Url Parse is missing fragment - Python

Question

I need to save a file with the name of the given acquisition path's file.

Given an URL I would like to parse it and extract the name of the file, here's my code...

I read a JSON parameter and give it to the Parse Url function. The acquisition path is a string.

ParseUrl.py:

from urllib.parse import urlparse as up
a = up(jtp["AcquisitionPath"])    # => http://127.0.0.1:8000/Users/YodhResearch/Desktop/LongCtrl10min.tiff
print(a)
print(os.path.basename(a))

Result:

ParseResult(scheme='http', netloc='127.0.0.1:8000', path='/Users/YodhResearch/Desktop/LongCtrl10min.tiff', params='', query='', fragment='')
[....]
TypeError: expected str, bytes or os.PathLike object, not ParseResult

As you can see it Parse the URL but "LongCtrl10min.tiff" is not in the fragment section but is all on the path section. Why is that happening? Maybe because "AcquisitionPath" is a string and UrlParse recognize all as a unique path?

EDIT:

a.path WORKS, I would like to know why I don't get it into the fragment section.

Here's another example:

from urllib.parse import urlparse as up

string = "http://127.0.0.1:8000/GIULIO%20FERRARI%20FOLDER/Giulio%20_%20CSV/Py%20Script/sparse%20python/tiff_test.tiff_IDAnal#1_IDAcq#10_TEMP_.json"

a = up(string)
print(a)
print(os.path.basename(a))

Results:

ParseResult(scheme='http', netloc='127.0.0.1:8000', path='/GIULIO%20FERRARI%20FOLDER/Giulio%20_%20CSV/Py%20Script/sparse%20python/tiff_test.tiff_IDAnal', params='', query='', fragment='1_IDAcq#10_TEMP_.json')

See, Now it doesn't get the right fragment that should be: "tiff_test.tiff_IDAnal#1_IDAcq#10_TEMP_.json"

SOLUTION:

Fragment needs '#' symbol! Thanks to all.

It works but I would like to know why I dont get it into fragment section — Giulio, Feb 25 '20 at 15:59
Because fragment in the url has form `http://127.0.0.1:8000/Users/YodhResearch/Desktop/LongCtrl10min.tiff#fragment` — hurlenko, Feb 25 '20 at 16:02
get it into... i mean I want to see the file name in the fragment parameter of the urlparse function. I still don't get why this not happend. I made another example in the question — Giulio, Feb 25 '20 at 16:05
Ok Hurlenko I get it what you mean with #fragment. THANKS YOU. — Giulio, Feb 25 '20 at 16:08
You may *want* to see that, but you won't, because the file name is not part of the fragment. The fragment *begins* with the `#`. — chepner, Feb 25 '20 at 16:09

score 1 · Accepted Answer · edited Oct 07 '21 at 11:07

There are two issues here: how to identify the components of a URL, and how to create the desired path from those components.

First, you are confused over what the fragment actually is. From RFC 3986:

The following are two example URIs and their component parts:

         foo://example.com:8042/over/there?name=ferret#nose
         \_/   \______________/\_________/ \_________/ \__/
          |           |            |            |        |
       scheme     authority       path        query   fragment
          |   _____________________|__
         / \ /                        \
         urn:example:animal:ferret:nose

The fragment is only the portion following the #, not the entire final component of the path.

Second, the urlparse()function from urllib module returns a ParseResult object and the basename()-method from os.path wants a str as argument.

What you probably want is to get the path from the ParseResult-object. You will get this with a.path (the path you have given via urlparse is saved in the attribute path of the ParseResult-object).

from urllib.parse import urlparse as up
a = up("http://127.0.0.1:8000/Users/YodhResearch/Desktop/LongCtrl10min.tiff")
print(os.path.basename(a.path))

This will output:

LongCtrl10min.tiff

If you want to include also the fragments, you can do this by explicitly adding this. The fragments are saved in a separated attribute in the ParseResult object, i.e. a.fragment in your case:

from urllib.parse import urlparse as up 
a = up("http://127.0.0.1:8000/Users/YodhResearch/Desktop/LongCtrl10min.tiff#anyfragment") 
print(os.path.basename(a.path) + "#" + a.fragment)

will output:

LongCtrl10min.tiff#anyfragment

It works with a.path but I would like to know why i dont get it into the fragment section. — Giulio, Feb 25 '20 at 16:00
That was not clear from your original question. I've updated my answer as well. — colidyre, Feb 25 '20 at 16:11

Url Parse is missing fragment - Python

1 Answers1