我只字不提“大鼠:目标号没有快速方法”,我看不出错误来自何地。 我很新,可以 j笑Qu语和 Java语,而不是确定什么是错的。
我正在尝试这方面的实例:。
我拿到了这一代码,我掌握了指数档案,我可以看到它工作得当,因此,我刚刚将清单抄过并走了主。 j 内容与我已经必须履行这一职能:
$(document).ready(function () {
$("#myController").jFlow({ controller: ".jFlowControl", slideWrapper: "#jFlowSlider", slides: "#mySlides", selectedWrapper: "jFlowSelected", width: "960px", height: "350px", duration: 400, prev: ".jFlowPrev", next: ".jFlowNext", auto: true });
$().UItoTop({ easingType: easeOutQuart }); jQuery("a[data-gal^= prettyPhoto ]").prettyPhoto({ social_tools: false });
// get the action filter option item on page load
var $filterType = $( #filterOptions li.active a ).attr( class );
// get and assign the ourHolder element to the
// $holder varible for use later
var $dataholder = $( ul.ourHolder );
// clone all items within the pre-assigned $holder element
var $data = $dataholder.clone();
// attempt to call Quicksand when a filter option
// item is clicked
$( #filterOptions li a ).click(function(e) {
// reset the active class on all the buttons
$( #filterOptions li ).removeClass( active );
// assign the class of the clicked filter option
// element to our $filterType variable
var $filterType = $(this).attr( class );
$(this).parent().addClass( active );
if ($filterType == all ) {
// assign all li items to the $filteredData var when
// the All filter option is clicked
var $filteredData = $data.find( li );
}
else {
// find all li elements that have our required $filterType
// values for the data-type element
var $filteredData = $data.find( li[data-type= + $filterType + ] );
}
// call quicksand and assign transition parameters
$dataholder.quicksand($filteredData, {
duration: 800,
easing: easeInOutQuad ,
attribute: "data-id",
});
return false;
});
}
);
" But I get an error at : "$dataholder.quicksand($filteredData,duration: 800,easing: easeInOutQuad }); "
任何想法需要做些什么?