For instance I have the following url pattern:
url_pattern = {
"scheme": "http",
"netloc": "for.example.com",
"path": "/ex1",
"params": "",
"query": a=b&c=d,
"fragment": ""
}
It's just like the inverse of the output of urlparse.urlparse("http://for.example.com/ex1?a=b&c=d")
. Can I get the url from the dict?
urlparse
generates a ParseResult
object. Hence, just construct an object of ParseResult
and use the geturl()
method to generate the URL.
>>> url_pattern = {
"scheme": "http",
"netloc": "for.example.com",
"path": "/ex1",
"params": "",
"query": "a=b&c=d",
"fragment": ""
}
>>> from urlparse import ParseResult
>>> ParseResult(**url_pattern).geturl()
'http://for.example.com/ex1?a=b&c=d'