正则表达式 - 也允许空字符串

Question

I have the following regular expression... This one does not allow empty String. How do I have to manipulate?

var phoneReg = new RegExp(/^\+?[0-9]+(\([0-9]+\))?[0-9-]*[0-9]$/i);

Answer 1

Just use the "zero or one" operator ? , such as:

/^(?:regex)?$/
           ^

(?:regex) simply makes the concerned ground uncapturable (ie no need to change your $X , \\X or ?X indexes if any).

Answer 2

I believe that the following would work:

/^$|^\+?[0-9]+(\([0-9]+\))?[0-9-]*[0-9]$/i

I've inserted ^$| at the beginning of your regex. The pipe ( | ) is called an "alternation" indicator and says that either the expression before it or after it must match. ^$ will simply match a string with no characters between the start and end of input. The result: an empty string or the original expression.

Answer 3

You can put a ^$| in front which in which ^ = beginning, $ = end, | = or this other thing to the right.

var phoneReg = new RegExp(/^$|^\+?[0-9]+(\([0-9]+\))?[0-9-]*[0-9]$/i);