blocks|key|1295140|text|urllib和urlparse模块有一些奇怪的地方。下面是一个有效的示例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1295141|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))|code-block|syntax|javascript|1295142|ParseResult，urlparse()，is+read-only的结果，我们需要将它转换成list，然后才能尝试修改它的数据。|1295143|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/urllib.parse.html#urllib.parse.urlparse^0|0|6|7|8|0|0|0|B|C|A|1C|4|N|C|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|M|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

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

<pre><code>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))
</code></pre>

<code>ParseResult</code>, the result of <code>urlparse()</code>, <a href="https://docs.python.org/3/library/urllib.parse.html#urllib.parse.urlparse" rel="noreferrer">is read-only</a> and we need to convert it to a <code>list</code> before we can attempt to modify its data.

blocks|key|1295096|text|如果字符串可以包含任意数据(例如，需要对与号、斜杠等字符进行编码)，则需要使用URL编码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1295097|查看urllib.urlencode：|1295098|>>>+import+urllib
>>>+urllib.urlencode({'lang':'en','tag':'python'})
'lang=en&tag=python'|code-block|syntax|javascript|1295099|在python3中：|1295100|from+urllib+import+parse
parse.urlencode({'lang':'en','tag':'python'})|1295101|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You want to use URL encoding if the strings can have arbitrary data (for example, characters such as ampersands, slashes, etc. will need to be encoded).

Check out urllib.urlencode:

<pre><code>&gt;&gt;&gt; import urllib
&gt;&gt;&gt; urllib.urlencode({'lang':'en','tag':'python'})
'lang=en&amp;tag=python'
</code></pre>

In python3:

<pre><code>from urllib import parse
parse.urlencode({'lang':'en','tag':'python'})
</code></pre>

blocks|key|1080373|text|外包给经过战斗考验的requests+library。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1080374|这就是我要做的：|1080375|from+requests.models+import+PreparedRequest
url+=+'http://example.com/search?q=question'
params+=+{'lang':'en','tag':'python'}
req+=+PreparedRequest()
req.prepare_url(url,+params)
print(req.url)|code-block|syntax|javascript|1080376|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/requests/requests/blob/master/requests/models.py#L347^0|A|G|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Outsource it to the battle tested <a href="https://github.com/requests/requests/blob/master/requests/models.py#L347" rel="noreferrer">requests library</a>.

This is how I will do it:

<pre><code>from requests.models import PreparedRequest
url = 'http://example.com/search?q=question'
params = {'lang':'en','tag':'python'}
req = PreparedRequest()
req.prepare_url(url, params)
print(req.url)
</code></pre>

blocks|key|1080244|text|您还可以使用furl模块https://github.com/gruns/furl|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1080245|>>>+from+furl+import+furl
>>>+print+furl('http://example.com/search?q=question').add({'lang':'en','tag':'python'}).url
http://example.com/search?q=question&lang=en&tag=python|code-block|syntax|javascript|1080246|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/gruns/furl^0|C|T|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You can also use the furl module <a href="https://github.com/gruns/furl">https://github.com/gruns/furl</a>

<pre><code>&gt;&gt;&gt; from furl import furl
&gt;&gt;&gt; print furl('http://example.com/search?q=question').add({'lang':'en','tag':'python'}).url
http://example.com/search?q=question&amp;lang=en&amp;tag=python
</code></pre>

blocks|key|1080325|text|如果您正在使用requests+lib|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1080326|import+requests
...
params+=+{'tag':+'python'}
requests.get(url,+params=params)|code-block|syntax|javascript|1080327|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python-requests.org/en/master/^0|7|C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

If you are using the <a href="http://docs.python-requests.org/en/master/" rel="noreferrer">requests lib</a>:

<pre><code>import requests
...
params = {'tag': 'python'}
requests.get(url, params=params)
</code></pre>

blocks|key|1080275|text|基于this+answer，简单情况的一行代码(Python3代码)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1080276|from+urllib.parse+import+urlparse,+urlencode


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

url+%2B=+('&'+if+urlparse(url).query+else+'?')+%2B+urlencode(params)|code-block|syntax|javascript|1080277|或者：|1080278|url+%2B=+('&',+'?')[urlparse(url).query+==+'']+%2B+urlencode(params)|1080279|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/9538343/1113207^0|2|4|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Based on <a href="https://stackoverflow.com/a/9538343/1113207">this</a> answer, one-liner for simple cases (Python 3 code):

<pre><code>from urllib.parse import urlparse, urlencode


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

url += ('&amp;' if urlparse(url).query else '?') + urlencode(params)
</code></pre>

or:

<pre><code>url += ('&amp;', '?')[urlparse(url).query == ''] + urlencode(params)
</code></pre>

blocks|key|1079990|text|是:使用urllib。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1079991|从文档中的examples：|1079992|>>>+import+urllib
>>>+params+=+urllib.urlencode({'spam':+1,+'eggs':+2,+'bacon':+0})
>>>+f+=+urllib.urlopen("http://www.musi-cal.com/cgi-bin/query?%25s"+%25+params)
>>>+print+f.geturl()+#+Prints+the+final+URL+with+parameters.
>>>+print+f.read()+#+Prints+the+contents|code-block|syntax|javascript|1079993|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/urllib.html|1|http://docs.python.org/library/urllib.html#examples^0|4|6|0|0|5|8|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|F|3|G|5|H|7|12|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

Yes: use <a href="http://docs.python.org/library/urllib.html" rel="noreferrer">urllib</a>.

From the <a href="http://docs.python.org/library/urllib.html#examples" rel="noreferrer">examples</a> in the documentation:

<pre><code>&gt;&gt;&gt; import urllib
&gt;&gt;&gt; params = urllib.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
&gt;&gt;&gt; f = urllib.urlopen("http://www.musi-cal.com/cgi-bin/query?%s" % params)
&gt;&gt;&gt; print f.geturl() # Prints the final URL with parameters.
&gt;&gt;&gt; print f.read() # Prints the contents
</code></pre>

blocks|key|1295197|text|我喜欢Łukasz版本，但是由于urllib和urllparse函数在这种情况下使用起来有些笨拙，我认为这样做更简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1295198|params+=+urllib.urlencode(params)

if+urlparse.urlparse(url)[4]:
++++print+url+%2B+'&'+%2B+params
else:
++++print+url+%2B+'?'+%2B+params|code-block|syntax|javascript|1295199|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I liked Łukasz version, but since urllib and urllparse functions are somewhat awkward to use in this case, I think it's more straightforward to do something like this:

<pre><code>params = urllib.urlencode(params)

if urlparse.urlparse(url)[4]:
 print url + '&amp;' + params
else:
 print url + '?' + params
</code></pre>

blocks|key|1295436|text|python3，我想不言自明|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1295437|from+urllib.parse+import+urlparse,+urlencode,+parse_qsl

url+=+'https://www.linkedin.com/jobs/search?keywords=engineer'

parsed+=+urlparse(url)
current_params+=+dict(parse_qsl(parsed.query))
new_params+=+{'location':+'United+States'}
merged_params+=+urlencode({**current_params,+**new_params})
parsed+=+parsed._replace(query=merged_params)

print(parsed.geturl())
#+https://www.linkedin.com/jobs/search?keywords=engineer&location=United%2BStates|code-block|syntax|javascript|1295438|entityMap^0|0|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

<code>python3</code>, self explanatory I guess
<pre class="lang-py prettyprint-override"><code>from urllib.parse import urlparse, urlencode, parse_qsl

url = 'https://www.linkedin.com/jobs/search?keywords=engineer'

parsed = urlparse(url)
current_params = dict(parse_qsl(parsed.query))
new_params = {'location': 'United States'}
merged_params = urlencode({**current_params, **new_params})
parsed = parsed._replace(query=merged_params)

print(parsed.geturl())
# https://www.linkedin.com/jobs/search?keywords=engineer&amp;location=United+States
</code></pre>

blocks|key|1295045|text|使用各种urlparse函数拆分现有的URL，组合字典上的urllib.urlencode()，然后使用urlparse.urlunparse()将其重新组合在一起。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1295046|或者直接获取urllib.urlencode()的结果并将其适当地连接到URL。|1295047|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/urlparse.html|1|http://docs.python.org/library/urllib.html#urllib.urlencode^0|4|8|T|I|1G|L|4|8|0|T|I|1|0|6|I|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]|$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]|P|$5|K|L|M|E|$N|Q]]]]

Use the various <a href="http://docs.python.org/library/urlparse.html" rel="noreferrer"><code>urlparse</code></a> functions to tear apart the existing URL, <a href="http://docs.python.org/library/urllib.html#urllib.urlencode" rel="noreferrer"><code>urllib.urlencode()</code></a> on the combined dictionary, then <code>urlparse.urlunparse()</code> to put it all back together again.

Or just take the result of <code>urllib.urlencode()</code> and concatenate it to the URL appropriately.

blocks|key|1080221|text|还有另一个答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1080222|def+addGetParameters(url,+newParams):
++++(scheme,+netloc,+path,+params,+query,+fragment)+=+urlparse.urlparse(url)
++++queryList+=+urlparse.parse_qsl(query,+keep_blank_values=True)
++++for+key+in+newParams:
++++++++queryList.append((key,+newParams[key]))
++++return+urlparse.urlunparse((scheme,+netloc,+path,+params,+urllib.urlencode(queryList),+fragment))|code-block|syntax|javascript|1080223|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Yet another answer:

<pre><code>def addGetParameters(url, newParams):
 (scheme, netloc, path, params, query, fragment) = urlparse.urlparse(url)
 queryList = urlparse.parse_qsl(query, keep_blank_values=True)
 for key in newParams:
 queryList.append((key, newParams[key]))
 return urlparse.urlunparse((scheme, netloc, path, params, urllib.urlencode(queryList), fragment))
</code></pre>

blocks|key|1080131|text|在python+2.5中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1080132|import+cgi
import+urllib
import+urlparse

def+add_url_param(url,+**params):
++++n=3
++++parts+=+list(urlparse.urlsplit(url))
++++d+=+dict(cgi.parse_qsl(parts[n]))+#+use+cgi.parse_qs+for+list+values
++++d.update(params)
++++parts[n]=urllib.urlencode(d)
++++return+urlparse.urlunsplit(parts)

url+=+"http://stackoverflow.com/search?q=question"
add_url_param(url,+lang='en')+==+"http://stackoverflow.com/search?q=question&lang=en"|code-block|syntax|javascript|1080133|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In python 2.5

<pre><code>import cgi
import urllib
import urlparse

def add_url_param(url, **params):
 n=3
 parts = list(urlparse.urlsplit(url))
 d = dict(cgi.parse_qsl(parts[n])) # use cgi.parse_qs for list values
 d.update(params)
 parts[n]=urllib.urlencode(d)
 return urlparse.urlunsplit(parts)

url = "http://stackoverflow.com/search?q=question"
add_url_param(url, lang='en') == "http://stackoverflow.com/search?q=question&amp;lang=en"
</code></pre>

blocks|key|1080206|text|下面是我如何实现它的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1080207|import+urllib

params+=+urllib.urlencode({'lang':'en','tag':'python'})
url+=+''
if+request.GET:
+++url+=+request.url+%2B+'&'+%2B+params
else:
+++url+=+request.url+%2B+'?'+%2B+params++++|code-block|syntax|javascript|1080208|就像一个护身符。然而，我希望有一种更简洁的方式来实现这一点。|1080209|实现上述功能的另一种方式是将其放入一个方法中。|1080210|import+urllib

def+add_url_param(request,+**params):
+++new_url+=+''
+++_params+=+dict(**params)
+++_params+=+urllib.urlencode(_params)

+++if+_params:
++++++if+request.GET:
+++++++++new_url+=+request.url+%2B+'&'+%2B+_params
++++++else:
+++++++++new_url+=+request.url+%2B+'?'+%2B+_params
+++else:
++++++new_url+=+request.url

+++return+new_ur|1080211|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here is how I implemented it.

<pre><code>import urllib

params = urllib.urlencode({'lang':'en','tag':'python'})
url = ''
if request.GET:
 url = request.url + '&amp;' + params
else:
 url = request.url + '?' + params 
</code></pre>

Worked like a charm. However, I would have liked a more cleaner way to implement this.

Another way of implementing the above is put it in a method.

<pre><code>import urllib

def add_url_param(request, **params):
 new_url = ''
 _params = dict(**params)
 _params = urllib.urlencode(_params)

 if _params:
 if request.GET:
 new_url = request.url + '&amp;' + _params
 else:
 new_url = request.url + '?' + _params
 else:
 new_url = request.url

 return new_ur
</code></pre>

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

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

Is there any standard way to do this?

Add params to given URL in Python

Python

假设我得到了一个URL。它可能已经有了GET参数(例如http://example.com/search?q=question)，也可能没有(例如http://example.com/)。现在我需要给它添加一些参数，比如{'lang':'en','tag':'python'}。在第一种情况下，我将使用http://example.com/search?q=question&lang=en&tag=

问在Python中向给定的URL添加参数
EN

回答 13

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问在Python中向给定的URL添加参数
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