如何修复 ValueError: read of closed file 异常? [英] How do I fix a ValueError: read of closed file exception?

问题描述

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)

引发此异常:

Traceback (most recent call last):
  File "C:\Users\ricardo\Desktop\Google-Scholar\BibTex\test2.py", line 8, 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 1597, in retrieve
    block = fp.read(bs)
ValueError: read of closed file

我认为这可能是一个临时问题，所以我添加了一些简单的异常处理，如下所示:

I thought this might be a temporary problem, so I added some simple exception handling like so:

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

但这只是无限打印Downloading....

推荐答案

您的 URL 返回 403 代码错误并且显然 urllib.request.urlretrieve 并不擅长检测所有的 HTTP 错误，因为它使用的是 urllib.request.FancyURLopener 和这个最新的尝试通过返回一个 urlinfo 而不是引发错误.

Your URL return a 403 code error and apparently urllib.request.urlretrieve is not good at detecting all the HTTP errors, because it's using urllib.request.FancyURLopener and this latest try to swallow error by returning an urlinfo instead of raising an error.

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):

About the fix if you still want to use urlretrieve you can override FancyURLopener like this (code included to also show the error):

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，这就是我推荐的你可以使用 urllib.request.urlopen 像这样:

Else and this is what i recommend you can use urllib.request.urlopen like so:

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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