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How to find the mime type of a file in python?

Let's say you want to save a bunch of files somewhere, for instance in BLOBs. Let's say you want to dish these files out via a web page and have the client automatically open the correct application/viewer.

Assumption: The browser figures out which application/viewer to use by the mime-type (content-type?) header in the HTTP response.

Based on that assumption, in addition to the bytes of the file, you also want to save the MIME type.

How would you find the MIME type of a file? I'm currently on a Mac, but this should also work on Windows.

Does the browser add this information when posting the file to the web page?

Is there a neat python library for finding this information? A WebService or (even better) a downloadable database?


S
Smart Manoj

The python-magic method suggested by toivotuo is outdated. Python-magic's current trunk is at Github and based on the readme there, finding the MIME-type, is done like this.

# For MIME types
import magic
mime = magic.Magic(mime=True)
mime.from_file("testdata/test.pdf") # 'application/pdf'

thanks for the comment! please note, that "above" is a difficult concept in stackoverflow, since the ordering is grouped by votes and ordered randomly inside the groups. I am guessing you refer to @toivotuo's answer.
Yeh, I didn\t have enough "points" to create comments at the time of writing this reply. But I probably should have written it as a comment, so that the @toivotuo could have edited his question.
rpm -qf /usr/lib/python2.7/site-packages/magic.py -i URL : darwinsys.com/file Summary : Python bindings for the libmagic API rpm -qf /usr/bin/file -i Name : file URL : darwinsys.com/file python-magic from darwinsys.com/file and which comes with Linux Fedora works like @toivotuo's said. And seems more main stream.
Beware that the debian/ubuntu package called python-magic is different to the pip package of the same name. Both are import magic but have incompatible contents. See stackoverflow.com/a/16203777/3189 for more.
As I commented on toivotuo’s answer, it is not outdated! You are talking about a different library. Can you please remove or replace that statement in your answer? It currently makes finding the best solution really difficult.
j
jfs

The mimetypes module in the standard library will determine/guess the MIME type from a file extension.

If users are uploading files the HTTP post will contain the MIME type of the file alongside the data. For example, Django makes this data available as an attribute of the UploadedFile object.


If the files are stored in BLOBs, as specified in the question, you may not know the file extension.
File extensions are not a reliable way to determine mime type.
import mimetypes mimetypes.MimeTypes().guess_type(filename)[0]
in python 3.6 this works: mimetypes.guess_type(path_file_to_upload)[1]
While @cerin is right that file extensions are not reliable, I've just discovered that the accuracy of python-magic (as suggested in the top answer) to be even lower, as confirmed by github.com/s3tools/s3cmd/issues/198. So, mimetypes seems a better candidate for me.
t
toivotuo

More reliable way than to use the mimetypes library would be to use the python-magic package.

import magic
m = magic.open(magic.MAGIC_MIME)
m.load()
m.file("/tmp/document.pdf")

This would be equivalent to using file(1).

On Django one could also make sure that the MIME type matches that of UploadedFile.content_type.


See Simon Zimmermann's post for an updated use of python-magic
@DarenThomas: As mentioned in mammadori’s answer, this answer is not outdated and distinct from Simon Zimmermann’s solution. If you have the file utility installed, you can probably use this solution. It works for me with file-5.32. On gentoo you also have to have the python USE-flag enabled for the file package.
S
Smart Manoj

This seems to be very easy

>>> from mimetypes import MimeTypes
>>> import urllib 
>>> mime = MimeTypes()
>>> url = urllib.pathname2url('Upload.xml')
>>> mime_type = mime.guess_type(url)
>>> print mime_type
('application/xml', None)

Please refer Old Post

Update - In python 3+ version, it's more convenient now:

import mimetypes
print(mimetypes.guess_type("sample.html"))

I don't think the urllib is required in your example.
for Python 3.X replace import urllib with from urllib import request. And then use "request" instead of urllib
Works for python 2.7 also
@oetzi's solution uses this module, but is more simple.
P
Pedro Lobito

13 year later... Most of the answers on this page for python 3 were either outdated or incomplete. To get the mime type of a file I use:

import mimetypes

mt = mimetypes.guess_type("https://www.w3.org/WAI/ER/tests/xhtml/testfiles/resources/pdf/dummy.pdf")
if mt:
    print("Mime Type:", mt[0])
else:
    print("Cannot determine Mime Type")

# Mime Type: application/pdf

Live Demo

From Python docs:

mimetypes.guess_type(url, strict=True)

Guess the type of a file based on its filename, path or URL, given by url. URL can be a string or a path-like object.

The return value is a tuple (type, encoding) where type is None if the type can’t be guessed (missing or unknown suffix) or a string of the form 'type/subtype', usable for a MIME content-type header.

encoding is None for no encoding or the name of the program used to encode (e.g. compress or gzip). The encoding is suitable for use as a Content-Encoding header, not as a Content-Transfer-Encoding header. The mappings are table driven. Encoding suffixes are case sensitive; type suffixes are first tried case sensitively, then case insensitively.

The optional strict argument is a flag specifying whether the list of known MIME types is limited to only the official types registered with IANA. When strict is True (the default), only the IANA types are supported; when strict is False, some additional non-standard but commonly used MIME types are also recognized.

Changed in version 3.8: Added support for url being a path-like object.


if file path has no extension,then mime can not get the right result.
b
bodo

Python bindings to libmagic

All the different answers on this topic are very confusing, so I’m hoping to give a bit more clarity with this overview of the different bindings of libmagic. Previously mammadori gave a short answer listing the available option.

libmagic

module name: magic

pypi: file-magic

source: https://github.com/file/file/tree/master/python

When determining a files mime-type, the tool of choice is simply called file and its back-end is called libmagic. (See the Project home page.) The project is developed in a private cvs-repository, but there is a read-only git mirror on github.

Now this tool, which you will need if you want to use any of the libmagic bindings with python, already comes with its own python bindings called file-magic. There is not much dedicated documentation for them, but you can always have a look at the man page of the c-library: man libmagic. The basic usage is described in the readme file:

import magic

detected = magic.detect_from_filename('magic.py')
print 'Detected MIME type: {}'.format(detected.mime_type)
print 'Detected encoding: {}'.format(detected.encoding)
print 'Detected file type name: {}'.format(detected.name)

Apart from this, you can also use the library by creating a Magic object using magic.open(flags) as shown in the example file.

Both toivotuo and ewr2san use these file-magic bindings included in the file tool. They mistakenly assume, they are using the python-magic package. This seems to indicate, that if both file and python-magic are installed, the python module magic refers to the former one.

python-magic

module name: magic

pypi: python-magic

source: https://github.com/ahupp/python-magic

This is the library that Simon Zimmermann talks about in his answer and which is also employed by Claude COULOMBE as well as Gringo Suave.

filemagic

module name: magic

pypi: filemagic

source: https://github.com/aliles/filemagic

Note: This project was last updated in 2013!

Due to being based on the same c-api, this library has some similarity with file-magic included in libmagic. It is only mentioned by mammadori and no other answer employs it.


G
Gringo Suave

2017 Update

No need to go to github, it is on PyPi under a different name:

pip3 install --user python-magic
# or:
sudo apt install python3-magic  # Ubuntu distro package

The code can be simplified as well:

>>> import magic

>>> magic.from_file('/tmp/img_3304.jpg', mime=True)
'image/jpeg'

can you do same for js or css file ?
Sure, why not??
m
mammadori

There are 3 different libraries that wraps libmagic.

2 of them are available on pypi (so pip install will work):

filemagic

python-magic

And another, similar to python-magic is available directly in the latest libmagic sources, and it is the one you probably have in your linux distribution.

In Debian the package python-magic is about this one and it is used as toivotuo said and it is not obsoleted as Simon Zimmermann said (IMHO).

It seems to me another take (by the original author of libmagic).

Too bad is not available directly on pypi.


I added a repo for convenience: github.com/mammadori/magic-python that way you can: pip install -e git://github.com/mammadori/magic-python.git#egg=Magic_file_extensions
a
apito

in python 2.6:

import shlex
import subprocess
mime = subprocess.Popen("/usr/bin/file --mime " + shlex.quote(PATH), shell=True, \
    stdout=subprocess.PIPE).communicate()[0]

This is unnecessary, since the file command is basically just a wrapper around libmagic. You may as well just use the python binding (python-magic), as in Simon's answer.
That depends on the operating system. On Mac OS X, for example, you have "file" but not libmagic in the normal environment.
this looks unsafe to me, PATH should be escaped. idk what the Python equivalent is, but php devs would use Popen("/usr/bin/file --mime ".escapeshellarg(PATH)); - for example your code would fail on files containing newlines or quotes, probably also $dollarsign, but it would also protect you against hackers doing PATH='; rm -rfv / and such shell injection
C
Community

You didn't state what web server you were using, but Apache has a nice little module called Mime Magic which it uses to determine the type of a file when told to do so. It reads some of the file's content and tries to figure out what type it is based on the characters found. And as Dave Webb Mentioned the MimeTypes Module under python will work, provided an extension is handy.

Alternatively, if you are sitting on a UNIX box you can use sys.popen('file -i ' + fileName, mode='r') to grab the MIME type. Windows should have an equivalent command, but I'm unsure as to what it is.


Nowdays you can just do subprocess.check_output(['file', '-b', '--mime', filename])
There is really no reason to resort to using an external tool when python-magic does the equivalent thing, all wrapped and cozy.
e
ewr2san

@toivotuo 's method worked best and most reliably for me under python3. My goal was to identify gzipped files which do not have a reliable .gz extension. I installed python3-magic.

import magic

filename = "./datasets/test"

def file_mime_type(filename):
    m = magic.open(magic.MAGIC_MIME)
    m.load()
    return(m.file(filename))

print(file_mime_type(filename))

for a gzipped file it returns: application/gzip; charset=binary

for an unzipped txt file (iostat data): text/plain; charset=us-ascii

for a tar file: application/x-tar; charset=binary

for a bz2 file: application/x-bzip2; charset=binary

and last but not least for me a .zip file: application/zip; charset=binary


o
oetzi

python 3 ref: https://docs.python.org/3.2/library/mimetypes.html

mimetypes.guess_type(url, strict=True) Guess the type of a file based on its filename or URL, given by url. The return value is a tuple (type, encoding) where type is None if the type can’t be guessed (missing or unknown suffix) or a string of the form 'type/subtype', usable for a MIME content-type header. encoding is None for no encoding or the name of the program used to encode (e.g. compress or gzip). The encoding is suitable for use as a Content-Encoding header, not as a Content-Transfer-Encoding header. The mappings are table driven. Encoding suffixes are case sensitive; type suffixes are first tried case sensitively, then case insensitively. The optional strict argument is a flag specifying whether the list of known MIME types is limited to only the official types registered with IANA. When strict is True (the default), only the IANA types are supported; when strict is False, some additional non-standard but commonly used MIME types are also recognized.

import mimetypes
print(mimetypes.guess_type("sample.html"))

C
Claude COULOMBE

In Python 3.x and webapp with url to the file which couldn't have an extension or a fake extension. You should install python-magic, using

pip3 install python-magic

For Mac OS X, you should also install libmagic using

brew install libmagic

Code snippet

import urllib
import magic
from urllib.request import urlopen

url = "http://...url to the file ..."
request = urllib.request.Request(url)
response = urlopen(request)
mime_type = magic.from_buffer(response.readline())
print(mime_type)

alternatively you could put a size into the read

import urllib
import magic
from urllib.request import urlopen

url = "http://...url to the file ..."
request = urllib.request.Request(url)
response = urlopen(request)
mime_type = magic.from_buffer(response.read(128))
print(mime_type)

Will it be load whole file?
No, it's a stream, so normally just few bytes.
I've edited by response.readline() or response.read(128) Thank you!
J
Jak Liao

I try mimetypes library first. If it's not working, I use python-magic libary instead.

import mimetypes
def guess_type(filename, buffer=None):
mimetype, encoding = mimetypes.guess_type(filename)
if mimetype is None:
    try:
        import magic
        if buffer:
            mimetype = magic.from_buffer(buffer, mime=True)
        else:
            mimetype = magic.from_file(filename, mime=True)
    except ImportError:
        pass
return mimetype

E
Eric McLachlan

I'm surprised that nobody has mentioned it but Pygments is able to make an educated guess about the mime-type of, particularly, text documents.

Pygments is actually a Python syntax highlighting library but is has a method that will make an educated guess about which of 500 supported document types your document is. i.e. c++ vs C# vs Python vs etc

import inspect

def _test(text: str):
    from pygments.lexers import guess_lexer
    lexer = guess_lexer(text)
    mimetype = lexer.mimetypes[0] if lexer.mimetypes else None
    print(mimetype)

if __name__ == "__main__":
    # Set the text to the actual defintion of _test(...) above
    text = inspect.getsource(_test)
    print('Text:')
    print(text)
    print()
    print('Result:')
    _test(text)

Output:

Text:
def _test(text: str):
    from pygments.lexers import guess_lexer
    lexer = guess_lexer(text)
    mimetype = lexer.mimetypes[0] if lexer.mimetypes else None
    print(mimetype)


Result:
text/x-python

Now, it's not perfect, but if you need to be able to tell which of 500 document formats are being used, this is pretty darn useful.


H
Helder

The mimetypes module just recognise an file type based on file extension. If you will try to recover a file type of a file without extension, the mimetypes will not works.


I don't think that's true. The MIME type is about how to tell others about a data format, not about how to find out the data format yourself. If you use a tool that guesses the format only based on the extension and prints out MIME types then you can't use that tool if there are no file extensions. But other ways to guess the format are possible as well, e.g., by checking with a parser.
A
Ajay

For byte Array type data you can use magic.from_buffer(_byte_array,mime=True)


A
Artem Bernatskyi

I 've tried a lot of examples but with Django mutagen plays nicely.

Example checking if files is mp3

from mutagen.mp3 import MP3, HeaderNotFoundError  

try:
    audio = MP3(file)
except HeaderNotFoundError:
    raise ValidationError('This file should be mp3')

The downside is that your ability to check file types is limited, but it's a great way if you want not only check for file type but also to access additional information.


i need to check security also