A regular expression matching at least n of m groups

Question

I was going to write a regular expression that would only match if the string contains at least n different character classes. I was going to use this to get my users to create strong passwords and wanted to check if the password contains at least 3 of the following:

• Characters
• Characters of capital
• The numbers
• Special symbols

Writing a regular expression that matches if all of these classes are present is trivial using lookaheads. However, I cannot wrap my head around the “at least 3” part. Is this possible (in a nice, compact expression) or do I need to create a monster expression?

Answer 1

I think this will be more compact than listing each possible combination of 3 out of 4. It uses a negative scan to make sure that the entire string does not consist of just one or two of the character classes you listed:

(?!([a-zA-Z]*|[az\d]*|[^AZ\d]*|[AZ\d]*|[^az\d]*|[^a-zA-Z]*)$).* 

In order, the following groups are here:

• lower and / or upper
• below and / or numbers
• lower and / or special
• upper and / or numbers
• top and / or special
• numbers and / or special

This regular expression will fail if the entire string (due to $ in a negative image) contains only characters from any of the above groups.

Answer 2

You must write an expression for each possible combination of 3 out of 4 (a total of four expressions), and then | separate expressions together so that they pass if they execute at least one of the original expressions.

Answer 3

What do you think of this decision?

 var str='23khkS_s'; var countGroups=0; if(str.match(/[az]+/)) countGroups++; if(str.match(/[AZ]+/)) countGroups++; if(str.match(/[0-9]+/)) countGroups++; if(str.match(/[ -_!@ #$%*\(\)]+/)) countGroups++; console.log(countGroups); 

Instead of using one monster-like expression, you can use 4 small REs. And then work with the variable countGroups , which contains the number of marked character groups. I hope this is helpful