Question

<style>
    #container { 
    width: 320px; 
    height: 400px;
    border: 1px solid red;

    }
    #top { 
    width: 320px; 
    height: 80px;
    border: 1px solid blue;

    }
</style>
<script type="text/javascript" src="One.js"> </script>

</head>
<body>

<div id="top">
<button onclick="start()">Start</button>
<button onclick="stop()">Stop</button>
<button onclick="pause()">Pause</button>

</div>


<div style="position: relative;" id="container">

 <canvas id="myCanvas" width="320" height="400" 
   style="position: absolute; left: 0; top: 0; z-index: 0;"></canvas>
    <canvas id="plane" width="320" height="400" 
   style="position: absolute; left: 0; top: 0; z-index: 0;"></canvas>
    <canvas id="buildings" width="320" height="400" 
   style="position: absolute; left: 0; top: 0; z-index: 0;"></canvas>
       <canvas id="cloud" width="320" height="400" 
   style="position: absolute; left: 0; top: 0; z-index: 0;"></canvas>

    <canvas id="buildings2" width="320" height="400" 
   style="position: absolute; left: 0; top: 0; z-index: 0;"></canvas>

 <canvas id="canvas2" width="100" height="100" 
   style="position: absolute; left: 0; top: 0; z-index: 1;"></canvas>
</div>

所以我希望能够调用将运行画布动画的start函数。当我在html页面上有js时，我的代码完全正常工作。但是当它作为外部脚本时，我无法让它工作。谢谢

Answer 1

将<script type="text/javascript" src="One.js"> </script>放在</body>标记之前。那会有所帮助。在运行Javascript文件之前必须加载DOM，这就是为什么最好把它放在最后，当你确定已经加载了所有东西时。

调用外部js函数

1 个答案: