处理python爬虫的timeout报错

最新推荐文章于 2025-06-18 03:35:38 发布

原创最新推荐文章于 2025-06-18 03:35:38 发布 · 1.8w 阅读

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CC 4.0 BY-SA版权

在执行Python爬虫过程中遇到'.timeout'错误，原因是未对请求设置合适的超时处理。错误发生在尝试读取HTTP响应内容时，提示socket超时。解决方法是在请求中添加超时参数，以防止因长时间等待响应导致的程序中断。

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尽管添加了

import socket
socket.setdefaulttimeout(timeout)

但是在爬虫过程中依然会出现

Traceback (most recent call last):
File "C:\Users\wenxue5\Documents\Tencent Files\910872628\FileRecv\zongheng.py", line 93, in <module>
search(line.strip())
File "C:\Users\wenxue5\Documents\Tencent Files\910872628\FileRecv\zongheng.py", line 60, in search
the_page = response.read() #读取反馈的内容
File "C:\Python27\lib\socket.py", line 349, in read
data = self._sock.recv(rbufsize)
File "C:\Python27\lib\httplib.py", line 532, in read
return self._read_chunked(amt)
File "C:\Python27\lib\httplib.py", line 590, in _read_chunked
value.append(self._safe_read(chunk_left))
File "C:\Python27\lib\httplib.py", line 634, in _safe_read
chunk = self.fp.read(min(amt, MAXAMOUNT))
File "C:\Python27\lib\socket.py", line 378, in read
data = self._sock.recv(left)
timeout: timed out

这是因为没有对timeout进行处理造成的

import urllib2
import socket
#In Python 2.7.3

try:
    urllib2.urlopen("www.baidu.com", timeout = 10)
except urllib2.URLError as e:
    print type(e)    #not catch
except socket.timeout as e:
    print type(e)    #catched