在页面已加载后加载脚本时重新初始化dom

2024-05-17 • 问答

在我的页面上，我阻止脚本并想在按下按钮后加载它们。

以Google Recaptcha为例，这已经很好用了。

当我运行myFunction时，脚本已加载。

jquery和popper.js一样

但是我认为要使用bootstrap下拉列表，不需要重新初始化它。有没有办法解决这个问题？我什么都找不到。

代码如下：

    <script>
        function blockScript(origin) {
            const scripts = Array.from(document.getElementsByTagName("SCRIPT"));
            if (scripts.length > 0) {
                scripts.forEach((script) => {
                    if (script.src.includes(origin)) {
                        script.setattribute('data-src',script.src);
                        script.removeAttribute("src");  
                    }
                })
            }
        }
    </script>
    <script>

      var json = [
{
    "name" : "Google Recaptcha","url"   : "code.jquery.com","desc" : "Recaptcha","rights": ""
},{
    "name" : "Google Recaptcha","url"   : "www.google.com","url"   : "cdnjs.cloudflare.com","rights": ""
}];
      if(!localStorage.getItem('cc-marketing') == 1){
            const observer = new MutationObserver((mutationsList,observer) => {
                for(let mutation of mutationsList) {
                    const addedNodes = Array.from(mutation.addedNodes);
                    if (addedNodes && addedNodes.some(n => n.nodename === 'SCRIPT')) {
                      for(var i = 0; i < json.length; i++) {
        var obj = json[i];
        blockScript(obj.url);
    }
                    }
                }
            });

            observer.observe(document,{ childList: true,subtree: true });
              }
    </script>

<script>
function myFunction() {
 // window.localStorage.setItem('cc-marketing',1);
var elements = document.querySelectorAll("[data-src]");

Object.keys(elements).forEach(function(key) {
  console.log(elements[key].getattribute('data-src'));
  elements[key].setattribute('src',elements[key].getattribute('data-src'));
  elements[key].removeAttribute("data-src");  
});
}
</script>

<script  integrity="sha384-q8i/X+965DzO0rT7abK41JStQIAqVgRVzpbzo5smXKp4YfRvH+8abtTE1Pi6jizo" crossorigin="anonymous" src="https://code.jquery.com/jquery-3.3.1.slim.min.js"></script>
<script  integrity="sha384-UO2eT0CpHqdSJQ6hJty5KVphtPhzWj9WO1clHTMGa3JDZwrnqq4sF86dIHNDz0W1" crossorigin="anonymous" src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.14.7/umd/popper.min.js"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.3.1/js/bootstrap.min.js" integrity="sha384-JjSmvgyd0p3pXB1rRibZUAYoIIy6OrQ6VrjIEaFf/nJGzIxFDsf4x0xIM+B07jRM" crossorigin="anonymous"></script>
<script src="https://www.google.com/recaptcha/api.js?render=MYKEY"></script>

问题不在于刷新dom。加载脚本时，它将向已加载的dom添加任何侦听器，因此可以正常工作。

问题是您的bootstrap.min.js在popper.min.js之前被加载，因为您阻止了popper但没有引导。根据引导程序文档，bootstrap.min.js需要最后加载。（依赖项）。只需在您的json中添加bootstrap.min.js来阻止它，它就会起作用。

   <script>
        function blockScript(origin) {
            const scripts = Array.from(document.getElementsByTagName("SCRIPT"));
            if (scripts.length > 0) {
                scripts.forEach((script) => {
                    if (script.src.includes(origin)) {
                        script.setAttribute('data-src',script.src);
                        script.removeAttribute("src");  
                    }
                })
            }
        }
    </script>
    <script>

      var json = [
{
    "name" : "Google Recaptcha","url"   : "code.jquery.com","desc" : "Recaptcha","rights": ""
},{
    "name" : "Google Recaptcha","url"   : "www.google.com","url"   : "cdnjs.cloudflare.com",{
    "name" : "Bootstrap","url"   : "stackpath.bootstrapcdn.com","desc" : "Bootstrap","rights": ""
}];
      if(!localStorage.getItem('cc-marketing') == 1){
            const observer = new MutationObserver((mutationsList,observer) => {
                for(let mutation of mutationsList) {
                    const addedNodes = Array.from(mutation.addedNodes);
                    if (addedNodes && addedNodes.some(n => n.nodeName === 'SCRIPT')) {
                      for(var i = 0; i < json.length; i++) {
        var obj = json[i];
        blockScript(obj.url);
    }
                    }
                }
            });

            observer.observe(document,{ childList: true,subtree: true });
              }
    </script>

<script>
function myFunction() {
 // window.localStorage.setItem('cc-marketing',1);
var elements = document.querySelectorAll("[data-src]");

Object.keys(elements).forEach(function(key) {
  console.log(elements[key].getAttribute('data-src'));
  elements[key].setAttribute('src',elements[key].getAttribute('data-src'));
  elements[key].removeAttribute("data-src");  
});
}
</script>

<script  integrity="sha384-q8i/X+965DzO0rT7abK41JStQIAqVgRVzpbzo5smXKp4YfRvH+8abtTE1Pi6jizo" crossorigin="anonymous" src="https://code.jquery.com/jquery-3.3.1.slim.min.js"></script>
<script  integrity="sha384-UO2eT0CpHqdSJQ6hJty5KVphtPhzWj9WO1clHTMGa3JDZwrnQq4sF86dIHNDz0W1" crossorigin="anonymous" src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.14.7/umd/popper.min.js"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.3.1/js/bootstrap.min.js" integrity="sha384-JjSmVgyd0p3pXB1rRibZUAYoIIy6OrQ6VrjIEaFf/nJGzIxFDsf4x0xIM+B07jRM" crossorigin="anonymous"></script>
<script src="https://www.google.com/recaptcha/api.js?render=MYKEY"></script>

在页面已加载后加载脚本时重新初始化dom

jianjunliqi 回答：在页面已加载后加载脚本时重新初始化dom

大家都在问