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suppose I have the following function:
def test():
...
if x['error']:
raise
This would raise an exception regardless if x['error'] is defined or not.
Instead if I try this, it doesn't throw any exception:
def test():
...
try:
if x['error']:
raise
except:
return
How can I test for a specific value and return an exception if it is defined, and to return successfully if it is not defined?
201
Use raise to throw exceptions in Python If you have a specific condition in your function that should loudly crash your program (if/when that condition is met) you can raise an exception by using the raise statement and providing an exception object to raise.
As a Python developer you can choose to throw an exception if a condition occurs. To throw (or raise) an exception, use the raise keyword.
Catching Exceptions in Python In Python, exceptions can be handled using a try statement. The critical operation which can raise an exception is placed inside the try clause. The code that handles the exceptions is written in the except clause.
You can't raise and return , but you could return multiple values, where the first is the same as what you're currently using, and the second indicates if an exception arose return True, sys.
If you want to return error as string:
>>> def test():
try:
if x['error']:raise
except Exception as err:
return err
>>> test()
NameError("name 'x' is not defined",)
If you want to an error to occur:
>>> def test():
try:
if x['error']:raise
except:
raise
>>> test()
Traceback (most recent call last):
File "<pyshell#20>", line 1, in <module>
test()
File "<pyshell#19>", line 3, in test
if x['error']:raise
NameError: name 'x' is not defined
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