烧瓶/ Python.从上传的文件中获取mimetype

Question

我正在使用Flask微框架0.6和Python 2.6

我需要从上传的文件中获取mimetype,以便我可以存储它.

这是相关的Python/Flask代码:

@app.route('/upload_file', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        file = request.files['file']
        mimetype = #FIXME
        if file:
            file.save(os.path.join(UPLOAD_FOLDER, 'File-Name')
            return redirect(url_for('uploaded_file'))
        else:
            return redirect(url_for('upload'))

这是网页的代码:

<form action="upload_file" method=post enctype=multipart/form-data> 
Select file to upload: <input type=file name=file> 
<input type=submit value=Upload> 
</form> 

代码有效,但我需要能够在上传时获取mimetype.我在这里看过Flask文档:http ://flask.pocoo.org/docs/api/#incoming-request-data
所以我知道它确实得到了mimetype,但我无法弄清楚如何检索它 - 作为文本字符串,例如'txt/plain'.

有任何想法吗？

谢谢.

Answer 1

从文档中,file.content_type包含带编码的完整类型,mimetype仅包含mime类型.

@app.route('/upload_file', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'POST':
        file = request.files.get('file')
        if file:
            mimetype = file.content_type
            filename = werkzeug.secure_filename(file.filename)
            file.save(os.path.join(UPLOAD_FOLDER, filename)
            return redirect(url_for('uploaded_file'))
        else:
            return redirect(url_for('upload'))

Answer 2

理论上您可以使用request.files['YOUR_FILE_KEY'].content_type，但实现（包含在下面，在werkzeug.datastructures 中找到）要么信任客户端提供的任何内容，要么mimetypes.guess_type只使用它检查文件扩展名（请参阅此处的Python 文档）。

class FileMultiDict(MultiDict):

    """A special :class:`MultiDict` that has convenience methods to add
    files to it.  This is used for :class:`EnvironBuilder` and generally
    useful for unittesting.
    .. versionadded:: 0.5
    """

    def add_file(self, name, file, filename=None, content_type=None):
        """Adds a new file to the dict.  `file` can be a file name or
        a :class:`file`-like or a :class:`FileStorage` object.
        :param name: the name of the field.
        :param file: a filename or :class:`file`-like object
        :param filename: an optional filename
        :param content_type: an optional content type
        """
        if isinstance(file, FileStorage):
            value = file
        else:
            if isinstance(file, string_types):
                if filename is None:
                    filename = file
                file = open(file, 'rb')
            if filename and content_type is None:
                content_type = mimetypes.guess_type(filename)[0] or \
                    'application/octet-stream'
            value = FileStorage(file, filename, name, content_type)

        self.add(name, value)

根据您的用例，您可能想要使用python-magic，它将使用实际文件来获取 mimetype。它会是这样的：

import magic


def get_mimetype(data: bytes) -> str:
    """Get the mimetype from file data."""
    f = magic.Magic(mime=True)
    return f.from_buffer(data)


get_mimetype(request.files['YOUR_FILE_KEY'].stream.read(MAX_LENGTH))