Question

I need a regex for 5 digits in increasing order, like 12345, 24579, 34680, and so on.

0 comes after 9.

Answer 1

You can try (as seen on rubular.com)

^(?=\d{5}$)1?2?3?4?5?6?7?8?9?0?$

Explanation

• ^ and $ are the beginning and end of string anchors respectively
• \d{5} is the digit character class \d repeated exactly {5} times
• (?=...) is a positive lookahead
• ? on each digit makes each optional

How it works

• First we use lookahead to assert that anchored at the beginning of the string, we can see \d{5} till the end of the string
• Now that we know that we have 5 digits, we simply match the digits in the order we want, but making each digit optional
  • The assertion ensures that we have the correct number of digits

regular-expressions.info

• Anchors, Character Classes, Finite Repetition, Lookarounds, and Optional

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Generalizing the technique

Let’s say that we need to match strings that consists of:

• between 1-3 vowels [aeiou]
• and the vowels must appear in order

Then the pattern is (as seen on rubular.com):

^(?=[aeiou]{1,3}$)a?e?i?o?u?$

Again, the way it works is that:

• Anchored at the beginning of the string, we first assert (?=[aeiou]{1,3}$)
  • So correct alphabet in the string, and correct length
• Then we test for each letter, in order, making each optional, until the end of the string

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Allowing repetition

If each digit can repeat, e.g. 11223 is a match, then:

• instead of ? (zero-or-one) on each digit,
• we use * (zero-or-more repetition)

That is, the pattern is (as seen on rubular.com):

^(?=\d{5}$)1*2*3*4*5*6*7*8*9*0*$