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我有一个函数可以生成并返回一个_GET包含 URI 查询字段的查询键/值对的字典。假设 URI 是http://127.0.0.1/path/to/query?foo=bar&bar=foo，该函数在派生的内部使用BaseHTTPServer.BaseHTTPRequestHandler如下KeyError：

class HTTPHandler(BaseHTTPServer.BaseHTTPRequestHandler):
    def do_GET(self):
        self.send_response(200)
        self.send_header("Content-Type", "text/plain")
        self.end_headers()

        _GET = query_parse(urlparse.urlparse(self.path).query)

        # No KeyError here..
        print "foo: %s\r\nbar: %s" % (GET["foo"], _GET["bar"])

        # KeyError on _GET["foo"]..
        self.wfile.write("foo: %s\r\nbar: %s\r\n" % (_GET["foo"], _GET["bar"]))

        # Still KeyError on _GET["foo"] even if commenting above line
        # and uncommenting below one!
        #self.wfile.write("bar: %s\r\nfoo: %s\r\n" % (_GET["bar"], _GET["foo"]))

追溯：

localhost - - [19/Oct/2011 18:21:18] "GET /path/to/query?foo=bar&bar=foo HTTP/1.1" 200 -
localhost - - [19/Oct/2011 18:21:18] "GET /favicon.ico HTTP/1.1" 200 -
Traceback (most recent call last):
  File "E:\Program\Python27\lib\SocketServer.py", line 284, in _handle_request_noblock
    self.process_request(request, client_address)
  File "E:\Program\Python27\lib\SocketServer.py", line 310, in process_request
    self.finish_request(request, client_address)
  File "E:\Program\Python27\lib\SocketServer.py", line 323, in finish_request
    self.RequestHandlerClass(request, client_address, self)
  File "E:\Program\Python27\lib\SocketServer.py", line 639, in __init__
    self.handle()
  File "E:\Program\Python27\lib\BaseHTTPServer.py", line 337, in handle
    self.handle_one_request()
  File "E:\Program\Python27\lib\BaseHTTPServer.py", line 325, in handle_one_request
    method()
  File "http-test.py", line 40, in do_GET
    print "foo: %s" % _GET["foo"]
KeyError: 'foo'
4

2 回答 2

1

I don't know what your query_parse function do but there is already a function that do that which is urlparse.parse_qs:

>>> query = urlparse.urlparse('http://127.0.0.1/path/to/query?foo=bar&bar=foo').query
>>> urlparse.parse_qs(query)
{'bar': ['foo'], 'foo': ['bar']}

Or if you don't like the value of the dictionary to be lists you can do:

>>> dict(urlparse.parse_qsl(query))
{'bar': 'foo', 'foo': 'bar'}

Hope this can help.

于 2011-10-19T16:39:59.060 回答
0

代替 query_parse()，使用 Python 自己的查询解析器：

>>> s = 'http://127.0.0.1/path/to/query?foo=bar&bar=foo'
>>> t = urlparse.urlparse(s)
>>> urlparse.parse_qs(t.query)
{'foo': ['bar'], 'bar': ['foo']}

使用 pdb 或插入打印语句来帮助调试很容易看到这一点。

于 2011-10-19T16:53:44.380 回答