Regex 101 Discussion I9 - Count the number of matches.

Regex 101 Exercise I9 - Count the number of matches Given a string like: # # 4 6 # # 7 # 45 # 43 # 65 56 2 # 4345 # # 23   Count how many numbers there are in this string ----- There are a few ways to approach this problem. In all of them, we need a way to capture the numeric parts of the string. We've done that before, using something like: \d+ We can then apply that repeatedly, with code that looks something like this: Regex regex = new Regex("\d+"); Match match = regex.Match(inputString);int count = 0; while (match.Success){ count++; match = match.NextMatch();} That's a bit ugly, however. There's a shortcut, however, using: MatchCollection matches = regex.Matches(inputString);int count = matches.Count; That gives the same result. There's another, more "advanced" approach we can use. The regex is more complex, and the code that you write is harder to understand, so I probably wouldn't prefer it over the last approach. To use it, you need to know a new piece of information about the Match class (where "new" means "something I haven't talked about"). In earlier examples, we used the "Groups" indexer to pull out values that we had captured. So, if we wrote something like: (?<Digits>\d+) We would use: match.Groups["Digits"].Value to get the string. It is possible to write a regex in which a given capture is used more that one time, and therefore corresponds to multiple strings. If we write: ( # Start up repeating section (?<Digits>\d+) # a sequence of digits (\D+|$) # not digits or end )+ # repeat match We have a single regex that will match a repeating series of digits follow by non-digits, and each match is stored using the "Digits" capture. To get at these captures, we use: match.Captures["Digits"] which is a collection of captures, with each one containing the value from one of the captures. To solve our problem, we'd be interested in the length, which is: match.Captures["Digits"].Length If you want extra credit, you can also do this by using Regex.Split(), though I've found that Regex.Match() is easier to use for this sort of problem.