How to make a variable inside a try/except block public?

Question

How can I make a variable inside the try/except block public?

import urllib.request

try:
    url = "http://www.google.com"
    page = urllib.request.urlopen(url)
    text = page.read().decode('utf8')
except (ValueError, RuntimeError, TypeError, NameError):
    print("Unable to process your request dude!!")

print(text)

This code returns an error

NameError: name 'text' is not defined

How can I make the variable text available outside of the try/except block?

Solution 1

try statements do not create a new scope, but text won't be set if the call to url lib.request.urlopen raises the exception. You probably want the print(text) line in an else clause, so that it is only executed when there is no exception.

try:
    url = "http://www.google.com"
    page = urllib.request.urlopen(url)
    text = page.read().decode('utf8')
except (ValueError, RuntimeError, TypeError, NameError):
    print("Unable to process your request dude!!")
else:
    print(text)

If text needs to be used later, you really need to think about what its value is supposed to be if the assignment to page fails and you can't call page.read(). You can give it an initial value prior to the try statement:

text = 'something'
try:
    url = "http://www.google.com"
    page = urllib.request.urlopen(url)
    text = page.read().decode('utf8')
except (ValueError, RuntimeError, TypeError, NameError):
    print("Unable to process your request dude!!")

print(text)

or in the else clause:

try:
    url = "http://www.google.com"
    page = urllib.request.urlopen(url)
    text = page.read().decode('utf8')
except (ValueError, RuntimeError, TypeError, NameError):
    print("Unable to process your request dude!!")
else:
    text = 'something'

print(text)

Solution 2

As answered before there is no new scope introduced by using try except clause, so if no exception occurs you should see your variable in locals list and it should be accessible in current (in your case global) scope.

print(locals())

In module scope (your case) locals() == globals()

Solution 3

Just declare the variable text outside try except block,

import urllib.request
text =None
try:
    url = "http://www.google.com"
    page = urllib.request.urlopen(url)
    text = page.read().decode('utf8')
except (ValueError, RuntimeError, TypeError, NameError):
    print("Unable to process your request dude!!")
if text is not None:
    print(text)

Comments

[Kobi K]

I think the OP wants a default value for text in case there is an exception. so maybe updating this answer with default value outside the try except scope or inside the else will help him.

[Nathan Basanese]

// , Why would this be necessary? Python does not have a separate block-level scope for this try and except, AFAIK, and according to stackoverflow.com/questions/17195569/….

[4xy]

In the last snippet text will always eventually be assigned 'something' if no exception

[Silidrone]

"try statements do not create a new scope" this cleared up a lot of things for me, +1

[santma]

the "else" clause didn't work for me, but the initial value did. :D

[Qiang Xu]

In the 3rd example, aside from the problem mentioned by @4xy, print(text) would still cause the original NameError if page.read().decode('utf8') raises an exception. It would only work if text = 'something' is moved into the except block.

[nabil.adnan1610]

Thank you! This one line "text won't be set if the call to url lib.request.urlopen raises the exception." cleared up the confusion I had for ages!