正则表达式用于多个字符实例

Question

In Java, using a regular expression, how would I check a string to see if it had a correct amount of instances of a character.

For example take the string hello.world.hello:world: . How could this string be checked to see if it contained two instances of a . or two instances of a : ?

I have tried

Pattern p = Pattern.compile("[:]{2}");
Matcher m = p.matcher(hello.world.hello:world:);
m.find();

but that failed.

Edit

First I would like to say thank you for all the answers. I noticed a lot of the answers said something along the lines of "This means: zero or more non-colons, followed by a single colon, followed by zero or more non-colons - matched exactly twice ". So if you were checking for 3 : in a string such as Hello::World: how would you do it?

Answer 1

Well, using matches you could use:

"([^:]*:[^:]*){2}"

This means: "zero or more non-colons, followed by a single colon, followed by zero or more non-colons - matched exactly twice".

Using find is not as good, as there may be additional : and it will just ignore them.

Answer 2

You can use this regex based on two lookaheads assertions:

^(?=(?:[^.]*\.){2}[^.]*$)(?=(?:[^:]*:){2}[^:]*$)

(?=(?:[^.]*\\.){2}[^.]*$) makes sure there are exactly 2 DOTS and (?=(?:[^:]*:){2}[^:]*$) asserts that there are exactly 2 colons in input string.

RegEx Demo

Answer 3

You can determine whether the string has exectly the given number of a certain character, say ':' , by attempting to match it against a pattern of this form:

^(?:[^:]*[:]){2}[^:]*$

That says exactly two non-capturing groups consisting of any number (including zero) of characters other than ':' followed by one colon, with the second group followed by any number of additional characters other than ':' .