用Python构建包含URL的字典的正确方法

我试图查询一个API，在这里我需要填写零件号和制造商

假设我需要填写零件号为bav99，NXP Semiconductors为制造商。最终结果应如下所示：

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

我不得不说，我真的很困惑，哪些字符应该编码，哪些不应该。根据我所看到的，双引号应编码为%22，空格应编码为%20

我所做的：

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

我想我可以把我的担忧概括为以下两个问题：

为什么某些字符应该被编码，而有些不应该

什么是正确的方法和实用的方法来编码的对象，而不是使用占位符硬编码它

这对你有帮助吗

urllib.urlencodequery[，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.