正確地對用戶代理進行網址編碼

Question

我是Python的新手，似乎遇到了問題。我試圖來urlencode用戶代理字符串...

import urllib 

UserAgent = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; en-GB; rv:1.9.0.3 Gecko/2008092417 Firefox/3.0.3' 
print 'Agent: ' + UserAgent 
print urllib.urlencode(UserAgent) 

導致...

Mozilla/5.0 (Windows; U; Windows NT 5.1; en-GB; rv:1.9.0.3 Gecko/2008092417 Firefox/3.0.3 
Traceback (most recent call last): 
    File "D:\Source\SomePath\test.py", line 7, in <module> 
    print urllib.urlencode(UserAgent) 
    File "C:\Python26\lib\urllib.py", line 1254, in urlencode 
    raise TypeError 
TypeError: not a valid non-string sequence or mapping object 
Press any key to continue . . . 

我只能假設，雖然UserAgent被正確打印，我要麼錯過了關於urllib.urlencode()的一些字符串轉義選項或根本性錯誤？

Answer 1

urllib.urlencode需要映射或序列，每個映射或序列有兩個項目。如可以在docs

可以看到在你的代碼，你需要做到以下幾點：

urllib.urlencode({'Agent': UserAgent}) 

Answer 2

Wessie打我給它。爲了將來的參考，你也可以這樣做：

>>> help(urllib.urlencode) 
Help on function urlencode in module urllib: 

urlencode(query, doseq=0) 
    Encode a sequence of two-element tuples or dictionary into a URL query string. 

    If any values in the query arg are sequences and doseq is true, each 
    sequence element is converted to a separate parameter. 

    If the query arg is a sequence of two-element tuples, the order of the 
    parameters in the output will match the order of parameters in the 
    input.