Develop Reference

Download GitLab file with gitlab-python

I am trying to download a file or folder from my gitlab repository, but they only way I have seen to do it is using CURL and command line. Is there any way to download files from the repository with just the python-gitlab API? I have read through the API and have not found anything, but other posts said it was possible, just gave no solution.

You can do like this:
import requests
response = requests.get('https://<your_path>/file.txt')
data = response.text
and then save the contents (data) as file...
Otherwise use the API:
f = project.files.get(path='<folder>/file.txt',ref='<branch or commit>')
and then decode using:
import base64
content = base64.b64decode(f.content)
and then save content as file...

Download a file without name using Python

I want to download a file, there is a hyper link in html page which does not include the file name and extension. How can I download the file using python?
For example the link is http://1.1.1.1:8080/tank-20/a/at_download/file,
but whenever I click on it the file will download and open with browser.

Use python requests to get the body of the response and write to file, this is essentially what the browser is doing when you click the link.
Try the below:
import requests
# define variables
request_url = "http://1.1.1.1:8080/tank-20/a/at_download/file"
output_file = "output.txt"
# send get request
response = requests.get(request_url)
# use 'with' to write to file
with open(output_file, 'w') as fh:
fh.write(response.content)
fh.close()

Downloaded Share Point Excel Not Opening with Open

I am re-framing an existing question for simplicity. I have the following code to download Excel files from a company Share Point site.
import requests
import pandas as pd
def download_file(url):
filename = url.split('/')[-1]
r = requests.get(url)
with open(filename, 'wb') as output_file:
output_file.write(r.content)
df = pd.read_excel(r'O:\Procurement Planning\QA\VSAF_test_macro.xlsm')
df['Name'] = 'share_point_file_path_documentName' #i'm appending the sp file path to the document name
file = df['Name'] #I only need the file path column, I don't need the rest of the dataframe
# for loop for download
for url in file:
download_file(url)
The downloads happen and I don't get any errors in Python, however when I try to open them I get an error from Excel saying Excel cannot open the file because the file format or extension is not valid. If I print the link in Jupyter Notebooks it does open correctly, the issue appears to be with the download.

Check r.status_code. This must be 200 or you have the wrong url or no permission.
Open the downloaded file in a text editor. It might be a HTML file (Office Online)
If the URL contains a web=1 query parameter, remove it or replace it by web=0.

Django: MP3 file being opened, not saved, on Apple devices

I am working with Python and Django 1.9.
I manage to serve an MP3 file for download and it works fine on my computer and on my Android phone. However, on my iPad, the file isn't downloaded and saved as a file; it is directly opened in both the Safari and Chrome browsers.
Is there any way to save the file on the iPad?
This is the code snippet that I'm using:
fsock = open(filename, 'r')
response = HttpResponse(fsock, content_type='audio/mpeg')
response['Content-Disposition'] = "attachment; filename=%s.mp3" % (filename)
return response
Is there something I should change in my code or is it rather a constraint from Apple's OS to avoid saving downloaded MP3 files directly from the browser?

How do I allow users to download a MIDI file with Flask without getting a 0 byte download?

I am writing an application that creates a midi file using the MIDIUtil library. When the user submits an HTML form, a midi file object is created with MIDIUtil. How do I allow the user to download this as a .mid file? I have tried the following code, but I end up downloading a file of 0 bytes.
return Response(myMIDIFile, mimetype='audio/midi')

I use a variant of the following code to allow my users to download images they generate. The below code should work for you. Please note that you will most likely need to specify the full server path to the file being downloaded.
from flask import send_file
download_filename = FULL_PATH_TO_YOUR_MIDI_FILE
return(send_file(filename_or_fp = download_filename,mimetype="audio/midi",as_attachment=True))

I ended up using this, and it worked.
new_file = open('test.mid', 'wb')
myMIDI.writeFile(new_file)
new_file.close()
new_file = open('test.mid', 'rb')
return send_file(new_file, mimetype='audio/midi')

Might want to just try using send_file
from flask import send_file
return send_file("yourmidifile.mid", as_attachement=True, mimetype="audio\midi")