2014-04-04 40 views
2

我有隐藏的图像,并显示它

var sanmigBottle = document.createElement('sanmigBottle'); 
sanmigBottle.id="sanmigBottle"; 
sanmigBottle.width="25%"; 
sanmigBottle.src='sanmigbottle.png'; 
sanmigBottle.onmousedown="showLibraryInfo()" ; 
sanmigBottle.ontouchstart="showLibraryInfo()" ; 
sanmigBottle.align = "middle"; 
sanmigBottle.style= "-webkit-transform: rotate(350.77259795507035deg) translateZ(0px);'"; 

它加载在我的HTML,但我想隐藏它,我希望有一个触发显现。当我输入这个时,

sanmigBottle.style.visibility="hidden"; 

它不起作用。我怎样才能做到这一点?我对JS很感兴趣。

回答

0

您应该使用

sanmigBottle.style.display = 'none'; 
+0

它仍然显示。 :( – Jeongbebs

+0

你可以让一个JSFiddle所以我们可以经过? –

0

你必须设置displaynone

sanmigBottle.style.display="none"; 

然后将其设置为block或重复的事情,使之再次可见:

sanmigBottle.style.display="block"; 
+0

它仍然显示。:( – Jeongbebs

+0

@MiguelRivera也许你可以设置一个[jsfiddle](http://jsfiddle.net/)?那样我们就可以理解为什么它不管用。 – RononDex

0

首先检查图像的状态:

function showhide() { 
    var img = document.getElementById('someimage'); 
    if (img.style.visibility === 'hidden') { 
     // Currently hidden, make it visible 
     img.style.visibility = "visible"; 
    } else { 
     // Currently visible, make it hidden 
     img.style.visibility = "hidden"; 
    } 
} 

然后应用可见性。

-1

尝试使用jQuery

//create element 
var sanmigBottle = $('<img id="sanmigBottle" width="25%" src='sanmigbottle.png' onmousedown="showLibraryInfo()" ontouchstart="showLibraryInfo()" align = "middle" style= "-webkit-transform: rotate(350.77259795507035deg) translateZ(0px);"/>'); 
//add to document and store object in context 
var context = sanmigBottle.appendTo($('Document')); 
//hide object 
context.fadeOut(); 

demo in jsfiddle