本文介绍了检测< link>资源?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
浏览器为 / code>在我的情况下不可行。
Unfortunately, I think using a sentinel style and detecting load from a computedStyle isn't workable in my situation.
推荐答案
可能有更简单的方法,
There may be a simpler way to do it, but this worked for me.
请确保您的< link> 标记有 属性:
Make sure your <link> tag has a title attribute:
<link title="myStyles" rel="stylesheet" href="style.css" />
然后使用这样的函数检查特定样式集中是否存在样式: p>
Then use a function like this to check for the presence of a style within a particular styleseet:
function linkLoaded(linkTitle, checkStyle) { for (var ix=0; ix<document.styleSheets.length; ix++) { try { if (document.styleSheets[ix].title == linkTitle) { var mySheet=document.styleSheets[ix]; var myRules = mySheet.cssRules? mySheet.cssRules: mySheet.rules for (var jx=0; jx<myRules.length; jx++) { var thisSelector = myRules[jx].selectorText.toLowerCase(); if (thisSelector.substring(0, checkStyle.length) == checkStyle) { alert("Found style!"); return; } } } } catch (err) {} } alert("Not loaded!"); return; }
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