Flask:获取邮递员发送的 gzip 文件名 [英] Flask: Get gzip filename sent from Postman

问题描述

我正在从 Postman 向 Flask 端点发送一个 gzip 文件.我可以使用 request.data 获取该二进制文件并读取、保存、上传等.

I am sending a gzip file from Postman to a Flask endpoint. I can take that binary file with request.data and read it, save it, upload it, etc.

我的问题是我不能取它的名字.我该怎么做?

My problem is that I can't take its name. How can I do that?

我的 gzip 文件名为test_file.json.gz"，我的文件名为test_file.json".

My gzip file is called "test_file.json.gz" and my file is called "test_file.json".

我怎么能取任何这些名字?

How can I take any of those names?

我正在使用 io.BytesIO() 获取流数据，但是这个库不包含 name 属性或其他东西，尽管我可以在字符串中看到文件名:

I'm taking the stream data with io.BytesIO(), but this library doesn't contain a name attribute or something, although I can see the file name into the string if I just:

>>>print(request.data)
>>>b'x1fx8bx08x08xcaxb1xd3]x00x03test_file.jsonx00xabxe6RPPxcaNxad4Txb2RP*Kxcc)M5Txe2xaax05x00xc2x8bxb6;x16x00x00x00'

推荐答案

进一步评论，我认为处理您的上传的代码在这里是相关的.

Further to the comment, I think the code which handles your upload is relevant here.

关于request.data，请参阅这个答案:

request.data 包含传入的请求数据作为字符串，以防它带有 Flask 无法处理的 mimetype.

request.data Contains the incoming request data as string in case it came with a mimetype Flask does not handle.

推荐方式处理烧瓶中的文件上传是使用:

The recommended way to handle file uploads in flask is to use:

file = request.files['file']

• file 然后是类型:werkzeug.datastructures.FileStorage.

  • file is then of type: werkzeug.datastructures.FileStorage.

    file.stream 是流，可以用 file.stream.read() 或简单的 file.read()<读取/code>

    file.stream is the stream, which can be read with file.stream.read() or simply file.read()

    file.filename 是客户端指定的文件名.

    file.filename is the filename as specified on the client.

    file.save(path) 一种将文件保存到磁盘的方法.path 应该是一个类似 '/some/location/file.ext'

    file.save(path) a method which saves the file to disk. path should be a string like '/some/location/file.ext'

    来源

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