How to Get the ASP.NET LinkButton CssClass Property to Work
Hopefully this will save someone a lot of time. Today I was working on a webpage in ASP.NET and made use of the LinkButton control. But when I applied a CSS class using the CssClass property, it just wouldn't take. After much searching, I found the solution.
First off, this is a bug -- you can't spin it any other way. Fortunately, there IS a way to get it to work: you need to specify one of the following qualifiers on your CSS class: visited, link, or active.
For example, when I tried to apply the following class, it wasn't reflected in the control:
.Pager
{
font-family:Verdana;
font-size:8pt;
font-weight:bold;
color:Orange;
}
However, when I added the link qualifier (":link" added to the class name), it worked like a charm:
.Pager:link
{
font-family:Verdana;
font-size:8pt;
font-weight:bold;
color:Orange;
}
When you apply the class name to the CssClass property of the control, leave out the qualifier. For example, using my class above, I set the value of the CssClass property like this:
linkbtn.CssClass = "Pager";
Notice that I used "Pager" and not "Pager:link".
To recap:
1. Apply one of the aforementioned qualifiers (link, active, or visited) to the end of your class name, preceeded by a colon (there may also be other qualifiers, but these are the three I know about).
2. Specify your class name without the qualifier as the value for the LinkButton's CssClass property.
I hope this helps!
First off, this is a bug -- you can't spin it any other way. Fortunately, there IS a way to get it to work: you need to specify one of the following qualifiers on your CSS class: visited, link, or active.
For example, when I tried to apply the following class, it wasn't reflected in the control:
.Pager
{
font-family:Verdana;
font-size:8pt;
font-weight:bold;
color:Orange;
}
However, when I added the link qualifier (":link" added to the class name), it worked like a charm:
.Pager:link
{
font-family:Verdana;
font-size:8pt;
font-weight:bold;
color:Orange;
}
When you apply the class name to the CssClass property of the control, leave out the qualifier. For example, using my class above, I set the value of the CssClass property like this:
linkbtn.CssClass = "Pager";
Notice that I used "Pager" and not "Pager:link".
To recap:
1. Apply one of the aforementioned qualifiers (link, active, or visited) to the end of your class name, preceeded by a colon (there may also be other qualifiers, but these are the three I know about).
2. Specify your class name without the qualifier as the value for the LinkButton's CssClass property.
I hope this helps!
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