What is regex for space

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In the following code, the regular expression throws error on selection of a value from DDL.
There is some wrong in validation expression. How to use the correct validation expression for the data in this DDL. In DDL there is small characters, an -, brackets and space.

XML

<asp:DropDownList ID="DropDownList7" runat="server"
        style="z-index: 1;top: 246px; left: 225px; position: absolute; height: 18px; width:148px"
               AutoPostBack="True"
             onselectedindexchanged="DropDownList7_SelectedIndexChanged">
         <asp:ListItem>SELECT</asp:ListItem>
         <asp:ListItem>(7-14 days)</asp:ListItem>
         <asp:ListItem>(4-8 days)</asp:ListItem>
    </asp:DropDownList>
    <asp:RegularExpressionValidator ID="RegularExpressionValidator5" runat="server"  Display='None'
            ControlToValidate="DropDownList7" ErrorMessage="Select Delivery Days"
            ValidationExpression="[0-9 a-z \s \d-_ ][(][)]{1,50}" Font-Size="X-Small"></asp:RegularExpressionValidator>

Posted 18-Mar-14 20:47pm

S.Rajendran from Coimbatore

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Abhinav S · Answer 1 · 2014-03-18T21:04:00

Solution 1

The regex for space is \s.
For e.g. the regex for matching alphanumeric and spaces would be Regex.Replace(q, @"[^a-zA-Z0-9\s]", string.Empty);

Posted 18-Mar-14 21:04pm

Abhinav S

Comments

Maciej Los 19-Mar-14 3:06am

+5

Abhinav S 19-Mar-14 3:47am

Thank you.

S.Rajendran from Coimbatore 19-Mar-14 3:18am

I used like this : ValidationExpression= "[^a-zA-Z0-9\s][d\-_][(][)]" not working

Kan07 · Answer 2 · 2014-03-18T21:31:00

VB

^\s*(?=[a-zA-Z0-9'/\\&.-])([a-zA-Z0-9'/\\&.\s-]{3,})(?<=\S)\s*$


^                      # Start of string
\s*                    # Optional leading whitespace, don't capture that.
(?=                    # Assert that...
 [a-zA-Z0-9'/\\&.-]    # the next character is allowed and non-space
)
(                      # Match and capture...
 [a-zA-Z0-9'/\\&.\s-]{3,} # 
)
(?<=\S)                # Assert that the previous character is not a space
\s*                    # Optional trailing whitespace, don't capture that.
$                      # End of string

Bernhard Hiller · Answer 3 · 2014-03-19T00:02:00

Solution 3

Your problem is a different one: the brackets. () are used for catching some data, but here you want to match a bracket. You must use \( and \). [] stand for character ranges, if you want to match them, you must escape them also \[\]. I do not know if literal spaces are acceptable in a character range, better omit them.
I'd write

ValidationExpression="[0-9a-z\s\d-_\(\)\[\]]{1,50}"

Posted 19-Mar-14 0:02am

Bernhard Hiller

Comments

S.Rajendran from Coimbatore 19-Mar-14 6:42am

This worked for me: "[a-z 0-9 \s \d-_ ()]{1,50}" Thank you all.

Bernhard Hiller 19-Mar-14 6:47am

OK, then click "accept answer".