将参数添加到 Python 中的给定 URL [英] Add params to given URL in Python

问题描述

假设我得到了一个 URL.
它可能已经有 GET 参数(例如 http://example.com/search?q=question)，也可能没有(例如 http://example.com/).

Suppose I was given a URL.
It might already have GET parameters (e.g. http://example.com/search?q=question) or it might not (e.g. http://example.com/).

现在我需要向它添加一些参数，例如 {'lang':'en','tag':'python'}.在第一种情况下，我将使用 http://example.com/search?q=question&lang=en&tag=python，在第二种情况下 - http://example.com/search?lang=en&tag=python.

And now I need to add some parameters to it like {'lang':'en','tag':'python'}. In the first case I'm going to have http://example.com/search?q=question&lang=en&tag=python and in the second — http://example.com/search?lang=en&tag=python.

是否有任何标准方法可以做到这一点?

Is there any standard way to do this?

推荐答案

urllib 和 urlparse 模块有几个怪癖.这是一个工作示例:

There are a couple of quirks with the urllib and urlparse modules. Here's a working example:

try:
    import urlparse
    from urllib import urlencode
except: # For Python 3
    import urllib.parse as urlparse
    from urllib.parse import urlencode

url = "http://stackoverflow.com/search?q=question"
params = {'lang':'en','tag':'python'}

url_parts = list(urlparse.urlparse(url))
query = dict(urlparse.parse_qsl(url_parts[4]))
query.update(params)

url_parts[4] = urlencode(query)

print(urlparse.urlunparse(url_parts))

ParseResult，urlparse() 的结果，是只读的，我们需要将其转换为list，然后才能尝试修改其数据.

ParseResult, the result of urlparse(), is read-only and we need to convert it to a list before we can attempt to modify its data.

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