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    This is the community forum for my apps Pythonista and Editorial.

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    URL Scheme to Return to Safari

    Pythonista
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    • calebaharrison208
      calebaharrison208 last edited by

      Hi: I've got an action extension through the Workflow App that calls and runs the following URL-shortening (via Bitly) Pythonista script:

      import requests
      import json
      import clipboard
      import webbrowser
      import console
      
      LongURL = str(clipboard.get())
      
      query_params = {'access_token' : 'MYTOKEN',
                      'longUrl' : str(clipboard.get())}
                      
      endpoint = 'https://api-ssl.bitly.com/v3/shorten'
      response = requests.get(endpoint, params= query_params)
      
      data = json.loads(response.content)
      
      if data['status_code'] != 500:
      	clipboard.set(data['data']['url'])
      	webbrowser.open('safari-'+LongURL)
      
      if data['status_code'] == 500:
      	alert = console.alert('URL already shortened!', 'Return to Safari?', 'Safari', hide_cancel_button=True)
      	if alert == 0:
      		webbrowser.open('safari-'+LongURL)
      

      Is there a URL scheme that will just open Safari in iOS to the last opened tab/page? That is, I currently can't figure out a way to return to Safari after running the Pythonista action without telling Safari to open either a blank page or a specific page (in this case, the initial URL that I shortened, to prevent my returning to Safari from counting as a link view).

      I've tried safari-http://a to see if that would work, to no avail. Any help is appreciated!

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