如何从多个URL中使用urllib.request.urlopen获取所有图像URL [英] How to get all image urls with urllib.request.urlopen from multiple urls
本文介绍了如何从多个URL中使用urllib.request.urlopen获取所有图像URL的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
from bs4 import BeautifulSoup
import urllib.request
urls = [
"https://archillect.com/1",
"https://archillect.com/2",
"https://archillect.com/3",
]
soup = BeautifulSoup(urllib.request.urlopen(urls))
for u in urls:
for img in soup.find_all("img", src=True):
print(img["src"])
AttributeError:列表"对象没有属性超时"
AttributeError: 'list' object has no attribute 'timeout'
推荐答案
@krishna给出了答案.我给您另一种解决方案仅供参考.
@krishna has given you the answer. I'll give you another solution for reference only.
from simplified_scrapy import Spider, SimplifiedDoc, SimplifiedMain, utils
class ImageSpider(Spider):
name = 'archillect'
start_urls = ["https://archillect.com/1","https://archillect.com/2","https://archillect.com/3"]
def afterResponse(self, response, url, error=None, extra=None):
try:
# Create file name
end = url.find('?') if url.find('?')>0 else len(url)
name = 'data'+url[url.rindex('/',0,end):end]
# save image
if utils.saveResponseAsFile(response,name,'image'):
return None
else:
return Spider.afterResponse(self, response, url, error)
except Exception as err:
print (err)
def extract(self,url,html,models,modelNames):
doc = SimplifiedDoc(html)
urls = doc.listImg(url=url.url)
return {'Urls':urls}
SimplifiedMain.startThread(ImageSpider()) # Start
还有更多示例: https://github.com/yiyedata/simplified-scrapy-demo/tree/master/spider_examples
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