用django将文件复制到另一个文件夹中？

Question

我需要将配置文件图像上传到Django的不同文件夹中。因此，我为每个帐户都有一个文件夹，并且配置文件图像必须转到特定文件夹。我怎样才能做到这一点？

这是我的uploadprofile.html

<form action="{% url 'uploadimage' %}" enctype="multipart/form-data" method="POST">
  {% csrf_token %}
  <input type="file" name="avatar" accept="image/gif, image/jpeg, image/png">
  <button type="submit">Upload</button>
</form>

这是我在views.py中的观点

def uploadimage(request):
    img = request.FILES['avatar'] #Here I get the file name, THIS WORKS

    #Here is where I create the folder to the specified profile using the user id, THIS WORKS TOO
    if not os.path.exists('static/profile/' + str(request.session['user_id'])):
        os.mkdir('static/profile/' + str(request.session['user_id']))


    #Here is where I create the name of the path to save as a VARCHAR field, THIS WORKS TOO
    avatar = "../../static/profile/" + str(request.session['user_id']) + "/" + str(img)

    #THEN I HAVE TO COPY THE FILE IN img TO THE CREATED FOLDER

    return redirect(request, 'myapp/upload.html')

Answer 1

你可以将一个callable传递给upload_to。基本上，它意味着可调用返回的任何值，图像将被上传到该路径中。

例：

def get_upload_path(instance, filename):
    return "%s/%s" % (instance.user.id, filename)

class MyModel:
    user = ...
    image = models.FileField(upload_to=get_upload_path)

---

docs中有更多信息，也是一个例子，尽管与我上面发布的相似。

Answer 2

通过查看Django docs，当你执行img = request.FILES['avatar']时，你会得到一个文件描述符，指向一个带有你图像的打开文件。

然后你应该将内容转储到你真正的avatar路径中，对吧？

#Here is where I create the name of the path to save as a VARCHAR field, THIS WORKS TOO
avatar = "../../static/profile/" + str(request.session['user_id']) + "/" + str(img)
# # # # # 
with open(avatar, 'wb') as actual_file:
    actual_file.write(img.read())
# # # # #    
return redirect(request, 'myapp/upload.html')

注意：代码未经测试。