Question

Regular Expression to exclude some special characters

Asked by: tia_kamakshi

Hi,

I am working on C# and ASP.Net1.1

I require a reqular expression which can accept any character other than

single quote('), double quote("), @, %,#

Can anyone help me in this

Thanks

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Asked On
2008-07-17 at 08:02:31ID23573442
Tags

C#, ASP.net 1.1, Regular Expression

Topics

Regular Expressions

,

C# Programming Language

Participating Experts
2
Points
500
Comments
12

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Answers

 

by: ddrudikPosted on 2008-07-17 at 08:14:08ID: 22026338

^[^'"@%#]+$

Double the ' or " depending on how you enclose the pattern in code.

 

by: abelPosted on 2008-07-17 at 08:15:07ID: 22026352

You could use a character class. The regex would look like:
['"@%#]

In C# that would be:

System.Text.RegularExpressions.Regex.IsMatch("test", "['\"@%#]")

                                              
1:

Select allOpen in new window

 

by: abelPosted on 2008-07-17 at 08:18:09ID: 22026400

Note that there's a slight difference between ddrudik's and my attempts: mine will test true if any of your mentioned characters hit, wherever in the string. Ddrudik's will match when all characters are not of the set you mentioned.

In terms of performance they may differ a bit, but in ASP, I wouldn't bother too much about that.

 

by: ddrudikPosted on 2008-07-17 at 08:22:04ID: 22026452

>I require a reqular expression which can accept any character other than...
That is what my offered solution was constructed to match.  It will only match true if all characters in the string tested are not a ',",@,% or #.

 

by: abelPosted on 2008-07-17 at 08:27:57ID: 22026517

:)
I didn't mean to say you were wrong. They are two approaches to the same requirement.

 

by: ddrudikPosted on 2008-07-17 at 08:41:19ID: 22026729

No worries here.

 

by: tia_kamakshiPosted on 2008-07-17 at 08:52:48ID: 22026859

Thanks all
ddrudik's expression works great

Can you please help me in understanding this

^[^'"@%#]+$

What is +$ at the end means

Thanks again to all for co-operation

 

by: ddrudikPosted on 2008-07-17 at 08:55:57ID: 22026897

Here's that:
----------------------------------------------------------------------
  ^                        the beginning of the string
----------------------------------------------------------------------
  [^'"@%#]+                any character except: ''', '"', '@', '%',
                           '#' (1 or more times (matching the most
                           amount possible))
----------------------------------------------------------------------
  $                        the end of the string

 

by: ddrudikPosted on 2008-07-17 at 08:57:07ID: 22026913

Using * instead of + would match 0 or more times, which would allow an empty string.  + requires that the string be at least 1 character in length.

 

by: abelPosted on 2008-07-17 at 10:08:23ID: 22027569

In case you want to know more, as a good starting point (and to a certain extend also for advanced users): http://www.regular-expressions.info/ is a good place to be...

 

by: tia_kamakshiPosted on 2008-07-17 at 11:11:38ID: 31477520

Thanks

 

by: ddrudikPosted on 2008-07-17 at 11:12:16ID: 22028291

Thanks for the question and the points.

20120131-EE-VQP-002

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