加载页面上的所有图像并等待完成

时间:2019-05-12 15:03:20

标签: javascript dom browser google-chrome-extension lazy-loading

我正在尝试从chrome扩展端获取页面上的所有图像,然后在这些图像上执行一些代码,有时网站会延迟加载图像(例如,当用户向下滚动时)

有什么方法可以确保全部加载?我尝试向下滚动到窗口的底部,但无法使用分页。

export function imagesHaveLoaded() {
  return Array.from(document.querySelectorAll("img")).every(img => img.complete && img.naturalWidth);
}

return Promise.resolve()
      .then(() => (document.scrollingElement.scrollTop = bottom))
      .then(
        new Promise((resolve, reject) => {
          let wait = setTimeout(() => {
            clearTimeout(wait);
            resolve();
          }, 400);
        })
      )
      .then(() => {
        console.log(document.scrollingElement.scrollTop);
        document.scrollingElement.scrollTop = currentScroll;
      })
      .then(
        new Promise((resolve, reject) => {
          let wait = setTimeout(() => {
            clearTimeout(wait);
            resolve();
          }, 400);
        })
      )
      .then(until(imagesHaveLoaded, 2000))
      .then(Promise.all(promises));

它会加载延迟加载的图像,但是如果有更多延迟加载的图像将不起作用(我需要加载图像本身而不是url,这样我才能读取其数据)

1 个答案:

答案 0 :(得分:1)

如果您想捕获加载到页面上的每个新图像,可以使用MutationObserver这样的代码片段:

    const targetNode = document.getElementById("root");

    // Options for the observer (which mutations to observe)
    let config = { attributes: true, childList: true, subtree: true };

    // Callback function to execute when mutations are observed
    const callback = function(mutationsList, observer) {
        for(let mutation of mutationsList) {
            if (mutation.addedNodes[0].tagName==="IMG") {
                console.log("New Image added in DOM!");
            }   
        }
    };

    // Create an observer instance linked to the callback function
    const observer = new MutationObserver(callback);

    // Start observing the target node for configured mutations
    observer.observe(targetNode, config);