字符串中任何位置至少 n 位的正则表达式？

Question

A regular expression for at least one digit in a string looks like this:

(?=.*[0-9]) 

So that would find a1a or a2a2 etc.

Suppose we wanted at least 2 digits in the string. So it should return true for a2a2 or a2a2a2 , but a2 would not pass.

What would that regex look like?

Answer 1

You could use match here with the regex pattern \d.*\d :

 var inputs = ["a2", "a2a2", "a2a2a2"]; for (var i=0; i < inputs.length; ++i) { if (inputs[i].match(/\d.*\d/)) { console.log("MATCH => " + inputs[i]); } else { console.log("NO MATCH => " + inputs[i]); } }

We can also use a length assertion, after removing all non digits characters:

 var inputs = ["a2", "a2a2", "a2a2a2"]; for (var i=0; i < inputs.length; ++i) { if (inputs[i].replace(/\D+/g, "").length >= 2) { console.log("MATCH => " + inputs[i]); } else { console.log("NO MATCH => " + inputs[i]); } }

Answer 2

You can use a quantifier. For example:

([a-z][0-9]){2,}

This says at least 2 occurrences of the pattern. See more here https://docs.microsoft.com/en-us/dotnet/standard/base-types/quantifiers-in-regular-expressions