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Python 如何在Django的视图中运行url?_Python_Django - Fatal编程技术网

Python 如何在Django的视图中运行url?

Python 如何在Django的视图中运行url?,python,django,Python,Django,我有一张PDF上传表格。当用户发送此表单时,我希望在视图中触发URL。我希望调用的URL地址只在后端打开一次。用户无法查看此页面。我希望这样做的原因是,我在这里调用的API使我能够执行另一个进程。 如何在我的视图后端运行URL 注意:使用此功能有效吗? r = requests.get(url) 视图.py def upload_pdf(request, id): if request.method == 'POST': customer = get_object_

我有一张PDF上传表格。当用户发送此表单时,我希望在视图中触发URL。我希望调用的URL地址只在后端打开一次。用户无法查看此页面。我希望这样做的原因是,我在这里调用的API使我能够执行另一个进程。 如何在我的视图后端运行URL

注意:使用此功能有效吗?

 r = requests.get(url)

视图.py

def upload_pdf(request, id):
    if request.method == 'POST':
        customer = get_object_or_404(Customer, id=id)
        form = PdfForm(request.POST, request.FILES)

        if form.is_valid():
            new_pdf = form.save(commit=False)
            new_pdf.owner = customer
            new_pdf.save()
            name = new_pdf.pdf.name.replace(".pdf", "")
            url = 'https://api..../perform/{0}'.format(name)
            return redirect('ocr', id=new_pdf.id)
    else:
        form = PdfForm()
        customer = get_object_or_404(Customer, id=id)

    return render(request, 'upload_pdf.html', {'form': form, 'customer': customer})
导入请求

url = request.get/post('https://api..../perform/{0}'.format(name))