【发布时间】:2020-11-05 09:26:03
【问题描述】:
我试图获取我们可以在the following site 或其他人的照片下方看到的元素,等效:
- https://www.nosetime.com/xiangshui/947895-oulong-xuecheng-atelier-cologne-orange.html
- https://www.nosetime.com/xiangshui/705357-pomelo-paradis.html
- https://www.nosetime.com/xiangshui/592260-cl-mentine-california.html
- https://www.nosetime.com/xiangshui/612353-oulong-atelier-cologne-trefle.html
- https://www.nosetime.com/xiangshui/911317-oulong-nimingmeigui-atelier-cologne.html
但我无法从源代码中获取它。它应该使用 javascript 脚本动态下载。事实上它似乎在一个 xhr 文档中:
那么如何获取访问页面时下载的xhr文档呢?
我试过了:
url = "https://www.nosetime.com/xiangshui/350870-oulong-atelier-cologne-oolang-infini.html"
r = requests.post(url, headers=headers)
data = r.json()
print(data)
Pero me develve:
---------------------------------------------------------------------------
JSONDecodeError Traceback (most recent call last)
<ipython-input-8-e72156ddb336> in <module>()
2
3 r = requests.post(url, headers=headers)
----> 4 data = r.json()
5
6 print(data)
3 frames
/usr/lib/python3.6/json/decoder.py in raw_decode(self, s, idx)
355 obj, end = self.scan_once(s, idx)
356 except StopIteration as err:
--> 357 raise JSONDecodeError("Expecting value", s, err.value) from None
358 return obj, end
JSONDecodeError: Expecting value: line 1 column 1 (char 0)
【问题讨论】:
标签: python-3.x ajax web-scraping xmlhttprequest