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Tooltip: Ignore disabled checks when closing. Fixes #8758 - Tooltip: Tooltip is shown, but not hidden if element has class=ui-state-disabled.

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This commit is contained in:
Scott González 2012-11-09 13:07:53 -05:00
parent 1cdeeccab0
commit 498aadf644
2 changed files with 14 additions and 1 deletions
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13
tests/unit/tooltip/tooltip_core.js
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@ -94,4 +94,17 @@ test( "form containing an input with name title", function() {
equal( $( ".ui-tooltip" ).length, 0, "no tooltip for form" );
});
test( "tooltip on .ui-state-disabled element", function() {
expect( 2 );
var container = $( "#contains-tooltipped" ).tooltip(),
element = $( "#contained-tooltipped" ).addClass( "ui-state-disabled" );
element.trigger( "mouseover" );
equal( $( ".ui-tooltip" ).length, 1 );
container.empty();
equal( $( ".ui-tooltip" ).length, 0 );
});
}( jQuery ) );

2
ui/jquery.ui.tooltip.js vendored
View File

@ -299,7 +299,7 @@ $.widget( "ui.tooltip", {
if ( !event || event.type === "focusin" ) {
events.focusout = "close";
}
this._on( target, events );
this._on( true, target, events );
},
close: function( event ) {