重复字符 X 次，但不计算介于两者之间的 Y 个字符（正则表达式）

Question

I need a regex that does the following:

• Repeat any number x amount of times
• Ignore characters and do not count them (example: whitespace and '/')

This is the regex I have right now: [0-9]{0,5}

However, it does not cover the following legit strings:

1 2 3 4 5
123/45
1234 5
1 234 4
1/234/5
1/234 5
1 2/3 4/5
1     2345
1 ////23/////45

I tried:

[\s*\/*0-9]{0,5} //counts unwanted characters
[0-9\s\/]{0,5} //counts unwanted characters
[0-9-\s+\/]{0,5} //counts unwanted characters
[\s+\/-]{0,}[0-9]{0,5} // does not mix 

Is this even possible in regex?

The other solution, I can do is to remove these characters and then comparing to the pattern.

Answer 1

The regex you can use is

(?:[ \/]*\d){0,5}

See the regex demo . Details :

• (?: - start of a non-capturing group:
  • [ \/]* - zero or more spaces or slashes
  • \d - a digit
• ){0,5} - end of the group, match zero to five times.

If you need to match any spaces or slashes after the last digit, add [ \/]* at the end:

(?:[ \/]*\d){0,5}[ \/]*