Question

我觉得我在这里遗漏了一些关于python进程限制的基本内容。我有一个屏幕刮板，应该每周一次进入受密码保护的站点，填写表格以更新现有记录，然后获取新记录。（我正在使用Django实际插入记录，如果这很重要）。

我正在抓取的数据在一年中逐渐形成。所以在1月，这个过程相对较快。到8月份，除了添加了任何新记录外，还有数千行需要更新。

它今年的工作就像一个梦，但最近开始遇到这个追溯的连接错误：

Traceback (most recent call last):
  File "douglasdivorces.py", line 42, in <module>
    forms = [f for f in br.forms()]
  File "/usr/local/lib/python2.6/dist-packages/mechanize-0.2.4.py2.6.egg/mechanize/_mechanize.py", line 420, in forms
return self._factory.forms()
File "/usr/local/lib/python2.6/dist-packages/mechanize-0.2.4-py2.6.egg/mechanize/_html.py", line 557, in forms
self._forms_factory.forms())
File "/usr/local/lib/python2.6/dist-packages/mechanize-0.2.4-py2.6.egg/mechanize/_html.py", line 237, in forms
_urlunparse=_rfc3986.urlunsplit,
File "/usr/local/lib/python2.6/dist-packages/mechanize-0.2.4-py2.6.egg/mechanize/_form.py", line 844, in ParseResponseEx
_urlunparse=_urlunparse,
File "/usr/local/lib/python2.6/dist-packages/mechanize-0.2.4-py2.6.egg/mechanize/_form.py", line 979, in _ParseFileEx
data = file.read(CHUNK)
File "/usr/local/lib/python2.6/dist-packages/mechanize-0.2.4-py2.6.egg/mechanize/_response.py", line 195, in read
data = self.wrapped.read(to_read)
File "/usr/lib/python2.6/socket.py", line 353, in read
    data = self._sock.recv(left)
    File "/usr/lib/python2.6/httplib.py", line 518, in read
    return self._read_chunked(amt)
  File "/usr/lib/python2.6/httplib.py", line 551, in _read_chunked
    line = self.fp.readline()
  File "/usr/lib/python2.6/socket.py", line 397, in readline
    data = recv(1)
  File "/usr/lib/python2.6/ssl.py", line 96, in <lambda>
    self.recv = lambda buflen=1024, flags=0: SSLSocket.recv(self, buflen, flags)
  File "/usr/lib/python2.6/ssl.py", line 217, in recv
    return self.read(buflen)
  File "/usr/lib/python2.6/ssl.py", line 136, in read
    return self._sslobj.read(len)
socket.error: [Errno 104] Connection reset by peer

你是否可以解决这个错误，保持我的循环到位，直到问题解决？或者我应该采取另一种方法吗？

再一次，我的希望是我缺少学前班级的东西，所以我会免除你发布我的代码所带来的痛苦。如果不那么简单，请说出这个词，我会编辑问题以包含脚本。

非常感谢！非常好奇，听到什么导致我适合！

Answer 1

socket.error：[Errno 104]通过对等方重置连接

即服务器不喜欢您的请求。也许他们改变了一些东西。

Python站点scraper因socket.error而失败104

1 个答案: