Question

我需要检查（在Javascript中）是否加载了CSS文件，如果没有，则加载它。 jQuery很好。

Answer 1

只需检查是否存在<link>元素，并将href属性设置为CSS文件的网址：

if (!$("link[href='/path/to.css']").length)
    $('<link href="/path/to.css" rel="stylesheet">').appendTo("head");

使用document.styleSheets collection：

，简单的JS方法也很简单

function loadCSSIfNotAlreadyLoadedForSomeReason () {
    var ss = document.styleSheets;
    for (var i = 0, max = ss.length; i < max; i++) {
        if (ss[i].href == "/path/to.css")
            return;
    }
    var link = document.createElement("link");
    link.rel = "stylesheet";
    link.href = "/path/to.css";

    document.getElementsByTagName("head")[0].appendChild(link);
}
loadCSSIfNotAlreadyLoadedForSomeReason();

Answer 2

我只需要编写类似的东西，我想分享它。这个是为多个案件准备的。

如果没有css文件请求（css文件没有链接......）
如果有css文件请求，但是如果失败（css文件不再可用......）

var styles = document.styleSheets;
for (var i = 0; i < styles.length; i++) {
    // checking if there is a request for template.css
    if (styles[i].href.match("template")) {
        console.log("(Iteration: " + i + ") Request for template.css is found.");
        // checking if the request is not successful
        // when it is successful .cssRules property is set to null
        if (styles[i].cssRules != null && styles[i].cssRules.length == 0) {
            console.log("(Iteration: " + i + ") Request for template.css failed.");
            // fallback, make your modification
            // since the request failed, we don't need to iterate through other stylesheets
            break;
        } else {
            console.log("(Iteration: " + i + ") Request for template.css is successful.");
            // template.css is loaded successfully, we don't need to iterate through other stylesheets
            break;
        }
    }
    // if there isn't a request, we fallback
    // but we need to fallback when the iteration is done
    // because we don't want to apply the fallback each iteration
    // it's not like our css file is the first css to be loaded
    else if (i == styles.length-1) {
        console.log("(Iteration: " + i + ") There is no request for template.css.");
        // fallback, make your modification
    }
}

TL; DR版

var styles = document.styleSheets;
for (var i = 0; i < styles.length; i++) {
    if (styles[i].href.match("css-file-name-here")) {
        if (styles[i].cssRules != null && styles[i].cssRules.length == 0) {
            // request for css file failed, make modification
            break;
        }
    } else if (i == styles.length-1) {
        // there is no request for the css file, make modification
    }
}

更新：由于我的回答得到了一些upvotes，这导致我修改了代码，我决定更新它。

// document.styleSheets holds the style sheets from LINK and STYLE elements
for (var i = 0; i < document.styleSheets.length; i++) {

    // Checking if there is a request for the css file
    // We iterate the style sheets with href attribute that are created from LINK elements
    // STYLE elements don't have href attribute, so we ignore them
    // We also check if the href contains the css file name
    if (document.styleSheets[i].href && document.styleSheets[i].href.match("/template.css")) {

        console.log("There is a request for the css file.");

        // Checking if the request is unsuccessful
        // There is a request for the css file, but is it loaded?
        // If it is, the length of styleSheets.cssRules should be greater than 0
        // styleSheets.cssRules contains all of the rules in the css file
        // E.g. b { color: red; } that's a rule
        if (document.styleSheets[i].cssRules.length == 0) {

            // There is no rule in styleSheets.cssRules, this suggests two things
            // Either the browser couldn't load the css file, that the request failed
            // or the css file is empty. Browser might have loaded the css file,
            // but if it's empty, .cssRules will be empty. I couldn't find a way to
            // detect if the request for the css file failed or if the css file is empty

            console.log("Request for the css file failed.");

            // There is a request for the css file, but it failed. Fallback
            // We don't need to check other sheets, so we break;
            break;
        } else {
            // If styleSheets.cssRules.length is not 0 (>0), this means 
            // rules from css file is loaded and the request is successful
            console.log("Request for the css file is successful.");
            break;
        }
    }
    // If there isn't a request for the css file, we fallback
    // But only when the iteration is done
    // Because we don't want to apply the fallback at each iteration
    else if (i == document.styleSheets.length - 1) {
        // Fallback
        console.log("There is no request for the css file.");
    }
}

TL; DR

for (var i = 0; i < document.styleSheets.length; i++) {
    if (document.styleSheets[i].href && document.styleSheets[i].href.match("/template.css")) {
        if (document.styleSheets[i].cssRules.length == 0) {
            // Fallback. There is a request for the css file, but it failed.
            break;
        }
    } else if (i == document.styleSheets.length - 1) {
        // Fallback. There is no request for the css file.
    }
}

Answer 3

与JFK就接受的答案所作的评论表示同意：

我将问题理解为＆＃34;如何检查css文件是否是   是否加载＆＃34;而不是＆＃34;如何检查元素   存在＆＃34;

元素可能存在（并且路径也可能是正确的），但它   并不意味着css文件已成功加载。

如果您通过getElementById访问链接元素，则无法检查/读取CSS文件中定义的样式。

为了检查样式是否已成功加载，我们必须使用getComputedStyle（或currentStyle用于IE）。

<强> HTML

//somewhere in your html document

<div id="css_anchor"></div>

<强> CSS

//somewhere in your main stylesheet

#css_anchor{display:none;}

<强> JAVASCRIPT

//js function to check the computed value of a style element

function get_computed_style(id, name){

        var element = document.getElementById(id);

        return element.currentStyle ? element.currentStyle[name] : window.getComputedStyle ? window.getComputedStyle(element, null).getPropertyValue(name) : null;

}

 //on document ready check if #css_anchor has been loaded

    $(document).ready( function() {

            if(get_computed_style('css_anchor', 'display')!='none'){

            //if #css_anchor style doesn't exist append an alternate stylesheet

                var alternateCssUrl = 'http://example.com/my_alternate_stylesheet.css';

                var stylesheet = document.createElement('link');

                stylesheet.href = alternateCssUrl;
                stylesheet.rel = 'stylesheet';
                stylesheet.type = 'text/css';
                document.getElementsByTagName('head')[0].appendChild(stylesheet);

            }
    });

部分答案来自：myDiv.style.display returns blank when set in master stylesheet

在这里演示：http://jsfiddle.net/R9F7R/

Answer 4

这样的事情（使用jQuery）：

function checkStyleSheet(url){
   var found = false;
   for(var i = 0; i < document.styleSheets.length; i++){
      if(document.styleSheets[i].href==url){
          found=true;
          break;
      }
   }
   if(!found){
       $('head').append(
           $('<link rel="stylesheet" type="text/css" href="' + url + '" />')
       );
   }
}

Answer 5

除了上面所有的好答案之外，您可以简单地在您的标记和css文件中放置一个虚拟元素，为其提供除默认之外的任何样式。然后在代码中检查属性是否应用于虚拟元素，如果不是，则加载css。只是一个想法，而不是一个干净的方式做你想做的事情。

Answer 6

我的2美分。这将检查是否在css上设置了任何规则，这意味着它已成功加载

if(jQuery("link[href='/style.css']").prop('sheet').cssRules.length == 0){
    //Load the css you want
}

Answer 7

文档对象包含一个样式表集合，其中包含所有已加载的样式表。

有关参考，请参阅http://www.javascriptkit.com/domref/stylesheet.shtml

您可以循环此集合以验证您要验证的样式表是否在其中并由浏览器加载。

document.styleSheets[0] //access the first external style sheet on the page

但是您应该注意一些浏览器不兼容性。

Answer 8

一种方法：使用document.getElementsByTagName("link")迭代每个方法并检查其href是否等于您检查的CSS文件。

另一种方式：如果您知道某些CSS规则仅在该文件中设置，请检查此规则是否真的适用于检查某物的背景是否真的是红色。

Answer 9

var links = document.getElementsByTagName('link');
var file  = 'my/file.css';
var found = false;

for ( var i in links )
{
    if ( links[i].type == 'text/css' && file == links[i].href ) {
        found = true; break;
    }
}

if ( !( found ) ) {
    var styles = document.getElementsByTagName('style');
    var regexp = new RegExp('/\@import url\("?' + file + '"?\);/');

    for ( var i in styles )
    {
        if ( styles[i].src == file ) {
            found = true; break;
        } else if ( styles[i].innerHTML.match(regexp) ) {
            found = true; break;
        }
    }
}

if ( !( found ) ) {
    var elm = document.createElement('link');
        elm.href = file;
    document.documentElement.appendChild(elm);
}

Answer 10

您可以检查文件名是否在您的标记内，例如：

var lnks    = document.getElementsByTagName('link'),
    loadcss = true;

for(var link in lnks) {
    href = link.getAttribute('href');

    if( href.indexOf('foooobar.css') > -1) ){
            loadcss = false;
            return false;
    }
});

if( loadcss ) {
        var lnk     = document.createElement('link'),
            head    = document.getElementsByTagName('head')[0] || document.documentElement;        

        lnk.rel     = 'stylesheet';
        lnk.type    = 'text/css';
        lnk.href    = '//' + location.host + 'foooobar.css';            

        head.insertBefore(lnk, head.firstChild);
}

或者如果加载了样式表，您可以检查应该可用的特定className。这可能更接近于特征检测。

Answer 11

为了获得良好的一致性和可重复性，我编写了两个模仿$.getScript(url, callback) jQuery方法的jQuery插件（但是它们不会强制从服务器重新加载$.getScript()。有两种方法：一个将在调用时加载CSS文件，一个只加载一次的文件。我在开发过程中发现前者在进行更改时非常方便，而后者非常适合快速部署。

/**
 * An AJAX method to asynchronously load a CACHED CSS resource
 * Note: This removes the jQuery default behaviour of forcing a refresh by means
 * of appending a datestamp to the request URL. Actual caching WILL be subject to
 * server/browser policies
 */
$.getCachedCss = function getCachedCss(url, callback)
{
    $('<link>',{rel:'stylesheet', type:'text/css', 'href':url, media:'screen'}).appendTo('head');

    if (typeof callback == 'function')
        callback();
}

/**
 * An AJAX method to asynchronously load a CACHED CSS resource Only ONCE.
 * Note: This removes the jQuery default behaviour of forcing a refresh by means
 * of appending a datestamp to the request URL. Actual caching WILL be subject to
 * server/browser policies
 */
$.getCachedCssOnce = function getCachedCssOnce(url, callback)
{
    if (!$("link[href='" + url + "']").length) {
        $.getCachedCss(url, callback);

        if (typeof callback == 'function')
            callback();
    }
}

用法示例：

$(function() {
    $.getCachedCssOnce("pathToMyCss/main.css");
)}

回调的用法示例：

$(function() {
    $.getCachedCssOnce("pathToMyCss/main.css", function() {
        // Do something once the CSS is loaded
});

Answer 12

在jQuery中使用.sheet：

HTML：

<link rel="stylesheet" href="custom.css">

jQuery的：

if($("link[href='custom.css']")[0].sheet.cssRules.length==0){
//custom.css was not loaded, do your backup loading here
}

Answer 13

使用javascript ..

的简单方法

loadCssIfNotLoaded('https://maxcdn.bootstrapcdn.com/bootstrap/4.1.3/css/bootstrap.min.css');
loadCssIfNotLoaded('https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.7.0/css/font-awesome.min.css');

function loadCssIfNotLoaded(url) {
    var element=document.querySelectorAll('link[href="' + url + '"]');
    if (element.length == 0)
    {
        var link = document.createElement('link');
        link.rel = 'stylesheet';
        link.href = url;
        document.getElementsByTagName("head")[0].appendChild(link);
    }
}

Answer 14

使用jQuery的一行。如果可见#witness div，则必须加载css文件。

在HTML中，我们有一个：

<div id="witness"></div>

在要加载的CSS文件中，我们有：

  #witness{display:none;}

因此，如果加载了CSS文件，则#witness div不可见。我们可以使用jQuery进行检查并做出决定。

!$('#witness').is(':visible') || loadCss() ;

作为摘要：

function loadCss(){
  //...
  console.log('Css file required');
};


!$('#witness').is(':visible') || loadCss();

#witness{display:none;}

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>

<div id="witness"></div>

如果没有加载，如何使用Javascript检查和加载CSS？

14 个答案: