我正在使用codrops slide& amp;的修改版本。推送菜单(http://tympanus.net/codrops/2013/04/17/slide-and-push-menus/)以获取网页上的叠加层。但我无法用另一个链接再次关闭它。 任何帮助,将不胜感激。 jsfiddle:http://jsfiddle.net/qu80jto2/
HTML:
<nav class="modal modal-vertical modal-right" id="modal">
<h1>CONTENT</h1>
</nav>
<h1 id="showRight">OPEN</h1>
<h1 id="hideRight">CLOSE</h1>
JQUERY:
<script>
var
menuRight = document.getElementById( 'modal' ),
body = document.body;
showRight.onclick = function() {
classie.toggle( this, 'active' );
classie.toggle( menuRight, 'modal-open' );
disableOther( 'showRight' );
};
function disableOther( button ) {
if( button !== 'showRight' ) {
classie.toggle( showRight, 'disabled' );
}
}
</script>
CSS:
.modal {
background: yellow;
position: fixed;
}
.modal-vertical {
width: 60%;
height: 100%;
top: 0;
z-index: 1000;
}
.modal-right {
right: -60%;
}
.modal-open {
right: 0px;
}
.modal-push {
overflow-x: hidden;
position: relative;
left: 0;
}
.modal,
.modal-push {
-webkit-transition: all 0.6s ease;
-moz-transition: all 0.6s ease;
transition: all 0.6s ease;
}
答案 0 :(得分:0)
你可以尝试这个:
<nav class="modal modal-vertical modal-right" id="modal">
<h1>CONTENT</h1>
</nav>
<h1 id="showRight">OPEN</h1>
<h1 id="hideRight">CLOSE</h1>
<script>
var
menuRight = document.getElementById( 'modal' ),
body = document.body;
showRight.onclick = function() {
classie.toggle( this, 'active' );
classie.toggle( menuRight, 'modal-open' );
disableOther( 'showRight' );
};
function disableOther( button ) {
if( button !== 'showRight' ) {
classie.toggle( showRight, 'disabled' );
}
}
hideRight.onclick = function() {
classie.toggle( this, 'active' );
classie.toggle( menuRight, 'modal-open' );
disableOther( 'hideRight' );
};
function disableOther( button ) {
if( button !== 'hideRight' ) {
classie.toggle( hideRight, 'disabled' );
}
}
</script>
答案 1 :(得分:0)
根据您的问题,当您点击打开并关闭<时,您希望打开或滑入叠加层/ strong>点击关闭时,滑出。
然后这将为你做到:
rest = --- # i am splitting the string in to three(L- as left number cmpr- as compare and R- as right number )
class_name.calc(rest[0],rest[1],rest[2])
def self.calc(L,cmpr,R)
case cmpr
when "<"
if L.to_i < R.to_i
return true
end
....
....
....
end
end
这是一个有效的JSFiddle