Python - Not recognizing Exception Handling

Question

So I am having problems with exception handling, I am running Python 3.6.3. this is my code:

txt = ""
txtFile = input("gimme a file.")
f = open(txtFile, "r")
try:    
    for line in f:
        cleanedLine = line.strip() 
        txt += cleanedLine
except FileNotFoundError:
    print("!")

So if I try and get an error with a bad input Instead of printing ! I still get the error:

Traceback (most recent call last):
File "cleaner.py", line 11, in <module>
    f = open(txtFile, "r")
FileNotFoundError: [Errno 2] No such file or directory: 'nonexistentfile'

I have tried swapping in OSError, I have also tried just except:, which tells me that I am doing something wrong(because I shouldn't do that in the first place) and since I understand that except: should catch all of the exceptions.

The error occurs before your try statement. Hence the error will not be caught. — error
– error, Commented Feb 1, 2018 at 15:12

Cleve Green · Accepted Answer · 2018-02-01 15:15:01Z

1

Simple, you're opening something outside of the exception.

txt = []
txtFile = input("gimme a file.")
try:        
    f = open(txtFile, "r")
    for line in f.read().split('\n'):
        cleanedLine = line.strip()
        txt.append(cleanedLine)
except FileNotFoundError:
    print("!")

answered Feb 1, 2018 at 15:15

Cleve Green

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Comments

Matt_G · Accepted Answer · 2018-02-01 15:14:17Z

0

Your try catch is encapsulating the loop through the lines.

The error is occurring when you are trying to open the file, outside of your try block.

answered Feb 1, 2018 at 15:14

Matt_G

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1 Comment

AyO Over a year ago

thanks, I am new to this, and I was misinterpreting the traceback .

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