RegEx for pattern in c#

Question

I am still learning RegEx so I need some help on this one.

Rules are as below:

Min/Max Length of string: 1-5
String has to be a whole word
String can contain: A-Z and/or a-z and/or 0-9 or combination of both only (e.g. _ - . * are not allowed)

Examples:

12345 – allowed
Os342 – allowed
2O3d3 – allowed
3sdfds dsfsdf – not allowed
Sdfdf.sdfdf –now allowed

Because every language has slightly different regexp syntax. Which is one reason I probably can't help you much, I don't know C#. — Carl Smotricz
– Carl Smotricz, Commented Jan 13, 2010 at 1:17
There are slightly different regex implementations in different languages. Even Visual Studio itself is using non-.NET regex implementation in it's find/replace dialog: codinghorror.com/blog/archives/000633.html — Miroslav Popovic
– Miroslav Popovic, Commented Jan 13, 2010 at 1:19

Rubens Farias · Accepted Answer · 2010-01-13 01:19:07Z

3

What about:

string input = "12345";
bool match = Regex.IsMatch(input, "^[a-z0-9]{1,5}$", RegexOptions.IgnoreCase);

answered Jan 13, 2010 at 1:19

Rubens Farias

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3 Comments

Carl Smotricz Over a year ago

Without ^ and $, the regexp tester he gave me matched "12345" of "123456"; apparently that should be rejected though.

Carl Smotricz Over a year ago

And you were right about those other chars. I deleted my answer, thanks.

Rubens Farias Over a year ago

@Jeffrey: I used RegexOptions.IgnoreCase, so will not differ a-z from A-Z (as OP said is working with C#). But you can remove it and to add A-Z if you wish

YOU · Accepted Answer · 2010-01-13 01:22:37Z

3

How about

^[A-Za-z0-9]{1,5}$

answered Jan 13, 2010 at 1:22

YOU

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Comments

Darko · Accepted Answer · 2010-01-13 01:22:38Z

3

^[a-zA-Z0-9]{1,5}$

[a-zA-Z0-9] specifies the ranges allowed
^ specifies start of string
$ specifies end of string
{1,5} indicates minimum and maximum number of characters for the range

answered Jan 13, 2010 at 1:22

Darko

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1 Comment

Carl Smotricz Over a year ago

You want to do 1,5 to catch strings shorter than 5 chars too, I think.

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