问题
preg_match ("/[-_^%&().A-Z0-9]/", $file)
By the looks of it, I can see that the tail end is Anything A-Z and 0-9, but not sure for the beginning. Because ^ is not escaped so it could be a literal "^" or evaluated to a NOT, and same goes with ".", "(" and ")". So I am just a bit confused. Does having things inside the optional [] means things are literalized?
回答1:
It just looks for one of the characters in the character class. I.e. one of:
- Hyphen-minus, underscore, caret, percent, ampersant, parentheses, dot
- Capital Latin letters
- Arabic numerals
^ is only interpreted as negating a character class when at the very beginning of the class. . loses its special meaning inside character classes completely. So yes, in character classes many special regex characters are used literally. With exceptions, although the following list isn't exhaustive:
- Hyphen-minus (
-) indicates a range when between two other characters (i.e. not at the start or end). - The caret (
^) as detailed above. - Predefined character classes (e.g.
\w) can be used as well, which will form a union with the characters from that class. So your character class could be shortened to[-^%&().\w].
回答2:
^ only has a special meaning inside [] if it is the first character.- only has a special meaning inside [] if is between two literal characters.., ( and ) do not have a special meaning inside [].
Just reading the basic information in the docs would tell you all this.
The regex just matches one of the characters in the class.
来源：https://stackoverflow.com/questions/9466768/what-is-this-regular-expression-looking-for