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I currently have a regex that pulls up a 16 digit number from a file e.g.:
Regex:
Regex.Match(l, @"\d{16}")
This would work well for a number as follows:
1234567891234567
Although how could I also include numbers in the regex such as:
1234 5678 9123 4567
and
1234-5678-9123-4567
343
Example: Regex Number Range 1-20 Range 1-20 has both single digit numbers (1-9) and two digit numbers (10-20). For double digit numbers we have to split the group in two 10-19 (or in regex: "1[0-9]") and 20. Then we can join all these with an alternation operator to get "([1-9]|1[0-9]|20)".
To get a string contains only numbers (0-9) we use a regular expression (/^[0-9]+$/) which allows only numbers.
The regex [0-9] matches single-digit numbers 0 to 9. [1-9][0-9] matches double-digit numbers 10 to 99. That's the easy part. Matching the three-digit numbers is a little more complicated, since we need to exclude numbers 256 through 999.
+: one or more ( 1+ ), e.g., [0-9]+ matches one or more digits such as '123' , '000' . *: zero or more ( 0+ ), e.g., [0-9]* matches zero or more digits.
If all groups are always 4 digit long:
\b\d{4}[ -]?\d{4}[ -]?\d{4}[ -]?\d{4}\b
to be sure the delimiter is the same between groups:
\b\d{4}(| |-)\d{4}\1\d{4}\1\d{4}\b
If it's always all together or groups of fours, then one way to do this with a single regex is something like:
Regex.Match(l, @"\d{16}|\d{4}[- ]\d{4}[- ]\d{4}[- ]\d{4}")
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