如何打开动态创建内容的jquery菜单。 即我不想得到ul或任何其他div元素的id作为菜单项。 代码就像......
var mennnu = "<ul id='menu'>" +
"<li><span class='ui-icon ui-icon-disk'></span>Save</li>"+
"<li><span class='ui-icon ui-icon-zoomin'></span>Zoom In</li>"+
"<li><span class='ui-icon ui-icon-zoomout'></span>Zoom Out</li>"+
"<li class='ui-state-disabled'><span class='ui-icon ui-icon-print'></span>Print...</li>"+
"<li>"+
"Playback"+
"<ul>"+
"<li><span class='ui-icon ui-icon-seek-start'></span>Prev</li>"+
"<li><span class='ui-icon ui-icon-stop'></span>Stop</li>"+
"<li><span class='ui-icon ui-icon-play'></span>Play</li>"+
"<li><span class='ui-icon ui-icon-seek-end'></span>Next</li>"+
"</ul>"+
"</li>"+
"<li>Learn more about this menu</li>"+
"</ul>";
$(mennnu).menu({
width: 350,
height: 300,
position : [clickX+248,clickY+63]
}
);
答案 0 :(得分:0)
请参阅 DEMO
JS代码:
$(function() {
var mennnu = "<ul class='my_menu'>" +
"<li><span class='ui-icon ui-icon-disk'></span>Save</li>"+
"<li><span class='ui-icon ui-icon-zoomin'></span>Zoom In</li>"+
"<li><span class='ui-icon ui-icon-zoomout'></span>Zoom Out</li>"+
"<li class='ui-state-disabled'><span class='ui-icon ui-icon-print'></span>Print...</li>"+
"<li>"+
"Playback"+
"<ul>"+
"<li><span class='ui-icon ui-icon-seek-start'></span>Prev</li>"+
"<li><span class='ui-icon ui-icon-stop'></span>Stop</li>"+
"<li><span class='ui-icon ui-icon-play'></span>Play</li>"+
"<li><span class='ui-icon ui-icon-seek-end'></span>Next</li>"+
"</ul>"+
"</li>"+
"<li>Learn more about this menu</li>"+
"</ul>";
$('body').append(mennnu);
//$('.my_menu') //use this if you want to create menu by adding "class" attribute to <ul>
$('ul') //and with even "class" attribute added you can use this
.menu({
width: 350,
height: 300//,
//position : [clickX+248,clickY+63]
}
);
});