So what I've done is created an HTML form with which I can submit CSV files that are then handled by a Python script.
Further, I will very likely be creating the filenames of each CSV file dynamically in the following format: YYYY-MM-DD HH.MM.SS.csv.
Is there any way to retrieve the filename of the submitted CSV file so that I can open the file in a manner depicted below?
with open('????-??-?? ??.??.??.csv', newline='') as csvfile:
reader = csv.DictReader(csvfile)
for row in reader:
print(row['first_name'], row['last_name'])
Below is what the form looks like, and I know I can pass the filename forward using get, but there has to be a way to do it with post...
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<form action="process-excel.py" method="post" enctype="multipart/form-data">
<input type="file" name="myFile">
</form>
</body>
</html>
process-excel.py
Bear in mind that the script below is just an old template I kept for using PyMySQL to insert data into a database.
# -*- coding: utf-8 -*-
import pymysql.cursors
import csv
""" HOW DO I RETRIEVE THE FILENAME OF A FILE THAT IS SUBMITTED
''' FROM <input type="file" name="myFile"> IN THE FORM ABOVE?
"""
with open('names.csv', newline='') as csvfile:
reader = csv.DictReader(csvfile)
for row in reader:
print(row['first_name'], row['last_name'])
con = pymysql.Connect(host='localhost',
user='',
password='',
db='',
charset='utf8mb4',
cursorclass=pymysql.cursors.DictCursor)
try:
with con.cursor() as cursor:
sql = "INSERT INTO `test` (`title`, `description`, `category`) VALUES (%s, %s, %s)"
cursor.execute(sql, ('a', 'b', 'c'))
con.commit()
except Exception:
raise Exception('INSERT FAILED..................................')
