我正在尝试创建一个小部件。像google-adsense块之类的东西,所以我需要从页面的不同位置调用相同的脚本。是否可以在外部js脚本中创建一个元素并将其插入我调用脚本的地方?
我这样做:
<div><script type="text/javascript" src="loader.js"></div>
<div><script type="text/javascript" src="loader.js"></div>
Loader.js:
var div = document.createElement('div');
div.setAttribute('class', 'inner-div');
document.body.appendChild(div);
我有:
<div><script type="text/javascript" src="loader.js"></div>
<div><script type="text/javascript" src="loader.js"></div>
<div class="inner-div"></div>
<div class="inner-div"></div>
但我想得到这个结果:
<div>
<script type="text/javascript" src="loader.js">
<div class="inner-div"></div>
</div>
<div>
<script type="text/javascript" src="loader.js">
<div class="inner-div"></div>
</div>
答案 0 :(得分:0)
我想我找到了解决方案。查找当前脚本标记:
<div>
<script type="text/javascript">
var scripts = document.getElementsByTagName('script');
var script = scripts[scripts.length - 1];
loadScript('loader.js', function () {
loader().init({script: script});
});
</script>
</div>
loadScript函数:
function loadScript(url, callback){
var script = document.createElement("script")
script.type = "text/javascript";
if (script.readyState){ //IE
script.onreadystatechange = function(){
if (script.readyState == "loaded" ||
script.readyState == "complete"){
script.onreadystatechange = null;
callback();
}
};
} else { //Others
script.onload = function(){
callback();
};
}
script.src = url;
document.getElementsByTagName("head")[0].appendChild(script);
}
然后在loader.js中:
init: function(args) {
...
var div = document.createElement('div');
div.setAttribute('class', 'inner-div');
args.script.parentNode.appendChild(div);
...
}