正则表达式接受 0 到 100 之间的十进制数

Question

My requirement is a regular expression it accepts decimal values between 0 to 100 (like 1,2,3,....,99, 0.1,0.2,0.3,.....,99.9, 0.01,0.02,0.03,.....,99.99, 00.01 to 99.99). I found one solution

/^(?!0?0\.00$)\d{1,2}\.\d{2}$/ 

but it accepts only decimal values like 00.01 to 99.99.

Answer 1

How about:

^(?!0+(?:\.0+)?$)\d?\d(?:\.\d\d?)?$

Explanation:

^           : begining of string
  (?!       : negative lookahead, assumes there is no:
    0+      : 1 or more zero
    (?:     : non capture group
      \.0+  : a dot then 1 or more zeros
    )?$     : end of group, optional, until end of string
  )         : end of lookahead
  \d?\d     : 1 or 2 digit
  (?:       : non capture group
    \.\d\d? : a dot followed by 1 or 2 digit
  )?        : end of group, optional
$           : end of string

Answer 2

How about the following: ^(100|([0-9][0-9]?(\\.[0-9]+)?))$

First handle the outer case of the 100 number, then tend to the remaining combinations.

Make sure to escape the backward slash when using this in Java.

Answer 3

试试这个正则表达式。

/^(\d{1,2}\.\d{1,2}|\d{1,2})$/

Answer 4

/^(100|([0-9][0-9]?(\.\d{1,2})?))/

this will return the decimal number between 0 to 100 with match 2 digits and optional two decimal. eg out put would be like one of these

3
33
33.3
33.33