如何解析用户在Google评论中发布的图像？

Question

我正在研究一个刮擦项目来刮擦Google Maps评论，但是当我必须解析用户发布的图像时，我感到震惊。我尝试了这种方法，它仅给我提供了用户发布的第一张图片：

    $('.gws-localreviews__google-review').each((i,el) => {
    images[i] = $(el)
    .find(".EDblX .JrO5Xe").attr("style")
    })

我正在通过此URL刮擦Google评论：https://www.google.com/async/reviewdialog?hl = en＆assamp = enc; ：0x8DDDCA9EE380EF7E0，next_page_token：，sort_by：，start_index：，start_index：，ression_topic：，_ fmt：pc

这是我的响应：

[
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipPgClEw3JwTLJOuf-DqC2xtZRodoavkpYVFBYqu=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipOs9TSNoyYmW1GL4SH9PlkAihvWsUbMTn-8O2Sj=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipMBRGdJb3zL1rME20osajG-bosdIV8U82VTYS1n=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipOBuGDXFDhJP69LNo6yI9cZWcjSVHpVfPBNoKyL=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipP7wBZt8Kilm8VF75T8amjMrZ7ZkOpmtb0nHChF=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipOixJabuSd4mSHnveU5JSQ1ZszHJ6Hn-pkeosiY=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipPyOXO1vnyTXVnlkPJNLlnoHYHEna36vYnrqwE=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipPReBboes7S7lNklRT21pwn096JUQVJbTX3VRRA=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipNPLvARJu1vDk03r_y4fp8f7aDDvzRX-7yJklW8=w100-h100-p-n-k-no)',
  'background-image:url(https://lh5.googleusercontent.com/p/AF1QipMW3jp20hjKwuvhogH9ZC8IeH8QhQTUESH_ycNX=w100-h100-p-n-k-no)'
 ]

但是我想要一个包含一个用户映像的对象，然后是一个包含第二个用户图像的对象，然后是一个a的对象在。

我想要这样的结果：

[
{
 "All image's links posted by user 1"
},
{
 "All images links posted by user 2"
}
{
 "All images links posted by user 3"
},
{
 "All images links posted by user 4"
},
{
........
}
]

Answer 1

所以我找到了答案，这有点棘手，但是是的，它将像这样解决：

 $(".gws-localreviews__google-review").each((i, el) => {
     images[i] = $(el)
    .find(".EDblX .JrO5Xe")
    .toArray()
    .map($)
    .map(d => d.attr("style").substring(21 , d.attr("style").lastIndexOf(")")))
     });
 })

它将像这样返回响应：

[
 [
  All images posted by user 1,
 ],
 [
   All images posted by user 2,
 ],
 [.........
]