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I'm trying to match lines that doesn't end with either . or ! and doesn't end with either ." or !" so it should match both
say "bye"say "byebut shouldn't match:
say "bye.say "bye!say "bye."say "bye!"I tried using positive and negative lookahead, trying to use them as AND as suggested in Regex AND operator, but I can't make it work, nor I'm sure it's feasible with lookaheads.
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Similarly, the negation variant of the character class is defined as "[^ ]" (with ^ within the square braces), it matches a single character which is not in the specified or set of possible characters. For example the regular expression [^abc] matches a single character except a or, b or, c.
In other words, square brackets match exactly one character. (a-z0-9) will match two characters, the first is one of abcdefghijklmnopqrstuvwxyz , the second is one of 0123456789 , just as if the parenthesis weren't there. The () will allow you to read exactly which characters were matched.
3.2. A negative look-ahead, on the other hand, is when you want to find an expression A that does not have an expression B (i.e., the pattern) after it. Its syntax is: A(?!B) . In a way, it is the opposite of a positive look-ahead.
'?' matches/verifies the zero or single occurrence of the group preceding it. Check Mobile number example. Same goes with '*' . It will check zero or more occurrences of group preceding it.
You can use
^(?!.*[.!]"?$).*$
Regex Demo
Note:- This matches empty line too as we use * which means match anything zero or more time, if you want to avoid empty lines to match you can use + quantifier which means match one or more time
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Just use a negative lookbehind.
This matches exactly what you asked for: ^.*+(?<![.!]"?)$
^ - beginning of line.*+ - any amount of characters, not giving up for backtracking(?<! + ) - not preceded by[.!] - dot or exclamation mark"? - optional double-quote$ - end of line
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