How to Regex a string

Is it possible to check with regex:

1. the complete string are numbers AND
2. the first character is a 7 or 8 then the complete length of the string must be 11 OR
3. the first character is a 1 then the complete length of the string must be 10
4. OR the first character is a 0 then the complete length of the string must be 18 AND on character 8 must be a 8 or 7 OR on character 9 must be a 1

I hope you can see what I mean. Hope the examples will help you to know what I mean.

Here is my solution(not working completely-> I don't know how to check if in case it starts with a 0 and it is 18 characters long the character on position 8 must be 7or8 or on position 9 the character must be 1):

^(?:[78]\d{10}|[1-69]\d{9}|[0]/d{18})$

For example:

• 85556987456 -> starts with 8 and length is 11 -> match
• 75556987456 -> starts with 7 and length is 11 -> match
• 1555698745 -> starts with 1 and length is 10 -> match
• 000000085556987456 -> starts with 0 and length is 18 and on pos 8 is a 8 -> match
• 000000075556987456 -> starts with 0 and length is 18 and on pos 8 is a 7 -> match
• 000000001556987456 -> starts with 0 and length is 18 and on pos 9 is a 1 -> match

Thank you!

You can use

^(?:[78]\d{10}|1\d{9}|0\d{6}(?:[87]\d|\d1)\d{9})$

See the regex demo

Details

• ^ - start of string
• (?:[78]\d{10}|1\d{9}|0\d{6}(?:[87]\d|\d1)\d{9}) - one of the three alternatives:
  • [78]\d{10} - 7 or 8 and then 10 digits (11 all in all)
  • | - or
  • 1\d{9} - 1 and then 9 digits (10 all in all)
  • | - or
  • 0\d{6}(?:[87]\d|\d1)\d{9} - 0, then 6 digits, then the 8th digit equal to 8 or 7 and any one digit, or any one digit and the 9th digit equal to 1, and then 9 more digits (=18 digits)
• $ - end of string.