Django 将文件上传到取决于 POST URI 的特定目录

发布于 2024-08-21 09:24:49 字数 257 浏览 6 评论 0原文

我想将上传的文件存储到取决于 POST 请求的 URI 的特定目录中。也许，我还想将文件重命名为固定的名称（例如文件输入的名称），这样我就有一种简单的方法来 grep 文件系统等，并避免可能的安全问题。

在 Django 中执行此操作的首选方法是什么？

编辑：我应该澄清，我有兴趣可能将其作为文件上传处理程序来执行，以避免将大文件写入文件系统两次。

Edit2：我想人们可以将 tmp 文件“mv”到新位置。如果在同一文件系统上，这是一个便宜的操作。

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玩心态 2024-08-28 09:24:49

修复了 olooney 示例。现在正在工作

@csrf_exempt
def upload_video_file(request):
    folder = 'tmp_dir2/' #request.path.replace("/", "_")
    uploaded_filename = request.FILES['file'].name
    BASE_PATH = '/home/'
    # create the folder if it doesn't exist.
    try:
        os.mkdir(os.path.join(BASE_PATH, folder))
    except:
        pass

    # save the uploaded file inside that folder.
    full_filename = os.path.join(BASE_PATH, folder, uploaded_filename)
    fout = open(full_filename, 'wb+')

    file_content = ContentFile( request.FILES['file'].read() )

    try:
        # Iterate through the chunks.
        for chunk in file_content.chunks():
            fout.write(chunk)
        fout.close()
        html = "<html><body>SAVED</body></html>"
        return HttpResponse(html)
    except:
        html = "<html><body>NOT SAVED</body></html>"
        return HttpResponse(html)

Fixed olooney example. It is working now

@csrf_exempt
def upload_video_file(request):
    folder = 'tmp_dir2/' #request.path.replace("/", "_")
    uploaded_filename = request.FILES['file'].name
    BASE_PATH = '/home/'
    # create the folder if it doesn't exist.
    try:
        os.mkdir(os.path.join(BASE_PATH, folder))
    except:
        pass

    # save the uploaded file inside that folder.
    full_filename = os.path.join(BASE_PATH, folder, uploaded_filename)
    fout = open(full_filename, 'wb+')

    file_content = ContentFile( request.FILES['file'].read() )

    try:
        # Iterate through the chunks.
        for chunk in file_content.chunks():
            fout.write(chunk)
        fout.close()
        html = "<html><body>SAVED</body></html>"
        return HttpResponse(html)
    except:
        html = "<html><body>NOT SAVED</body></html>"
        return HttpResponse(html)

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太阳哥哥 2024-08-28 09:24:49

Django 使您可以完全控制保存文件的位置（以及是否保存）。请参阅： http://docs.djangoproject.com/en/dev /topics/http/file-uploads/

下面的示例展示了如何组合 URL 和上传文件的名称并将文件写入磁盘：

def upload(request):
    folder = request.path.replace("/", "_")
    uploaded_filename = request.FILES['file'].name

    # create the folder if it doesn't exist.
    try:
        os.mkdir(os.path.join(BASE_PATH, folder))
    except:
        pass

    # save the uploaded file inside that folder.
    full_filename = os.path.join(BASE_PATH, folder, uploaded_filename)
    fout = open(full_filename, 'wb+')
    # Iterate through the chunks.
    for chunk in fout.chunks():
        fout.write(chunk)
    fout.close()

编辑：如何使用 FileUploadHandler 执行此操作？它追踪了代码，似乎您需要做四件事来重新调整 TemporaryFileUploadHandler 的用途以保存在 FILE_UPLOAD_TEMP_DIR 之外：

extend TemporaryUploadedFile 并覆盖 init() 以通过不同的目录传递给 NamedTemporaryFile。它可以使用 try mkdir 除了上面显示的 pass 之外。
扩展 TemporaryFileUploadHandler 和覆盖 new_file() 以使用上述类。
还扩展 init() 以接受您想要文件夹所在的目录。
动态添加请求处理程序，传递由 URL 确定的目录：
request.upload_handlers = [ProgressBarUploadHandler(request.path.replace('/', '_')]

虽然不简单，但它仍然比从头开始编写处理程序更容易：特别是，您不必编写一行容易出错的缓冲读取步骤 3 和 4 是必要的，因为我相信默认情况下 FileUploadHandler 不会传递请求信息，因此如果您想以某种方式使用 URL，

我真的不建议为此编写自定义 FileUploadHandler 。相对于通过互联网上传文件的速度来说，这确实是一个混合层。如果文件很小，Django 只会将其保留在内存中，而不会将其写入临时文件。我有一种不好的预感，当你完成所有这些工作后，会发现你甚至无法衡量性能差异。

Django gives you total control over where (and if) you save files. See: http://docs.djangoproject.com/en/dev/topics/http/file-uploads/

The below example shows how to combine the URL and the name of the uploaded file and write the file out to disk:

def upload(request):
    folder = request.path.replace("/", "_")
    uploaded_filename = request.FILES['file'].name

    # create the folder if it doesn't exist.
    try:
        os.mkdir(os.path.join(BASE_PATH, folder))
    except:
        pass

    # save the uploaded file inside that folder.
    full_filename = os.path.join(BASE_PATH, folder, uploaded_filename)
    fout = open(full_filename, 'wb+')
    # Iterate through the chunks.
    for chunk in fout.chunks():
        fout.write(chunk)
    fout.close()

Edit: How to do this with a FileUploadHandler? It traced down through the code and it seems like you need to do four things to repurpose the TemporaryFileUploadHandler to save outside of FILE_UPLOAD_TEMP_DIR:

extend TemporaryUploadedFile and override init() to pass through a different directory to NamedTemporaryFile. It can use the try mkdir except for pass I showed above.
extend TemporaryFileUploadHandler and override new_file() to use the above class.
also extend init() to accept the directory where you want the folder to go.
Dynamically add the request handler, passing through a directory determined from the URL:
request.upload_handlers = [ProgressBarUploadHandler(request.path.replace('/', '_')]

While non-trivial, it's still easier than writing a handler from scratch: In particular, you won't have to write a single line of error-prone buffered reading. Steps 3 and 4 are necessary because FileUploadHandlers are not passed request information by default, I believe, so you'll have to tell it separately if you want to use the URL somehow.

I can't really recommend writing a custom FileUploadHandler for this. It's really mixing layers of responsibility. Relative to the speed of uploading a file over the internet, doing a local file copy is insignificant. And if the file's small, Django will just keep it in memory without writing it out to a temp file. I have a bad feeling that you'll get all this working and find you can't even measure the performance difference.

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