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Diego
Asked: 2020-04-06 22:12:43 +0800 CST 2020-04-06 22:12:43 +0800 CST

Conditional regular expressions based on a character

  • 772

Is it possible to create regular expressions based on string characters?

Let me explain with an example, I am looking for a regular expression that validates that in a string, if the first digit is 0, the second cannot be. But the rest of the digits can be the number they want, examples:

0102 - Bien
0123 - Bien
0012 - Mal

But, on the contrary, if the string begins with a number other than 0, that second digit can be 0 or the number you want, examples:

1234 - Bien
1000 - Bien
1089 - Bien

Let's see if you can give me a hand with the validation based on a digit, because I don't finish finding the key.

regex

3 Answers

  1. Best Answer
    ^(0[1-9]\d+|[1-9]\d+)$
  ~~~~~~~~~~ ~~~~~~~~~
   1er grupo  2º grupo

    With this we create two possible match groups:

    • The first controls that numbers starting with 0 cannot have a 0 in the second digit.
    • The second group, which validates any number that does not start with 0, accepts any digit as a second figure.

    Of course this regular expression could be simplified:

    ^0?[1-9]\d+$

    So starting with 0is optional, but if you do, the second number cannot be a 0, since the group [1-9]does not allow it.

    I would need the expression to start "acting" from the first character, not that it needs 3

    ^(0(?:[1-9]\d*)?|[1-9]\d*)$

    (?:..)Allows you to create a group but without capturing it. the idea is that everything that is not the 0initial one is optional (which is why after this non-capture group there is a question mark: . In addition, the by (?:..)?have been replaced to match 0 or more occurrences instead of necessarily one.+*

    This expression, like the first, can be easily simplified:

    ^0?(?:[1-9]\d*)?$
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  2. Based on the comments, the following can be deduced:

    1. The isolated number 0 is a solution, the expression would be:^0$
    2. In the event that 0 appears, it must necessarily be followed by a digit from the range 1 to 9, in case there are more than 2 digits, these will all be numeric and may appear zero or more times, the expression would be:^0?[1-9]\d*$

    Grouping the 2 cases we would have:^(0|0?[1-9]\d*)$

    Important Note : If what is being evaluated is a string that internally contains the patterns to locate, for example: 0 00 0102 0123 0012 03 Casa 120 Puerta 2516 Caracas

    In this case, it is delimited by blocks of words, leaving the regular expression:\b(0|0?[1-9]\d*)\b

    Remember to add the global and multi-line parameters in case the string has multiple lines.

    • 1

  3. Reading your question literally, the answer is simpler than it seems:

    ^00

    That is all. If it matches ^00, then it's wrong, if it doesn't, it's ok. It's just a matter of negating the final result of the regex.

    Of course, if you are also interested in validating that it is indeed a number, I would go with linuxexp's answer that solves it.

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