如何在 django 中上传保存的文件

Question

在我的例子中，用户在前端输入一个词，在后端，我将调用我编写的一些函数，最后，我将在/media/reports 目录中构建一个 .PDF 文件. 如何使这些文件特定于用户并将它们保存在数据库中？

在此之前，我实现了没有模型和表单的代码，只是将文件保存在/media/reports/ 目录中。并且用户可以在被重定向到下载页面后立即下载该文件。

但是现在，我想将这些文件保存到数据库中，以便每个用户都可以访问其个人资料中的新文件。我该怎么做？
这是我的代码：

views.py:

@login_required(login_url='/login/')
def dothat(request):
    if request.method == 'GET':
        return render(request, 'app/dothat.html')

    else:
        try:
            global word, user_name, function_name
            function_name = dothat.__name__
            word = request.POST.get("word")
            user_name = request.user
            full_name = request.user.get_full_name()
            myscript.script_one(word, user_name,full_name, function_name)
            

            # And in the called myscript, after doing some things,
            # the PDF file will be saved in /media/reports/ directory  



        except ValueError:
            return render(request, 'app/dashboard.html', {'error':'Bad data passed in. Try again.'})


    # And then, the user will be redirected to the download page to download that single file

    return render(request, 'app/download.html')

还有views.py中的download_file函数

@login_required(login_url='/login/')
def download_file(request):
    filename = f"{function_name}-{word}.pdf"
    # Define the full file path
    filepath = f"{BASE_DIR}/app/media/app/reports/{user_name}/{filename}"
    # Open the file for reading content
    if os.path.exists(filepath):
        # Set the return value of the HttpResponse
        response = HttpResponse(open(filepath, 'rb'))
    # Set the HTTP header for sending to browser
        response['Content-Disposition'] = "attachment; filename=%s" % filename
        return response
    # Return the response value
    else:
        raise HTTP404

这里是models.py，这是基于我的新需求，我不确定它是否正确：

from django.db import models
from django.contrib.auth.models import User


    class Report(models.Model):
        word = models.CharField(max_length=100)
        title = models.CharField(max_length=100) # i want this title be the file name that i built in the dothat() function.

        report_file = models.FileField(upload_to='reports/%Y/%m/%d')
        report_date = models.DateTimeField(auto_now=True)
        owner = models.ForeignKey(User, on_delete = models.CASCADE)

forms.py

from django.forms import ModelForm
from .models import Report
from django import forms


class IpscanForm(ModelForm):
    class Meta:
        model = Report
        fields = ['word'] # user just enter the word

我想在表单和模型中实现用户单一输入，将在后端处理的文件保存在数据库中，并收集所有用户特定的文件以在下载部分页面上显示给用户。我只是不知道如何将这些东西联系在一起。你有什么想法可以帮到我吗？

Answer 1

@login_required(login_url='/login/')
def user_reports(request):
   user_reports = models.Report.objects.filter(owner=request.user)
   return render(request, 'yourhtmlfile.html', {'user_reports':user_reports})

这将允许您显示发送请求的用户的所有报告。基本上，我们用请求用户过滤报表模型，得到所有与所有者用户的pk匹配的报表。