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我有 100 个 CSV 格式的网站列表。所有站点都具有相同的一般格式，包括一个有 7 列的大表。我编写了这个脚本来从每个网站的第 7 列中提取数据，然后将这些数据写入文件。然而，下面的脚本部分工作：打开输出文件（在运行脚本之后）显示正在跳过某些内容，因为它只显示 98 次写入（显然脚本还记录了许多异常）。非常感谢有关如何在这种情况下实现“捕获异常”的指导。谢谢！ 

import csv, urllib2, re
def replace(variab): return variab.replace(",", " ")

urls = csv.reader(open('input100.txt', 'rb'))  #access list of 100 URLs
for url in urls:
    html = urllib2.urlopen(url[0]).read()  #get HTML starting with the first URL
    col7 = re.findall('td7.*?td', html)  #use regex to get data from column 7
    string = str(col7)  #stringify data
    neat = re.findall('div3.*?div', string)  #use regex to get target text  
    result = map(replace, neat)  #apply function to remove','s from elements
    string2 = ", ".join(result)  #separate list elements with ', ' for export to csv
    output = open('output.csv', 'ab') #open file for writing 
    output.write(string2 + '\n') #append output to file and create new line
    output.close()

返回：

Traceback (most recent call last):
 File "C:\Python26\supertest3.py", line 6, in <module>
  html = urllib2.urlopen(url[0]).read()
 File "C:\Python26\lib\urllib2.py", line 124, in urlopen
  return _opener.open(url, data, timeout)
 File "C:\Python26\lib\urllib2.py", line 383, in open
  response = self._open(req, data)
 File "C:\Python26\lib\urllib2.py", line 401, in _open
  '_open', req)
 File "C:\Python26\lib\urllib2.py", line 361, in _call_chain
  result = func(*args)
 File "C:\Python26\lib\urllib2.py", line 1130, in http_open
  return self.do_open(httplib.HTTPConnection, req)
 File "C:\Python26\lib\urllib2.py", line 1103, in do_open
  r = h.getresponse()
 File "C:\Python26\lib\httplib.py", line 950, in getresponse
  response.begin()
 File "C:\Python26\lib\httplib.py", line 390, in begin
  version, status, reason = self._read_status()
 File "C:\Python26\lib\httplib.py", line 354, in _read_status
  raise BadStatusLine(line)
BadStatusLine
>>>>
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2 回答 2

2

将for循环体变为：

for url in urls:
  try:
    ...the body you have now...
  except Exception, e:
    print>>sys.stderr, "Url %r not processed: error (%s) % (url, e)

（或者，如果您已经在使用标准库的模块，请使用logging.error而不是 goofy [并且您应该；-)]）。print>>logging

于 2009-10-05T23:19:37.233 回答
1

我建议阅读错误和异常Python 文档，尤其是第 8.3 节——处理异常。

于 2009-10-05T22:58:21.457 回答