正则表达式 - 匹配除+之外的任何字符，空字符串也应匹配

Question

I am having a bit of trouble with one part of a regular expression that will be used in JavaScript. I need a way to match any character other than the + character, an empty string should also match.

[^+] is almost what I want except it does not match an empty string. I have tried [^+]* thinking: "any character other than + , zero or more times", but this matches everything including + .

Answer 1

• [^+] means "match any single character that is not a + "
• [^+]* means "match any number of characters that are not a + " - which almost seems like what I think you want, except that it will match zero characters if the first character (or even all of the characters) are + .

use anchors to make sure that the expression validates the ENTIRE STRING:

^[^+]*$

means:

^       # assert at the beginning of the string
[^+]*   # any character that is not '+', zero or more times
$       # assert at the end of the string

Answer 2

Add a {0,1} to it so that it will only match zero or one times, no more no less:

[^+]{0,1}

Or, as FailedDev pointed out, ? works too:

[^+]?

As expected, testing with Chrome's JavaScript console shows no match for "+" but does match other characters:

x = "+"
y = "A"

x.match(/[^+]{0,1}/)
[""]

y.match(/[^+]{0,1}/)
["A"]

x.match(/[^+]?/)
[""]

y.match(/[^+]?/)
["A"]

Answer 3

If you're just testing the string to see if it doesn't contain a + , then you should use:

^[^+]*$

This will match only if the ENTIRE string has no + .