如何修复ValueError：读取关闭的文件异常？

这个简单的Python 3脚本：

import urllib.request

host = "scholar.google.com"
link = "/scholar.bib?q=info:K7uZdMSvdQ0J:scholar.google.com/&output=citation&hl=en&as_sdt=1,14&ct=citation&cd=0"
url = "http://" + host + link
filename = "cite0.bib"
print(url)
urllib.request.urlretrieve(url, filename)

引发了这个例外：

Traceback (most recent call last):
  File "C:UsersricardoDesktopGoogle-ScholarBibTextest2.py", line 8, in <module>
    urllib.request.urlretrieve(url, filename)
  File "C:Python32liburllibrequest.py", line 150, in urlretrieve
    return _urlopener.retrieve(url, filename, reporthook, data)
  File "C:Python32liburllibrequest.py", line 1597, in retrieve
    block = fp.read(bs)
ValueError: read of closed file

我认为这可能是一个暂时的问题，所以我添加了一些简单的异常处理，如下所示：

import random
import time
import urllib.request

host = "scholar.google.com"
link = "/scholar.bib?q=info:K7uZdMSvdQ0J:scholar.google.com/&output=citation&hl=en&as_sdt=1,14&ct=citation&cd=0"
url = "http://" + host + link
filename = "cite0.bib"
print(url)
while True:
    try:
        print("Downloading...")
        time.sleep(random.randint(0, 5))
        urllib.request.urlretrieve(url, filename)
        break
    except ValueError:
        pass

但这只是打印Downloading...无限。

---

你的URL返回一个403代码错误，显然urllib.request.urlretrieve不擅长检测所有的HTTP错误，因为它使用了urllib.request.FancyURLopener并且这个最新的尝试通过返回一个urlinfo而不是引发错误来吞服错误。

关于修复，如果你仍然想使用urlretrieve，你可以像这样覆盖FancyURLopener（包含的代码也显示错误）：

import urllib.request
from urllib.request import FancyURLopener


class FixFancyURLOpener(FancyURLopener):

    def http_error_default(self, url, fp, errcode, errmsg, headers):
        if errcode == 403:
            raise ValueError("403")
        return super(FixFancyURLOpener, self).http_error_default(
            url, fp, errcode, errmsg, headers
        )

# Monkey Patch
urllib.request.FancyURLopener = FixFancyURLOpener

url = "http://scholar.google.com/scholar.bib?q=info:K7uZdMSvdQ0J:scholar.google.com/&output=citation&hl=en&as_sdt=1,14&ct=citation&cd=0"
urllib.request.urlretrieve(url, "cite0.bib")

否则，这是我建议你可以像这样使用urllib.request.urlopen ：

fp = urllib.request.urlopen('http://scholar.google.com/scholar.bib?q=info:K7uZdMSvdQ0J:scholar.google.com/&output=citation&hl=en&as_sdt=1,14&ct=citation&cd=0')
with open("citi0.bib", "w") as fo:
    fo.write(fp.read())

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