How do I isolate a problem in a Python script when the terminal output is the print statement from the except block?

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When I run this program with an actual gmail address and the correct password, the terminal output is:

➜  Program python3 mail.py
Something went wrong...
Something went wrong...
Something went wrong...

Here is the entire program file mail.py:

import smtplib
import csv

sender_email = '[EMAIL WILL GO HERE]'
sender_password = '[PASSWORD WILL GO HERE]'
body = 'This email was sent using Python.'

with open('emails.csv', 'r') as csv_file:
    read_file = csv.reader(csv_file)
    for row in read_file:
        if len(row[2]) > 1:
            person_email = row[2]
            try:
                server_ssl = smtplib.SMTP_SSL('smtp.gmail.com', 465)
                server_ssl.login(sender_email, sender_password)
                server_ssl.sendmail(sender_email, person_email, body)
                server_ssl.close()
                print('Emails sent!')
            except:
                print('Something went wrong...')

What I have tried:

I get the same terminal output when I run the program with an actual gmail account address and the correct password. I've tried searching Stack Overflow for answers, I've tried searching online for answers, but since I'm not getting a traceback, I'm getting the message from the except block, there is no information provided about which step is failing. How do I isolate a problem in a Python script when the terminal output is the print statement from the except block?

Posted 12-Sep-21 11:39am

Member 15355548

Updated 12-Sep-21 13:40pm

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Patrice T · Accepted Answer · 2021-09-12T13:40:00

Quote:
How do I isolate a problem in a Python script when the terminal output is the print statement from the except block?

By reading Python documentation for 'try' : 8. Errors and Exceptions — Python 3.9.7 documentation[^]
If you dare to read it until 'except', you will see how to get detail on the exception.

Python

... except Exception as inst:
...     print(type(inst))    # the exception instance
...     print(inst.args)     # arguments stored in .args
...     print(inst)          # __str__ allows args to be printed directly,
...                          # but may be overridden in exception subclasses

Also removing the 'try' will have your program crash with full error message from Python.