'Regex ignore first x characters and then match pattern

String = '11111111111110000000000000000000110000000000000011111111111111111111111111111111110011111111111110000011110000011111111111110000000000011111111111111111010001111111111111111111110011111111111111111111111111110111112111121111111111111111111000011000001011111111111101022111101111001111111111110000001000000111111111111111000000000000011111111111111100011111111001011111111100000000000000000000000000000000100111001000000000000000000011000000000000001111111000000000000000000000000000000000001111100000000000000000000011000000000000000000000010000000000333333333'

I want a pattern to take out 10 characters after the first 100 so i want to have 100 - 110 then I want to compare that one and see if that string with a length of 10 have 4 zeros in a row.

How can I do this with only Regex? I have been using substring before.

regex


Solution 1:[1]

You could use this:

^.{100}(?=.{0,6}0000)(.{10})

Explanation:

  • ^: matches the start of the string to avoid that the pattern is used anywhere in the input
  • .{100}: match 100 characters
  • (?= ): look ahead. This does not capture, but just verifies something that is still ahead.
  • .{0,6}: 0 to 6 characters
  • 0000: literally 4 zeroes
  • (.{10}): 10 characters, this time they are captured and can be referenced back with \1 or $1 depending on the flavour of regex.

Solution 2:[2]

The above answer is perfect. But that matches all the characters including first 100. In case of ignoring first 100, we can use 

(?<=.{100})

To check the required pattern in last 10 characters after first 100 only, we can use 

(?<=.{100})(?=.{0,6}0000)(.{10})

You can test it here

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 trincot
Solution 2 Tahir Hussain Mir

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