在循环中使用 urllib

Question

我正在尝试编写一个简单的 python 脚本来检查特定 URL 的状态代码并根据 return 代码采取必要的操作。我正在使用 urllib 模块来实现这一点。这段代码的问题是我在一个列表中定义了不同的部门（我需要在 urllib 中迭代这个列表）我似乎无法在 for 循环中找到 substitute/insert 列表元素的方法。

import getopt, sys
import urllib.request
depts =  [ 'support', 'edelivery', 'docs']

for dept in depts:
    res = urllib.request.urlopen('https://dept.oracle.com').getcode()
    print(res)

我收到以下错误。我们真的可以在循环中迭代 urllib 模块中的列表吗？

#python3 reg_c_with_all.py 
Traceback (most recent call last):
  File "/usr/lib64/python3.7/urllib/request.py", line 1350, in do_open
    encode_chunked=req.has_header('Transfer-encoding'))
  File "/usr/lib64/python3.7/http/client.py", line 1277, in request
    self._send_request(method, url, body, headers, encode_chunked)
  File "/usr/lib64/python3.7/http/client.py", line 1323, in _send_request
    self.endheaders(body, encode_chunked=encode_chunked)
  File "/usr/lib64/python3.7/http/client.py", line 1272, in endheaders
    self._send_output(message_body, encode_chunked=encode_chunked)
  File "/usr/lib64/python3.7/http/client.py", line 1032, in _send_output
    self.send(msg)
  File "/usr/lib64/python3.7/http/client.py", line 972, in send
    self.connect()
  File "/usr/lib64/python3.7/http/client.py", line 1439, in connect
    super().connect()
  File "/usr/lib64/python3.7/http/client.py", line 944, in connect
    (self.host,self.port), self.timeout, self.source_address)
  File "/usr/lib64/python3.7/socket.py", line 707, in create_connection
    for res in getaddrinfo(host, port, 0, SOCK_STREAM):
  File "/usr/lib64/python3.7/socket.py", line 752, in getaddrinfo
    for res in _socket.getaddrinfo(host, port, family, type, proto, flags):
socket.gaierror: [Errno -2] Name or service not known

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "reg_c_with_all.py", line 6, in <module>
    res = urllib.request.urlopen('https://dept.oracle.com').getcode()
  File "/usr/lib64/python3.7/urllib/request.py", line 222, in urlopen
    return opener.open(url, data, timeout)
  File "/usr/lib64/python3.7/urllib/request.py", line 525, in open
    response = self._open(req, data)
  File "/usr/lib64/python3.7/urllib/request.py", line 543, in _open
    '_open', req)
  File "/usr/lib64/python3.7/urllib/request.py", line 503, in _call_chain
    result = func(*args)
  File "/usr/lib64/python3.7/urllib/request.py", line 1393, in https_open
    context=self._context, check_hostname=self._check_hostname)
  File "/usr/lib64/python3.7/urllib/request.py", line 1352, in do_open
    raise URLError(err)
urllib.error.URLError: <urlopen error [Errno -2] Name or service not known>

当我运行它单独 return 是预期的响应代码。

    >>> urllib.request.urlopen('https://support.oracle.com').getcode()
    200
    >>> urllib.request.urlopen('https://docs.oracle.com').getcode()
    200
    >>> urllib.request.urlopen('https://edelivery.oracle.com').getcode()
    200
    >>>

Answer 1

您的代码遍历部门，但不使用它们来修改 URL。尝试使用 fstring 将 dept 的值插入 URL:

import getopt, sys
import urllib.request

depts =  [ 'support', 'edelivery', 'docs']

for dept in depts:
    url = f'https://{dept}.oracle.com'
    res = urllib.request.urlopen(url).getcode()
    print(f'{url} => {res}')

输出：

https://support.oracle.com => 200
https://edelivery.oracle.com => 200
https://docs.oracle.com => 200

在循环中使用 urllib

Using urlib in a loop

urllib

python-3.x