You need to make a regular expression to check the string denoting the version of the application. The string has the form XYZ, where X, Y, Z can be any integers, as well as * (except X). After the * character, there should be no further line. Numbers should not be 0. Examples of valid strings:
Examples of invalid lines:
I did not do such a regular season.
This is not a learning task.
Naive implementation of what you want:
^(0|[1-9]\d*)\.(\*|(0|[1-9]\d*)\.(\*|(0|[1-9]\d*)))(\r)?$ I got it by following such conclusions. First we create a regular checker that checks the string against the ABC pattern, it is simple:
^\d+\.\d+\.\d+(\r)?$ here \d+ is any number sequence, \. - point (must be escaped, yes), ^ - the beginning of the line, (\r)?$ - the end of the line, taking into account both \r\n , and just \n
Further, instead of the last block of digits, there can be a single asterisk, replacing \d+ with (\*|\d+) (the asterisk must also be escaped):
^\d+\.\d+\.(\*|\d+)(\r)?$ Also, the asterisk can be instead of the last two blocks, similarly to the previous one, change \d+\.(\*|\d+) to (\*|\d+\.(\*|\d+)) :
^\d+\.(\*|\d+\.(\*|\d+))(\r)?$ Well, it remains to exclude numbers with leading zeros, i.e. This is either a separate zero 0 , or non zero + any number of digits [1-9]\d* . Replace all three blocks \d+ with a construction (0|[1-9]\d*) , we get the final version.
^(0|[1-9]\d*)\.(\*|((0|[1-9]\d*)\.(\*|(0|[1-9]\d*))))$ A little bit of explanation:
the leading zero is cut off with the following expression 0|[1-9]\d* . Those. We allow either one zero or 1-9, and any number of any digits behind it.
* so controlled \*|((0|[1-9]\d*)\.(\*|(0|[1-9]\d*))) or asterisk, or everything else.
Test for regex101 https://regex101.com/r/41aBtQ/1
Source: https://ru.stackoverflow.com/questions/819659/
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