BeautifulSoup如何從.. img src獲取網址../ ..？

Question

因此，可以說，我試圖讓鏈接到一個特定的形象，就像這樣：

from bs4 import BeautfiulSoup 
import urlparse 

soup = BeautifulSoup("http://examplesite.com") 
for image in soup.findAll("img"): 
    srcd = urlparse.urlparse(src) 
    path = srcd.path # gets the path 
    fn = os.path.basename(path) # gets filename 

# lets say the webpage i was scraping had their images like this: 
# <img src="../..someimage.jpg" /> 

有沒有簡單的方法來得到的完整網址？或者我將不得不使用正則表達式？

Answer 1

使用urlparse.urljoin：

>>> import urlparse 
>>> base_url = "http://example.com/foo/" 
>>> urlparse.urljoin(base_url, "../bar") 
'http://example.com/bar' 
>>> urlparse.urljoin(base_url, "/baz") 
'http://example.com/baz'