Get Django views.py to return and execute javascript

So I'm working with django and file uploads and I need a javascript function to execute after the file has been uploaded. I have a file upload handler in my views.py which looks like this:

def upload_file(request):   
    form = UploadFileForm(request.POST, request.FILES)
    if form.is_valid():        
        for f in request.FILES.getlist('fileAttachments'):     
            handle_uploaded_file(f)
        return HttpJavascriptResponse('parent.Response_OK();')
    else:
        return HttpResponse("Failed to upload attachment.")

And I found a django snippet from http://djangosnippets.org/snippets/341/ and I put the HttpJavascriptResponse class in my views.py code. It looks as follows:

class HttpJavascriptResponse(HttpResponse):
    def __init__(self,content):
       HttpResponse.__init__(self,content,mimetype="text/java开发者_StackOverflow中文版script")

However, when I upload a file the browser simple renders "parent.Response_OK();" on the screen instead of actually executing the javascript. And Chrome gives me the warning: "Resource interpreted as Document but transferred with MIME type text/javascript"

Is there anyway to get views.py to execute the script?

A better way is to pass the mime to the HttpResponse Object.

See documentation: https://docs.djangoproject.com/en/3.2/ref/request-response/#django.http.HttpRequest.content_type.

    return HttpResponse("parent.Response_OK()", content_type="application/x-javascript")

Note - Some previous versions of Django used mimetype instead of content_type:

    return HttpResponse("parent.Response_OK()", mimetype="application/x-javascript")

I believe this will work.

return HttpResponse("<script>parent.Response_OK();</script>")

However, you might think about returning a success (200) status code in this case, and then having some javascript in parent attach to the load event of this child, and branch based on return status code. That way you have a separation of view rendering code and view behavior code.

Chase's solution worked for me, though I need to execute more javascript than I care to put in a python string:

from django.http import HttpResponse
from django.contrib.staticfiles.templatetags import staticfiles

...
    return HttpResponse("<script src='{src}'></script>".format(
        src = staticfiles.static('/path/to/something.js')))

I ended up here looking for a way to serve a dynamic js file using django.

Here is my solution :

return render(request, 'myscript.js', {'foo':'bar'},
                        content_type="application/x-javascript")