CODEWITHSUNDEEP
javaregex
Bogdan
Bogdan

Reputation: 53

Allow only left aligned zeros using regex

I am fairly new to using regex. I have a serial number which can take the following forms: VV-XXXXXX-P or VVXXXXXXP
If the hyphen variant is used, then the number of 'X' can be variable. For example 01-162-8 is equivalent to 010001628.

In order to identify the 2 formats, I have created the following regex's:

String HYPHENS = ([0]{1}[13]{1}-[1-9]{1,6}-[0-9]{1})
String NO_HYPHENS = ([0]{1}[13]{1}[0-9]{6}[0-9]{1})

However the issue with the NO_HYPHENS variant is that it allows 0 anywhere in the sequence For example: 010010628 should not be a valid sequence because there's a non leading 0.

Additionally, how would I create a regex that I can use to replace all 0 from the sequence but the first one? I have tried the following but it also replaces the first 0.

String code = 010001234;
code = code.replaceAll("0+", "");

How could I modify the regex's to achieve this?

Upvotes: 2

Views: 126

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626794

You can use

String NO_HYPHENS = "0[13](?!0*[1-9]0)[0-9]{6}[0-9]";
code = code.replaceAll("(?!^)0(?!$)", "");

See the regex demo. The 0[13](?!0*[1-9]0)[0-9]{6}[0-9] regex matches:

  • 0 - zero
  • [13] - one or three
  • (?!0*[1-9]0) - any zeros followed with a non-zero digit and then a zero is not allowed at this location
  • [0-9]{6} - six digits
  • [0-9] - a digit.

I understand you use it in Java with .matches, else, add ^ at the start and $ at the end.

Also, the (?!^)0(?!$) regex will match any zero that is not at the string start/end position.

Upvotes: 2

mateuszl1995
mateuszl1995

Reputation: 173

^0[13]0*[1-9]*[0-9]$

^ - beginning of string
0 - first sign must be zero
[13] - second sign must be one or three
0* - sequence of zeros of variable length
[1-9] - sequence of non-zeros of variable length
[0-9] - finally one digit (it can be replaced with \d also)
$ - end of string

This regex has one problem: it doesn't check how many digits are in the XXXXXX section of serial number. But you can check it with length function:

      String code = "010000230";
      if (code.matches("^0[13]0*[1-9]*[0-9]$") && code.length() == 9) {
         // code is valid
      }
      // replacement
      code = code.replaceAll("^(0[13])0*([1-9]*[0-9])$", "$1$2");

Explanation of the replacement:
(0[13]) group number 1 (groups are in bracket)
0* some zeros
([1-9]*[0-9]) group number 2

This will be replaced with:
$1$2 group number 1 and group number 2 ($1 means group number 1)

Upvotes: 1

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