问题
I try to find a regex that matches the string only if the string does not end with at least three '0' or more. Intuitively, I tried:
.*[^0]{3,}$
But this does not match when there one or two zeroes at the end of the string.
回答1:
If you have to do it without lookbehind assertions (i. e. in JavaScript):
^(?:.{0,2}|.*(?!000).{3})$
Otherwise, use hsz's answer.
Explanation:
^ # Start of string
(?: # Either match...
.{0,2} # a string of up to two characters
| # or
.* # any string
(?!000) # (unless followed by three zeroes)
.{3} # followed by three characters
) # End of alternation
$ # End of string
回答2:
You can try using a negative look-behind, i.e.:
(?<!000)$
Tests:
Test Target String Matches
1 654153640 Yes
2 5646549800 Yes
3 848461158000 No
4 84681840000 No
5 35450008748 Yes
Please keep in mind that negative look-behinds aren't supported in every language, however.
回答3:
What wrong with the no-look-behind, more general-purpose ^(.(?!.*0{3,}$))*$?
The general pattern is ^(.(?!.* + not-ending-with-pattern + $))*$. You don't have to reverse engineer the state machine like Tim's answer does; you just insert the pattern you don't want to match at the end.
回答4:
This is one of those things that RegExes aren't that great at, because the string isn't very regular (whatever that means). The only way I could come up with was to give it every possibility.
.*[^0]..$|.*.[^0].$|.*..[^0]$
which simplifies to
.*([^0]|[^0].|[^0]..)$
That's fine if you only want strings not ending in three 0s, but strings not ending in ten 0s would be long. But thankfully, this string is a bit more regular than some of these sorts of combinations, and you can simplify it further.
.*[^0].{0,2}$
来源：https://stackoverflow.com/questions/11431295/regex-to-match-string-not-ending-with-pattern