CODEWITHSUNDEEP
regex
Magus
Magus

Reputation: 1312

Matching percentages

I've been trying to enhance some code which determines whether a string is a valid percentage. I decided that it was time to finally have a hundred problems, and learned regex. I've been using this web regex tester to build my pattern.

I'm trying to do this rather loosely, such that valid percentages may be integer or decimal, positive or negative, include commas or not, and have any amount of whitespace at the beginning and end, as well as around the optional negative sign and the required percentage sign.

So far, I have \s*-?\s*\d+(,\d+)*(?:\.\d*)?\s*%\s*, which matches almost all of my test cases correctly:

0
 0 
  0  
% 0
        -              0                %            
  20948.924780% 
315%
2,456,875 %
2,104.86%
      89fqyf0gp948y1-%ghghpq98fy92,.?><
 , , , ,,,, 0,0,000,00,00,,,0
 , , , ,,,, 0,0,000,00,00,,,0%
000000000,00000000000 %
000000000,00000000000,00000000000 %
000000000,00000000000,00000000000,00000000000.00000000000 %

These are not in any particular order, some pass and some fail, but only one is incorrect. In , , , ,,,, 0,0,000,00,00,,,0%, the last 0%\n is a match, but the whole line should be invalid. Start and end indicators do not seem to have the effect I had assumed, as a $ makes only the last example match, while a ^ at the beginning makes no matches register.

It may be something small, but as someone who only learned regex yesterday, it's far beyond my reach.

Thanks!

Upvotes: 0

Views: 85

Answers (2)

user557597
user557597

Reputation:

It could be done like this.

Edit: So after realizing its a line thing, this is the regex now.

Note(s) -
Uses multiline mode line Bergi's.
Also, you CANNOT just use \s wihitespace class in this. It doesn't matter what mode used, \s will WILL match CRLF if it can, which means
-
000,000000.22
%
will match because it satisfies all the conditions.
[^\S\r\n] means match whitespace except CRLF characters. It could be replaced with
[^\S\n] in the real world. The initial input on that tester used \r\n linebreaks.

Good Luck!!

 #  ^[^\S\r\n]*-?[^\S\r\n]*(?:(?:\.\d+)|(?:\d+(?:,\d+)*(?:\.\d*)?))[^\S\r\n]*%[^\S\r\n]*$

 ^                   # BOL
 [^\S\r\n]* 
 -?                  # optional -
 [^\S\r\n]*              
 (?:                 # group
      (?: \. \d+ )        # .number
   |                      # or
      (?:                 # group
           \d+                 # number 
           (?: , \d+ )*        # optional many ,number
           (?: \. \d* )?       # optional . optional number
      )                   # end group
 )                   # end group
 [^\S\r\n]* 
 %                   # % 
 [^\S\r\n]* 
 $                   # EOL

Upvotes: 1

Bergi
Bergi

Reputation: 664650

Start and end indicators do not seem to have the effect I had assumed, as a $ makes only the last example match, while a ^ at the beginning makes no matches register.

Those anchors should be working. However, it does depend on the regex engine and the options whether they match line begins/ends or file begins/ends. On RegExr, you'd have to check the multiline option: http://regexr.com?380p9 - in programming, use the m flag.

Upvotes: 1

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