如何使用RegEx忽略第一个句点并匹配所有后续句点？

Question

How to use RegEx to ignore the first period and match all subsequent periods?

For example:

• 1.23 (no match)
• 1.23.45 (matches the second period)
• 1.23.45.56 (matches the second and third periods)

I am trying to limit users from entering invalid numbers. So I will be using this RegEx to replace matches with empty strings.

I currently have /[^.0-9]+/ but it is not enough to disallow . after an (optional) initial .

Answer 1

Constrain the number between the start ^ and end anchor $ , then specify the number pattern you require. Such as:

/^\\d+\\.?\\d+?$/

Which allows 1 or more numbers, followed by an optional period, then optional numbers.

Answer 2

I suggest using a regex that will match 1+ digits, a period, and then any number of digits and periods capturing these 2 parts into separate groups. Then, inside a replace callback method, remove all periods with an additional replace :

 var ss = ['1.23', '1.23.45', '1.23.45.56']; var rx = /^(\\d+\\.)([\\d.]*)$/; for (var s of ss) { var res = s.replace(rx, function($0,$1,$2) { return $1+$2.replace(/\\./g, ''); }); console.log(s, "=>", res); } 

Pattern details :

• ^ - start of string
• (\\d+\\.) - Group 1 matching 1+ digits and a literal .
• ([\\d.]*) - zero or more chars other than digits and a literal dot
• $ - end of string.