在Python中構建包含字典的URL的正確方法

Question

我想查詢一個API，我需要填寫partNumber和manufacturer。

假設我需要填寫partNumber爲bav99和NXP Semiconductors爲製造商。最終的結果應該如下：

https://app.datafireball.com/SearchService/search/listPartSearch? 
partNumber=[{ 
%22partNumber%22:%22bav99wt%22, 
%22manufacturer%22:%22NXP%20Semiconductors%22}]& 
fmt=xml 

我不得不說，我真是由哪些字符應不編碼和事實混淆。根據我所見，雙引號"應編碼爲%22，空格編碼爲%20。

我做了什麼：

import urllib 
urltemplate = """https://app.datafireball.com/SearchService/search/listPartSearch?partNumber=[{{%22partNumber%22:%22{0}%22,%22manufacturer%22:%22{1}%22}}]&fmt=xml""" 
# I have to recode the curly brackets to be double curly brackets, otherwise place holder won't work. 
url = urltemplate.format(urllib.quote('my partnumber'), urllib.quote('my manufacturer')) 
print url 

我想我可以總結我的問題納入以下兩個問題：

1. 爲什麼某些字符進行編碼，有些不是。

2. 什麼是正確的方式和實際的方式來編碼的對象，而不是使用佔位符硬編碼它。

Answer 1

是否this幫助你......

urllib.urlencode（查詢[，doseq]）¶

Convert a mapping object or a sequence of two-element tuples 
to a 「percent-encoded」 string, suitable to pass to urlopen() above as the 
optional data argument. This is useful to pass a dictionary of form 
fields to a POST request.