Get size of a file before downloading in Python

The size of the file is sent as the Content-Length header. Here is how to get it with urllib:

>>> site = urllib.urlopen("http://python.org")
>>> meta = site.info()
>>> print meta.getheaders("Content-Length")
['16535']
>>>

Using the returned-urllib-object method info(), you can get various information on the retrieved document. Example of grabbing the current Google logo:

>>> import urllib
>>> d = urllib.urlopen("http://www.google.co.uk/logos/olympics08_opening.gif")
>>> print d.info()

Content-Type: image/gif
Last-Modified: Thu, 07 Aug 2008 16:20:19 GMT  
Expires: Sun, 17 Jan 2038 19:14:07 GMT 
Cache-Control: public 
Date: Fri, 08 Aug 2008 13:40:41 GMT 
Server: gws 
Content-Length: 20172 
Connection: Close

It’s a dict, so to get the size of the file, you do urllibobject.info()['Content-Length']

print f.info()['Content-Length']

And to get the size of the local file (for comparison), you can use the os.stat() command:

os.stat("/the/local/file.zip").st_size

Also if the server you are connecting to supports it, look at Etags and the If-Modified-Since and If-None-Match headers.

Using these will take advantage of the webserver’s caching rules and will return a 304 Not Modified status code if the content hasn’t changed.

In Python3:

>>> import urllib.request
>>> site = urllib.request.urlopen("http://python.org")
>>> print("FileSize: ", site.length)

A requests-based solution using HEAD instead of GET (also prints HTTP headers):

#!/usr/bin/python
# display size of a remote file without downloading

from __future__ import print_function
import sys
import requests

# number of bytes in a megabyte
MBFACTOR = float(1 << 20)

response = requests.head(sys.argv[1], allow_redirects=True)

print("n".join([('{:<40}: {}'.format(k, v)) for k, v in response.headers.items()]))
size = response.headers.get('content-length', 0)
print('{:<40}: {:.2f} MB'.format('FILE SIZE', int(size) / MBFACTOR))

Usage

$ python filesize-remote-url.py https://httpbin.org/image/jpeg
...
Content-Length                          : 35588
FILE SIZE (MB)                          : 0.03 MB

For a python3 (tested on 3.5) approach I’d recommend:

with urlopen(file_url) as in_file, open(local_file_address, 'wb') as out_file:
    print(in_file.getheader('Content-Length'))
    out_file.write(response.read())

@PabloG Regarding the local/server filesize difference

Following is high-level illustrative explanation of why it may occur:

The size on disk sometimes is different from the actual size of the data.
It depends on the underlying file-system and how it operates on data.
As you may have seen in Windows when formatting a flash drive you are asked to provide ‘block/cluster size’ and it varies [512b – 8kb].
When a file is written on the disk, it is stored in a ‘sort-of linked list’ of disk blocks.
When a certain block is used to store part of a file, no other file contents will be stored in the same blok, so even if the chunk is no occupuing the entire block space, the block is rendered unusable by other files.

Example:
When the filesystem is divided on 512b blocks, and we need to store 600b file, two blocks will be occupied. The first block will be fully utilized, while the second block will have only 88b utilized and the remaining (512-88)b will be unusable resulting in ‘file-size-on-disk’ being 1024b.
This is why Windows has different notations for ‘file size’ and ‘size on disk’.

NOTE:
There are different pros & cons that come with smaller/bigger FS block, so do a better research before playing with your filesystem.

For anyone using Python 3 and looking for a quick solution using the requests package:

import requests 
response = requests.head( 
    "https://website.com/yourfile.mp4",  # Example file 
    allow_redirects=True
)
print(response.headers['Content-Length']) 

Note: Not all responses will have a Content-Length so your application will want to check to see if it exists.

if 'Content-Length' in response.headers:
    ... # Do your stuff here 

Here is a much more safer way for Python 3:

import urllib.request
site = urllib.request.urlopen("http://python.org")
meta = site.info()
meta.get('Content-Length') 

Returns:

'49829'

meta.get('Content-Length') will return the "Content-Length" header if exists. Otherwise it will be blank

Quick and reliable one-liner for Python3 using urllib:

import urllib

url = 'https://<your url here>'

size = urllib.request.urlopen(url).info().get('Content-Length', 0)

.get(<dict key>, 0) gets the key from dict and if the key is absent returns 0 (or whatever the 2nd argument is)

you can use requests to pull this data

File_Name=requests.head(LINK).headers["X-File-Name"]

#And other useful info** like the size of the file from this dict (headers)
#like 

File_size=requests.head(LINK).headers["Content-Length"]