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

65
votes

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?

3
pythonpython-3.xscopelocal-variablestry-except
What will be text if page = urllib.request.urlopen(url) will raise an exception? If you want to set it with some value you can do it in the except scope or outside the try except block. - Kobi K
text is the be the markup of the url. - x0x
In the except case text is never assigned. You could set text = None in that block or before the try. This isn't a scope problem. - jonrsharpe
if the error mentioned in the question is raised without going through the print statement in the except block then, I think the exception is not getting caught - Ashoka Lella
// , According to stackoverflow.com/questions/17195569/…, Python does not have a block scope. - Nathan Basanese

3 Answers

89
votes

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)
4
votes

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()

2
votes

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)