T
tshad
Is there some way to set up something like a class inside of a class?
I have various tags and classes that use one color. The problem is that if
I change that color I need to change it for each one.
For example:
Inside of my css class I have the following:
a:visited {
color:#4AAABD;
}
a:visited.blueClass {
color:blue;
}
a:visited.redClass {
color:red;
}
a:visited.orangeClass {
color:#FF6600;
}
a:link {
color:#4AAABD;
}
a:link.blueClass {
color:blue;
}
a:link.redClass {
color:red;
}
a:link.orangeClass {
color:#FF6600;
}
a:active {
color:#4AAABD;
}
a:active.blueClass {
color:blue;
}
a:active.redClass {
color:red;
}
a:active.orangeClass {
color:#FF6600;
}
However, I want to change the color #4AAABD, to #3EA2BC. The problem is I
have about 20 places to change it.
Is it possible to do something like:
..backing {
color: #3EA2BC
}
and then use it in my other styles:
a:active {
class:backing
}
I could then change .backing to another color and have all my other styles
use it.
I know that is right. I am just trying to illustrate what I am trying to
do.
Thanks,
Tom
I have various tags and classes that use one color. The problem is that if
I change that color I need to change it for each one.
For example:
Inside of my css class I have the following:
a:visited {
color:#4AAABD;
}
a:visited.blueClass {
color:blue;
}
a:visited.redClass {
color:red;
}
a:visited.orangeClass {
color:#FF6600;
}
a:link {
color:#4AAABD;
}
a:link.blueClass {
color:blue;
}
a:link.redClass {
color:red;
}
a:link.orangeClass {
color:#FF6600;
}
a:active {
color:#4AAABD;
}
a:active.blueClass {
color:blue;
}
a:active.redClass {
color:red;
}
a:active.orangeClass {
color:#FF6600;
}
However, I want to change the color #4AAABD, to #3EA2BC. The problem is I
have about 20 places to change it.
Is it possible to do something like:
..backing {
color: #3EA2BC
}
and then use it in my other styles:
a:active {
class:backing
}
I could then change .backing to another color and have all my other styles
use it.
I know that is right. I am just trying to illustrate what I am trying to
do.
Thanks,
Tom