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Usually when my regex patterns look like this:
http://www.microsoft.com/ Then i have to escape it like this:
string.match(/http:\/\/www\.microsoft\.com\//) Is there another way instead of escaping it like that?
I want to be able to just use it like this http://www.microsoft.com, cause I don't want to escape all the special characters in all my patterns.
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In order to use a literal ^ at the start or a literal $ at the end of a regex, the character must be escaped. Some flavors only use ^ and $ as metacharacters when they are at the start or end of the regex respectively. In those flavors, no additional escaping is necessary. It's usually just best to escape them anyway.
The \ is known as the escape code, which restore the original literal meaning of the following character. Similarly, * , + , ? (occurrence indicators), ^ , $ (position anchors) have special meaning in regex. You need to use an escape code to match with these characters.
Escape CharactersUse the backslash character to escape a single character or symbol. Only the character immediately following the backslash is escaped. Note: If you use braces to escape an individual character within a word, the character is escaped, but the word is broken into three tokens.
To match any of the metacharacters literally, one needs to escape these characters using a backslash ( \ ) to suppress their special meaning. Similarly, ^ and $ are anchors that are also considered regex metacharacters.
Regexp.new(Regexp.quote('http://www.microsoft.com/')) Regexp.quote simply escapes any characters that have special regexp meaning; it takes and returns a string. Note that . is also special. After quoting, you can append to the regexp as needed before passing to the constructor. A simple example:
Regexp.new(Regexp.quote('http://www.microsoft.com/') + '(.*)') This adds a capturing group for the rest of the path.
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