python selenium, find out when a download has completed?


I’ve used selenium to initiate a download. After the download is complete, certain actions need to be taken, is there any simple method to find out when a download has complete? (I am using the FireFox driver)

Asked By: applecider



There is no built-in to selenium way to wait for the download to be completed.

The general idea here would be to wait until a file would appear in your “Downloads” directory.

This might either be achieved by looping over and over again checking for file existence:

Or, by using things like watchdog to monitor a directory:

Answered By: alecxe

With Chrome, files which have not finished downloading have the extension .crdownload. If you set your download directory properly, then you can wait until the file that you want no longer has this extension. In principle, this is not much different to waiting for file to exist (as suggested by alecxe) – but at least you can monitor progress in this way.

Answered By: kd88
while x1==0:
    li = os.listdir("directorypath")
    for x1 in li:
        if x1.endswith(".crdownload"):
             count = count+1        
    if count==0:

This works if you are trying to check if a set of files(more than one) have finished downloading.

Answered By: Veiledwhale

I came across this problem recently. I was downloading multiple files at once and had to build in a way to timeout if the downloads failed.

The code checks the filenames in some download directory every second and exits once they are complete or if it takes longer than 20 seconds to finish. The returned download time was used to check if the downloads were successful or if it timed out.

import time
import os

def download_wait(path_to_downloads):
    seconds = 0
    dl_wait = True
    while dl_wait and seconds < 20:
        dl_wait = False
        for fname in os.listdir(path_to_downloads):
            if fname.endswith('.crdownload'):
                dl_wait = True
        seconds += 1
    return seconds

I believe that this only works with chrome files as they end with the .crdownload extension. There may be a similar way to check in other browsers.

Edit: I recently changed the way that I use this function for times that .crdownload does not appear as the extension. Essentially this just waits for the correct number of files as well.

def download_wait(directory, timeout, nfiles=None):
    Wait for downloads to finish with a specified timeout.

    directory : str
        The path to the folder where the files will be downloaded.
    timeout : int
        How many seconds to wait until timing out.
    nfiles : int, defaults to None
        If provided, also wait for the expected number of files.

    seconds = 0
    dl_wait = True
    while dl_wait and seconds < timeout:
        dl_wait = False
        files = os.listdir(directory)
        if nfiles and len(files) != nfiles:
            dl_wait = True

        for fname in files:
            if fname.endswith('.crdownload'):
                dl_wait = True

        seconds += 1
    return seconds
Answered By: Austin Mackillop

As answered before, there is no native way to check if download is finished. So here is a helper function that does the job for Firefox and Chrome. One trick is to clear the temp download folder before start a new download. Also, use native pathlib for cross-platform usage.

from pathlib import Path

def is_download_finished(temp_folder):
    firefox_temp_file = sorted(Path(temp_folder).glob('*.part'))
    chrome_temp_file = sorted(Path(temp_folder).glob('*.crdownload'))
    downloaded_files = sorted(Path(temp_folder).glob('*.*'))
    if (len(firefox_temp_file) == 0) and 
       (len(chrome_temp_file) == 0) and 
       (len(downloaded_files) >= 1):
        return True
        return False
Answered By: C. Feng

I know its too late for the answer, though would like to share a hack for future readers.

You can create a thread say thread1 from main thread and initiate your download here.
Now, create some another thread, say thread2 and in there ,let it wait till thread1 completes using join() method.Now here,you can continue your flow of execution after download completes.

Still make sure you dont initiate your download using selenium, instead extract the link using selenium and use requests module to download.

Download using requests module

For eg:

def downloadit():
     #download code here    

def after_dwn():
     dwn_thread.join()           #waits till thread1 has completed executing
     #next chunk of code after download, goes here

dwn_thread = threading.Thread(target=downloadit)

metadata_thread = threading.Thread(target=after_dwn)
Answered By: Dhyey Shah

this worked for me:

fileends = "crdownload"
while "crdownload" in fileends:
    for fname in os.listdir(some_path): 
        if "crdownload" in fname:
            fileends = "crdownload"
            fileends = "None"
Answered By: greencode

Check for "Unconfirmed" key word in file name in download directory:

# wait for download complete
wait = True
    for fname in os.listdir('pathtodownload directory'):
        if ('Unconfirmed') in fname:
            print('downloading files ...')
print('finished downloading all files ...')

As soon as the the filed download is completed it exits the while loop.

Answered By: Jawad Ahmad Khan

I got a better one though:

So redirect the function that starts the download. e.g. download_function= lambda:

than check number of files in directory and wait for a new file that doesnt have the download extension. After that rename it. (can be change to move the file instead of renaming it in the same directory)

def save_download(self, directory, download_function, new_name, timeout=30):
    Download a file and rename it
    :param directory: download location that is set
    :param download_function: function to start download
    :param new_name: the name that the new download gets
    :param timeout: number of seconds to wait for download
    :return: path to downloaded file
    """"Downloading " + new_name)
    files_start = os.listdir(directory)
    wait = True
    i = 0
    while (wait or len(os.listdir(directory)) == len(files_start)) and i < timeout * 2:
        wait = False
        for file_name in os.listdir(directory):
            if file_name.endswith('.crdownload'):
                wait = True
    if i == timeout * 2:
        self.logger.warning("Documents not downloaded")
        raise TimeoutError("File not downloaded")
    else:"Downloading done")
        new_file = [name for name in os.listdir(directory) if name not in files_start][0]"New file found renaming " + new_file + " to " + new_name)
        while not os.access(directory + r"\" + new_file, os.W_OK):
  "Waiting for write permission")
        os.rename(directory + "\" + new_file, directory + "\" + new_name)
        return directory + "\" + new_file
Answered By: Martijn Witteveen
import os
import time

def latest_download_file():
      path = r'Downloads folder file path'
      files = sorted(os.listdir(os.getcwd()), key=os.path.getmtime)
      newest = files[-1]

      return newest

fileends = "crdownload"
while "crdownload" == fileends:
    newest_file = latest_download_file()
    if "crdownload" in newest_file:
        fileends = "crdownload"
        fileends = "none"

This is a combination of a few solutions. I didn’t like that I had to scan the entire downloads folder for a file ending in "crdownload". This code implements a function that pulls the newest file in downloads folder. Then it simply checks if that file is still being downloaded. Used it for a Selenium tool I am building worked very well.

Answered By: Red

If using Selenium and Chrome, you can write a custom wait condition such as:

class file_has_been_downloaded(object):
def __init__(self, dir, number):
    self.dir = dir
    self.number = number

def __call__(self, driver):
    print(count_files(dir), '->', self.number)
    return count_files(dir) > self.number

The function count_files just verifies that the file has been added to the folder

def count_files(direct):
for root, dirs, files in os.walk(direct):
    return len(list(f for f in files if f.startswith('MyPrefix') and (
            not f.endswith('.crdownload')) ))

Then to implement this in your code:

files = count_files(dir)
<< copy the file. Possibly use shutil >>
WebDriverWait(driver, 30).until(file_has_been_downloaded(dir, files))
Answered By: cookee

create a function that uses "requests" to get the file content and call that one, your program will not move forward unless the file is downloaded

import requests
from selenium import webdriver
driver = webdriver.Chrome()
# Open the website

x = driver.find_element_by_partial_link_text('download')
y = x.get_attribute("href")
fc = requests.get(y)
fname = x.text
with open(fname, 'wb') as f:
Answered By: Jay777

This is VERY SIMPLE and worked for me (and works fot any extention)

import os, glob and time (not truly needed)

# count how many files you have in Downloads folder before download

user = os.getlogin()
downloads_folder = (r"C:/Users/" + user + "/Downloads/")
files_path = os.path.join(downloads_folder, '*')
files = sorted(glob.iglob(files_path), key=os.path.getctime, reverse=True)
files_before_download = files

print(f'files before download: {len(files)}')
finished = False

# ...
# code
# to
# download
# file
# ...

# just for extra safety

# wait for the download to finish if there is +1 file in Downloads folder
while not finished:
    files = sorted(glob.iglob(files_path), key=os.path.getctime, reverse=True)
    if (len(files) == len(files_before_download)) or (len(files) == (len(files_before_download)+2)):
        print('not finished')
        finished = False
        finished = True

last_downloaded_file = files[0]
Answered By: Renato Andrade

well, what if you check the size of the file until it has x size? There must be an average (Too bored to buid code, build it. Ideas helps too)

Answered By: Carlost


  • poll for the existence of the file
  • poll for non-zero filesize of that file

Observed Behaviour

I noticed that there can be a lag between a downloaded file appearing in the filesystem and the contents of that file being fully written, especially noticeable with large files.

I did some experimenting, using stat_result from os.stat() on Linux, and found the following,

  • when a file is first opened for writing
    • st_size == 0
    • st_atime == st_mtime == st_ctime
  • while data is being written to the file
    • st_size == 0
    • st_atime == st_mtime == st_ctime
  • once the writing is complete and the file is closed
    • st_size > 0
    • st_atime < st_mtime == st_ctime


  • Poll for a file using glob with a configurable timeout
    • This is useful when you don’t know exactly what the name of the downloaded file will be
  • Poll for the filesize of a specific file to be above a threshold
import glob
import polling2
import os

def poll_for_file_glob(file_glob: str, step: int=1, timeout: int=20):
        polling2.poll(lambda: len(glob.glob(file_glob)), step=step, timeout=timeout)
    except polling2.TimeoutException:
        raise RuntimeError(f"Unable to find file matching glob '{file_glob}'")
    return glob.glob(file_glob)[0]

def poll_for_file_size(file_path: str, size_threshold: int=0, step: int=1, timeout: int=20):
        polling2.poll(lambda: os.stat(file_path).st_size > size_threshold, step=step, timeout=timeout)
    except polling2.TimeoutException:
        file_size = os.stat(file_path).st_size
        raise RuntimeError(f"File '{file_path}' has size {file_size}, which is not larger than threshold {size_threshold}")
    return os.stat(file_path).st_size

You might use these functions like this,

  file_glob = "file_*.csv"
  file_path = poll_for_file_glob(file_glob=file_glob)
  file_size = poll_for_file_size(file_path=file_path)
  print(f"Problem polling for file matching '{file_glob}'")
  print(f"File '{file_path}' ({file_size}B) is ready")
Answered By: commander.trout
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