我正在练习python,我有一个包含URL的列表,我需要转到URL并将页面内容保存在save_folder
list_name = ['www.xyz.com/abc/sample1.txt','www.xyz.com/abc/sample2.txt','www.xyz.com/abc/sample44.txt']
for i in list_name:
http = urllib3.PoolManager()
r = http.request('get', i)
with open('save_folder/' + i,'w') as f:
f.write(r.data)
f.close
运行上面的代码时,我得到一个错误没有这样的文件或目录:'save_folder / www.xyz.com / abc / sample1.txt
目标是读取列表中的三个项目,并使用urllib3转到列表中的每个页面,并将每个页面保存在save_folder中,文件名为sample1.txt,sample2.txt,sample44.txt
答案 0 :(得分:2)
仅使用文件名,并确保可访问URL。
for i in list_name:
http = urllib3.PoolManager()
r = http.request('get', i)
fname = i.split("/")[-1]
with open('./save_folder/' + fname,'wb') as f:
f.write(r.data)
f.close