Python，基本问题：如何使用urllib.request.urlreter下载多个url

Question

我有以下功能齐全的工作代码：

import urllib.request
import zipfile

url = "http://url.com/archive.zip?key=7UCxcuCzFpYeu7tz18JgGZFAAgXQ2sop"
filename = "C:/test/archive.zip"
destinationPath = "C:/test"

urllib.request.urlretrieve(url,filename)
sourceZip = zipfile.ZipFile(filename, 'r')

for name in sourceZip.namelist():
    sourceZip.extract(name, destinationPath)
sourceZip.close()

它可以完美地工作几次，但由于从中检索文件的服务器有一些限制，因此一旦达到每日限制，就会出现此错误：

Traceback (most recent call last):
  File "script.py", line 11, in <module>
    urllib.request.urlretrieve(url,filename)
  File "C:\Python32\lib\urllib\request.py", line 150, in urlretrieve
    return _urlopener.retrieve(url, filename, reporthook, data)
  File "C:\Python32\lib\urllib\request.py", line 1591, in retrieve
    block = fp.read(bs)
ValueError: read of closed file

如何更改脚本，使其包含多个url的列表，而不是一个url，并且脚本继续尝试从列表中下载，直到一个成功，然后继续解压缩。我只需要一个成功的下载。

很抱歉对Python很陌生，但我想不出这一点。我假设我必须改变变量，使其看起来像这样：

url = {
"http://url.com/archive.zip?key=7UCxcuCzFpYeu7tz18JgGZFAAgXQ2soe",
"http://url.com/archive.zip?key=7UCxcuCzFpYeu7tz18JgGZFAAgXQ2sod",
"http://url.com/archive.zip?key=7UCxcuCzFpYeu7tz18JgGZFAAgXQ2soc",
"http://url.com/archive.zip?key=7UCxcuCzFpYeu7tz18JgGZFAAgXQ2sob",
"http://url.com/archive.zip?key=7UCxcuCzFpYeu7tz18JgGZFAAgXQ2soa",
}

然后把这条线变成某种循环：

urllib.request.urlretrieve(url,filename)

Answer 1

你想把你的网址放在一个列表中，然后在这个列表中循环并尝试每一个。您捕获但忽略它们抛出的异常，并在一个异常成功后中断循环。试试这个：

import urllib.request
import zipfile

urls = ["http://url.com/archive.zip?key=7UCxcuCzFpYeu7tz18JgGZFAAgXQ2sop", "other url", "another url"]
filename = "C:/test/test.zip"
destinationPath = "C:/test"

for url in urls:
    try:
        urllib.request.urlretrieve(url,filename)
        sourceZip = zipfile.ZipFile(filename, 'r')
        break
    except ValueError:
        pass

for name in sourceZip.namelist():
    sourceZip.extract(name, destinationPath)
sourceZip.close()