python - 如何使用 python 跳过存储库中不存在的文件？

Question

我想从问题跟踪系统一个一个地下载 xml 文件。当存储库中不存在文件时，它会产生错误消息。我包含了 python 脚本以更好地理解我的问题。

我的代码:

import urllib.request
for i in range(0,1000):
    issue_id1='DERBY-'+str(i)
    url ="https://issues.apache.org/jira/si/jira.issueviews:issue-xml/"+issue_id1+'/'+issue_id1+'.xml'
    s=urllib.request.urlopen(url)
    contents = s.read()
    file = open(issue_id1+'.xml', 'wb')
    file.write(contents)

file.close()

堆栈跟踪:

Traceback (most recent call last):
  File "/PhP/Learning/xmldownlaod.py", line 10, in <module>
    s=urllib.request.urlopen(url)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 161, in urlopen
    return opener.open(url, data, timeout)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 469, in open
    response = meth(req, response)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 579, in http_response
    'http', request, response, code, msg, hdrs)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 507, in error
    return self._call_chain(*args)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 441, in _call_chain
    result = func(*args)
  File "/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/urllib/request.py", line 587, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 404: Not Found

Answer 1

Python 使用 "try except"错误处理 block

import urllib.request
from urllib.error import URLError # the docs say this is the base error you need to catch
for i in range(0,1000):
    issue_id1='DERBY-'+str(i)
    url ="https://issues.apache.org/jira/si/jira.issueviews:issue-xml/"+issue_id1+'/'+issue_id1+'.xml'
    try:
        s=urllib.request.urlopen(url)
        contents = s.read()
    except URLError:
        print('an error occurred while fetching: "{}"'.format(url))
        continue # skip this url and proceed to the next
    file = open(issue_id1+'.xml', 'wb')
    file.write(contents)