Find largest image in a website using Puppeteer

user_3392772 0
Viewed 1622

I was using Cheerio to find the largest image inside a webpage. Here is the code I used:

  const { src } = $('img')
      .map((i, el) => ({
        src: el.attribs.src,
        width: el.attribs.width ? Number(el.attribs.width.match(/\d+/)[0]) : -1,
      }))
      .toArray()
      .reduce((prev, current) => (prev.width > current.width ? prev : current));

However, it works only if with width is inline for img. If there is no width I'd to set it's width to -1 and consider it in sorting

Is there any way to find the largest image in a webpage without these hacks, using Puppeteer? Since the browser is rendering these all, it can easily figure out which one is the largest

cheerio puppeteer node.js javascript
2 Answers

You can use page.evaluate() to execute JavaScript within the Page DOM context, and return the src attribute of the largest image back to Node/Puppeteer:

const largest_image = await page.evaluate(() => {
  return [...document.getElementsByTagName('img')].sort((a, b) => b.naturalWidth * b.naturalHeight - a.naturalWidth * a.naturalHeight)[0].src;
});

console.log(largest_image);
user_42835810

You should use the naturalWidth and naturlaHeight properties.

const image = await page.evaluate(() => {

  function size(img) {
    if (!img) {
      return 0;
    }
    return img.naturalWith * img.naturalHeight;
  }

  function info(img) {
    if (!img) {
      return null;
    }
    return {
      src:  img.src,
      size: size(img)
    }
  }

  function largest() {
    let best = null;
    let images = document.getElementsByTagName("img");
    for (let img of images) {
      if (size(img) > size(best)) {
        best = img
      }
    }
    return best;
  }

  return info(largest());
});
user_2159690
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