图像不会在隐藏和显示之间切换 [英] Image will not switch between hidden and show

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本文介绍了图像不会在隐藏和显示之间切换的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我无法获取图像在'隐藏'和'显示'之间来回切换

我正在使用来自$的创意b $ b 如何创建隐藏的< img>在JavaScript?

I'm using ideas from How to create a hidden <img> in JavaScript?

我有两个不同的按钮,使用html尝试一个,另一个使用javascript - 如果我注释掉一行,灯泡照片显示

I have two different buttons, trying one using html and the other to use javascript - If I comment out one line, the light bulb photo is displayed

 //document.getElementById("light").style.visibility = "hidden";

这行代码在我的'init'函数中

如果我不对该线进行评论,那么无论我点击哪个按钮,灯都会保持隐藏状态

If I do not comment the line, the light stays 'hidden', no matter which of the buttons I click

我在Safari控制台中看不到任何错误登录Safari

I don't see any errors in the Web Console Log in Safari

 <!DOCTYPE html>
 <html>
 <body>

 <h1>Switch on the Light</h1>

 <img id="light" src="WebVuCoolOldBulb-2.jpg" style="width:100px" >

 <button type = button

 onclick="document.getElementById('light').src.show ='WebVuCoolOldBulb-2.jpg'"  >Switch On the Light

 </button>
   <input type="button" id="onButton" value="ON" />

 </body>

 <script>

 //document.images['light'].style.visibility = hidden;

  function init() {
     //document.getElementById("light").style.visibility = "hidden";
     var onButton = document.getElementById("onButton");
     onButton.onclick = function() {
        demoVisibility() ;

     }
   }

 function demoVisibility() {
     document.getElementById("light").style.visibility = "show";

 }
  document.addEventListener('readystatechange', function() {
   // Seems like a GOOD PRACTICE - keeps me from getting type error I was getting

     // https://stackoverflow.com/questions/14207922/javascript-error-null-is-not-an-object

     if (document.readyState === "complete") {
       init();
     }
   });



 </script>
 </html>


推荐答案

visibility style属性的值为 visible 隐藏

没有 show value。

function init() {
     document.getElementById("light").style.visibility = "hidden";
     var onButton = document.getElementById("onButton");
     onButton.onclick = function() {
        demoVisibility();
     }
   }

 function demoVisibility() {
     document.getElementById("light").style.visibility = "visible";
 }

这篇关于图像不会在隐藏和显示之间切换的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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