is there a Way to Grab the Filename and location from a File that is being downloaded by selenium and webdriver - selenium

Question

I have a script that automates downloading a file using selenium and webdriver as chrome
basically, it logs in to a work website, and clicks on some settings to prepare to download a report file, and then clicks the download button
Is there a way for selenium or any other library to grab the file location and name of that file that is downloading or has just downloaded so I can store it in a variable to use later in the script
I do not know if the relative path will work or if it needs to be a full windows path name for it to work... probably safe to assume full path is more guaranteed to work
example being C:\Users\FunnyUserName\Downloads\report.xls
Adding Code to show what is happening better
#For Report Pull
#-----------------------------------------------
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
import datetime
import os
import glob
###############################################################
#Pull Report #
###############################################################
#Open Web Driver
browser = webdriver.Chrome()
#Open Website and Log in
print('Open Website and Log In')
browser.get(('https://SomeWebsite.com'))
print('Working on Getting the QC Report')
print('Please Stand By')
#####
#I Removed a lot of stuff not necessary to this question
#Get the File
WebDriverWait(browser,10).until(EC.element_to_be_clickable((By.XPATH,'//*[#id="btnGenerateReport"]'))).click()
time.sleep(4)
#Working on getting the last downloaded Filename
# get the user download folder (dynamic so will work on any machine)
downLoadFolder =os.path.join( os.getenv('USERPROFILE'), 'Downloads')
print(downLoadFolder)
#This shows the correct folder....
#In My Case C:\Users\My UserName\Downloads
# get the list of files
list_of_files = glob.glob(downLoadFolder+"/*.*") # * means all if need specific formats (if you are looking for any specific format then specify eg: "/*.xls" to filter)
print (list_of_files)
#Always Shows ['C:\\Users\\My UserName\\Downloads\\desktop.ini']
# get the latest file name
#Forced the Folder and file type as a test
latest_file = max(glob.glob("C:/Users/My Username/Downloads/*.xls"), key=os.path.getctime)
#print the latest file name
print(latest_file)
#Returns:latest_file = max(glob.glob("C:/Users/My Username/Downloads/*.xls"), key=os.path.getctime)
#ValueError: max() arg is an empty sequence

Answer 1

I might have figured out where the problem is
I need it to grab the actual downloads directory, not the system default
I'm testing on my home computer, it default downloads to my dropbox download directory that I share with my work computer as well
so its not C:\Users\My Username\Downloads\
its actually D:\Dropbox\Downloads... which is where my chrome default is set currently
how do I get the chrome download directory?

Answer 2

Here is the solution in the python.
Imports Needed:
import glob
import os
Script:
# get the user download folder (dynamic so will work on any machine)
downLoadFolder =os.path.join( os.getenv('USERPROFILE'), 'Downloads')
# get the list of files
list_of_files = glob.glob(downLoadFolder+"/*") # * means all if need specific formats (if you are looking for any specific format then specify eg: "/*.xlsx" to filter)
# get the latest file name
latest_file = max(list_of_files, key=os.path.getctime)
#print the latest file name
print(latest_file)