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Applied arrix's fix for getting style values on elements that aren't in the DOM, in Safari. (Bug #1482)
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src/jquery/jquery.js
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src/jquery/jquery.js
vendored
@ -1509,7 +1509,7 @@ jQuery.extend({
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// then some display: none elements are involved
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else {
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// Locate all of the parent display: none elements
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for ( var a = elem; color(a); a = a.parentNode )
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for ( var a = elem; a && color(a); a = a.parentNode )
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stack.unshift(a);
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// Go through and make them visible, but in reverse
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