Question

I am displaying the list of objects in the html table.
i have the download link in front of every row which i want them to download the linked file.

I have made this function

def make_downloadable_link(path):
    #Prepare the form for downloading
    wrapper      = FileWrapper(open(mypath))
    response     = HttpResponse(wrapper,'application/pdf')
    response['Content-Length']      = os.path.getsize(mypath)  
    fname = mypath.split('/')[-1]  
    response['Content-Disposition'] = 'attachment; filename= fname'
    return response

This is working fine if i use it for hard coded path in view for single file. But i want to make a generic view so that it works on all the files in the table

I hav the path of the file avaiable in object.path variable but i am confused how can i pass the path object to the downlaod file view. because i want to hide that actual path from the user.

I don’t know what to write in the URLs.py file fo that download file view

Answer 1

What would you like to do is get actual file path from object. And as you have said the file path is stored in object.path that makes it easy.

For example:

urls.py

url(r'^download/(?P<object_id>\d+)/$', "yourapp.views.make_downloadable_link", name="downloadable")

In views.py:

def make_downloadable_link(object_id):

    # get object from object_id
    object = ObjectModel.objects.get(id=object_id)
    mypath = object.path

    #prepare to serve the file
    wrapper      = FileWrapper(open(mypath))
    response     = HttpResponse(wrapper,'application/pdf')
    response['Content-Length']      = os.path.getsize(mypath)  
    fname = mypath.split('/')[-1]  
    response['Content-Disposition'] = 'attachment; filename= fname'
    return response