final regex that I came up with, which is clearly not doing what I would expect it to: "^(?!(\d)\1{2}-(\d)\1{1}-(\d)\1{3})$|^(?!(\d)\1{8})$". Can be seen here: https://regex101.com/r/9eHhF5/1
At the end of the day, I want to combine these 2 expressions, with this one (that already works as intended): "^(?!000|666|9\d\d)\d{3}-(?!00)\d\d-(?!0000)\d\d\d\d$". Can be seen here: https://regex101.com/r/AdRI8i/1.
I am still pretty new to regex, and really want to understand why I can't simply wrap the condition in (?!...) in order to match the opposite condition.
Thank you in advance
What you want to do is not flip, but reverse the regex logic.
Yes, to reverse the pattern logic, you should use a negative lookahead, but there are caveats.
First, the $ end of string anchor: if it was at the end of the "positive" regex, it must also be moved to the lookahead in the reverse pattern. So, your ^(?!(\d)\1{8})$ regex must be written as ^(?!(\d)\1{8}$). Same goes for your second regex.
Next, mind that each subsequent capturing group gets an incremented ID number, so you cannot keep the same backreferences when you "join" patterns with OR | operator. You must adjust these IDs to reflect their new values in the new regex.
So, you want to match a string that matches ^(?!000|666|9\d\d)\d{3}-(?!00)\d\d-(?!0000)\d\d\d\d$ first (let's note \d\d\d\d = \d{4}), then you can add restrictions with lookaheads:
(?!(\d)\1{8}$) - fails the match if, immediately from the current position, it matches identical 9 digits and then the string end comes(?!(\d)\2\2-(\d)\2-(\d)\2{3}$) - (note the ID incrementing continuation) fails the match if, immediately from the current position, it matches identical to the first one 3 digits, -, identical 2 digits, -, identical 5 digits, and then the string end comes.So, to follow your logic, you can use
^(?!(\d)\1{8}$)(?!(\d)\2\2-(\d)\2-(\d)\2{3}$)(?!000|666|9\d\d)\d{3}-(?!00)\d\d-(?!0000)\d{4}$
See the regex demo
As the lookaheads are non-consuming patterns, i.e. the regex index remains at the same position after matching their pattern sequences where it was before, the 3 lookaheads will all be tried at the start of the string (see the ^ anchor). If any of the three negative lookaheads at the start fails, the whole string match will be failed right away.
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