First of all, I believe you mean the complement of a regular expression, not it’s inverse. The inverse of a regular expression doesn’t make much sense; but if viewed as a function, I suppose you could say that the inverse of the matcher is the generator which generates all matching strings – or something. On … Read more
Answer to the question: “A Regular Expression to match anything that isn’t “foo” and “bar”?” ^(?!foo$|bar$).* would do exactly that. ^ # Start of string (?! # Assert that it’s impossible to match… foo # foo, followed by $ # end of string | # bar$ # bar, followed by end of string. ) # … Read more
^(?:(?!boon\.ini|http).)*$\r?\n? (taken from RegexBuddy‘s library) will match any line that does not contain boon.ini and/or http. Is that what you wanted?
This should work: ^[^\s]+(\s+[^\s]+)*$ If you want to include character restrictions: ^[-a-zA-Z0-9-()]+(\s+[-a-zA-Z0-9-()]+)*$ Explanation: the starting ^ and ending $ denotes the string. considering the first regex I gave, [^\s]+ means at least one not whitespace and \s+ means at least one white space. Note also that parentheses () groups together the second and third fragments … Read more
Your first two sentences contradict each other. “in \w but is not in \d” includes underscore. I’m assuming from your third sentence that you don’t want underscore. Using a Venn diagram on the back of an envelope helps. Let’s look at what we DON’T want: (1) characters that are not matched by \w (i.e. don’t … Read more
It only will do this if you have the option selected at the bottom: Disallow insecure . and [^…] in /RegExp/ From the docs: true if . and [^…] should not be allowed in RegExp literals. These forms should not be used when validating in secure applications. So the answer your question, if you start … Read more
The other answers here show some ways you can contort different varieties of regex to do this, although I think it does turn out that the answer is, in general, “don’t do that”. Such regular expressions are much harder to read and probably slower to execute than just combining two regular expressions using the boolean … Read more
UPDATE: It fails “with two ff before oo” as @Ciantic pointed out in the comments. ^(f(o[^o]|[^o])|[^f])*$ NOTE: It is much much easier just to negate a match on the client side instead of using the above regex. The regex assumes that each line ends with a newline char if it is not then see C++’s … Read more
But I need the match result to be … not in a match group… For what you are trying to do, this should work. \K resets the starting point of the match. \bObject Name:\s+\K\S+ You can do the same for getting your Security ID matches. \bSecurity ID:\s+\K\S+
The caret inside of a character class [^ ] is the negation operator common to most regular expression implementations (Perl, .NET, Ruby, Javascript, etc). So I’d do it like this: [^\W\s\d] ^ – Matches anything NOT in the character class \W – matches non-word characters (a word character would be defined as a-z, A-Z, 0-9, … Read more