Regular Expression for "0.000%"

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I want to know the regular expression for the string ends with % symbol(Ex:"0.000%" this is my string ) could any one tell me the answer....

Posted 5-Feb-13 19:51pm

Shekhar Reddy.N

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4 solutions

Solution 4

Parsing numbers might fool you if you deal with software that must run under variing cultures.
The decimal separator is not the same on all cultures.

Depending on where you got your string from (e.g. is it localized or culture invariant?) then you must follow differing approaches.

If it is culture invariant, take one of the other suggested soulutions.
If localized, let the system parse the number, i.e. take the non-space text before the % symbol and pass it to the TryParse method.
E.g.

C#

string s = "...";
var match = Regex.Match(s, @"^\s*(.*?)\s*%\s*$");
double percentage;
NumberStyles style = NumberStyles.Currency; // allow all but exponential form
CultureInfo culture = CultureInf.CurrentCulture;
if (match.Success && double.TryParse(match.Groups[1].Value, style, culture, out percentage))
{
   // do something with percentage...
   ...
}
else
{
   // handle error
   ...
}

Cheers
Andi

Posted 6-Feb-13 2:15am

Andreas Gieriet

Solution 3

Try:

C#

.*%$

This will match with any string ending with %

Or

C#

[0-9]*.[0-9]*%$

This will match with any number with single decimal and ending with %

You can create your own or try this on http://regexhero.net/

Posted 6-Feb-13 1:51am

T Pat

Updated 6-Feb-13 1:52am

v2

Comments

OriginalGriff 6-Feb-13 8:35am

Um.
"[0-9]*.[0-9]*%$"
"This will match with any number with single decimal and ending with %"
Are you sure?
Have you tried "X%" - because it will match that as well...

T Pat 6-Feb-13 9:08am

Ya that matched...
Can you please tell me the reason behind that...

OriginalGriff 6-Feb-13 9:33am

Two reasons:
1) "*" is "any number of, including zero". So "[0-9]*" is "zero or more digits"
2) "." is "any character".
So "[0-9]*.[0-9]*%$" happily matches "nothing, any character, nothing, percent, end of string"
What you actually wanted was "[0-9]+\.[0-9]+%$" which requires "at least one digit, a literal decimal point, at least one digit, percent, end of string"

BTW: "[0-9]" can be replaced with "\d"

Pick up a copy of Expresso (the address is in my post) - it's free and it makes working with regexes a whole lot easier! I really wish I'd written it...:sigh:

Solution 1

Hi,

I am not sure this will help you, however thought of passing this link to you.

http://oreilly.com/windows/archive/csharp-regular-expressions.html[^]

The htmlcode for % = amb hash thirty seven (& # 3 7)
http://www.web-source.net/symbols.htm[^]

I will post once again if have an accurate answer.

Thanks,
B.S

Posted 5-Feb-13 20:15pm

Sriram Bala

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OriginalGriff · Accepted Answer · 2013-02-05T20:26:00

Solution 2

Try:

\d+\.\d{3}%$

It matches 1 or more digits, decimal point, 3 digits, percent at the end of the string.

Get a copy of Expresso [^] - it's free, and it examines and generates Regular expressions.

Posted 5-Feb-13 20:26pm

OriginalGriff

Comments

Herman<T>.Instance 6-Feb-13 8:16am

Good tip, My 5. Now installing Expresso

VahanHakobyan 4-Apr-16 10:33am

What about any amount of digits after point?

OriginalGriff 4-Apr-16 11:05am

Replace the "{3}" with "+" - it says "one or more"