Regular expression to count number of commas in a string

Question

How can I build a regular expression that will match a string of any length containing any characters but which must contain 21 commas?

Answer 1

if exactly 21:

/^[^,]*(,[^,]*){21}$/

if at least 21:

/(,[^,]*){21}/

However, I would suggest don't use regex for such simple task. Because it's slow.

Answer 2

Might be faster and more understandable to iterate through the string, count the number of commas found and then compare it to 21.

Answer 3

^(?:[^,]*)(?:,[^,]*){21}$

Answer 4

/^([^,]*,){21}[^,]*$/

That is:

^     Start of string
(     Start of group
[^,]* Any character except comma, zero or more times
,     A comma
){21} End and repeat the group 21 times
[^,]* Any character except comma, zero or more times again
$     End of string

Answer 5

var valid = ((" " + input + " ").split(",").length == 22);

or...

var valid = 21 == (function(input){
    var ret = 0;
    for (var i=0; i<input.length; i++)
        if (input.substr(i,1) == ",")
            ret++;
    return ret
})();

Will perform better than...

var valid = (/^([^,]*,){21}[^,]*$/).test(input);

Answer 6

If you're using a regex variety that supports the Possessive quantifier (e.g. Java), you can do:

^(?:[^,]*+,){21}[^,]*+$

The Possessive quantifier can be better performance than a Greedy quantifier.

Explanation:

(?x)    # enables comments, so this whole block can be used in a regex.
^       # start of string
(?:     # start non-capturing group
[^,]*+  # as many non-commas as possible, but none required
,       # a comma
)       # end non-capturing group
{21}    # 21 of previous entity (i.e. the group)
[^,]*+  # as many non-commas as possible, but none required
$       # end of string