c# regular expression to only allow 1 or 2 digits only

Question

User1332640826 posted

I will admit i know nothing about regular expressions. what I am trying to do is use a regular expression which expects only 1 or 2 digits.

Sample

1

23

Answer 1

User-1506965535 posted

If you are using asp.net then you can define something like this

<asp:RegularExpressionValidator ID="RegularExpressionValidator1" runat="server" ErrorMessage="Please enter only two digit" ControlToValidate="yourtextboxid" ValidationExpression="^[0-9]{2}$"></asp:RegularExpressionValidator>

Answer 2

User1332640826 posted

That's the one i am using , but currently it accepts 2 digits. I need expression which accepts one or two digits.

Answer 3

User-1506965535 posted

OK,

try with this:-

/^\d{1,2}$/;

Answer 4

User281315223 posted

If you want a very simple regular expression to match one or two digits only, you just need the following :

^\d{1,2}$

This will explicitly match a string that only contains 1 or two digits. If you wanted to use it to match multiple instances of one or two digits within a string, then you could just leave off the beginning '^' and end '$' characters as seen below :

\d{1,2}

You can see a working example here or demonstrated below :

// An example string
var example = "There will be 1 sentence behind this. The answer to life, the universe and everything is 42, but I am 29 years old.";
		
// Use a Regex to match any 1 or two digit numbers
var matches = Regex.Matches(example,@"\d{1,2}"); // This will contain "1", "42" and "29"
		

Answer 5

User1332640826 posted

nope it dint work

Answer 6

User-1506965535 posted

Can you post your html code ?

Answer 7

User281315223 posted

nope it dint work

Did you try the code that I provided earlier? Additionally, could you post an example of how you are currently using this? The expression can be further explained a bit more as follows :

^           # Beginning of expression
\d{1,2}     # Explicit match one or two ({1,2}) digits (\d is equivalent to [0-9])
$           # End of expression

Answer 8

User1332640826 posted

thanks ! worked perfectly