k.r*_*cki 19 python django url path python-2.7
嗨,我需要将网址转换为路径,我得到的是这个网址如下:
url = u'/static/media/uploads/gallery/Marrakech%2C%20Morocco_be3Ij2N.jpg'
什么看起来像这样:
path = u'/static/media/uploads/gallery/Marrakech, Morocco_be3Ij2N.jpg'
谢谢.
fal*_*tru 20
使用urllib.unquote解码%编码过的字符串:
>>> import urllib
>>> url = u'/static/media/uploads/gallery/Marrakech%2C%20Morocco_be3Ij2N.jpg'
>>> urllib.unquote(url)
u'/static/media/uploads/gallery/Marrakech, Morocco_be3Ij2N.jpg'
使用urllib.quote或urllib.quote_plus,你可以回来:
>>> urllib.quote(u'/static/media/uploads/gallery/Marrakech, Morocco_be3Ij2N.jpg')
'/static/media/uploads/gallery/Marrakech%2C%20Morocco_be3Ij2N.jpg'
dta*_*tar 14
如果您使用的是Python3，您可以编写
urllib.parse.unquote(url)
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