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Gradient helper prevents having a 'top()' mixin #56

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ianstormtaylor opened this issue Jan 12, 2012 · 1 comment
Open

Gradient helper prevents having a 'top()' mixin #56

ianstormtaylor opened this issue Jan 12, 2012 · 1 comment

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@ianstormtaylor
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I get this error when trying to declare a mixing with a name of top()

TypeError: expected string, ident or literal, but got function:top()

and declaring a gradient using a string doesn't work either linear-gradient('top', ...

Not sure if there is a solution that keeps both of the functions around...
@tj
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tj commented Jan 12, 2012

we could probably also check for a function of the same name just in case, shouldn't hurt

@kizu kizu mentioned this issue May 14, 2012
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@tj @ianstormtaylor