Does Python have a Java equivalent of throw new Exception("text here")

Question

I'm a Java developer who's new to Python and I'm rewriting a Java class as a Python class. I'm trying to mimic the flow of the original class in my Python class as much as possible. The Java class has a few lines with,

    if(condition)
       throw new Exception("text here")

I've been looking over the Python documentation for exceptions and have not been able to find a Python equivalent to the Java syntax.

I've tried something (I think is close) with raise Exception("text here") by reading this StackOverflow post but it seems as if this is for use inside a try except block and will cause a jump from the try block to the except block; And I'm trying to avoid the try except blocks and just throw an exception.

A solution I think could work is this,

    try:
        if(condition):
           raise Exception("text here")
    except:
        ...

But I would like to know if there is an approach more closely related to the Java approach so that I can maintain as much of the flow as possible (have them look similar).

`raise` is like `throw`; `try` is like `try`. You don’t *have* to add a `try` block to use exceptions in Python. — Ry-, Jul 14 '14 at 19:57
Java exception do exactly the same: they either jump to the nearest suitable catch block or produce the stack trace :) Try to say `except: pass` and see what happens. — Ashalynd, Jul 14 '14 at 19:58
Note that you may derive your own Exceptions: https://docs.python.org/2/tutorial/errors.html#user-defined-exceptions — Julien Palard, Jul 14 '14 at 20:11
refer to http://stackoverflow.com/a/24065533/541136 for additional informatoin on exceptoins — Dwight J. Browne, Jan 09 '15 at 15:53

score 5 · Answer 1 · answered Jul 14 '14 at 19:57

Exception handling is perhaps the one non-trivial aspect aspect of Python that differs the least from Java, both in syntax and semantics. It's really just raise Exception("text here"). No, it doesn't have to be lexically within a try block. As in Java, it propagates up the call stack until it finally encounters a try block (with a matching except clause), or if there is no such block, it terminates the program and prints an error message.

score 2 · Accepted Answer · edited Jul 14 '14 at 20:46

2

Forget the try, your code would be exactly the equivalent without it

As others pointed it out: this throws (and doesn't catch or handle the exception):

condition = "Foo"
if(condition is "Foo"):
      raise Exception("FooException")

However if you want to handle it as you would if you had thrown in the java method then:

As explained in this documentation you will be throwing them by a method and then all you'll have to do is try the method and not every single condition in the method.

#!/usr/local/bin/python2.7

def foo(num):
    if (num == 2):
       raise Exception("2Exception")
    if (num == 3):
       raise Exception("Numception")

def handleError(e):
    print(e)

def main():
  try:
    foo(3)
    print("no errors")
  except Exception, e:
    handleError(e)


if __name__ == "__main__":
    main()

edited Jul 14 '14 at 20:46

Kyle Bridenstine

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answered Jul 14 '14 at 20:35

Mansueli

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This is very clear and concise which is why I accepted it as the answer. One quick question that's not a big deal, when I use your code my print out has a strange comma after the text Exception('Something Happened',) Do you know why? (Again, this answer was exactly what I needed!) – Kyle Bridenstine Jul 14 '14 at 20:52
1

@2g7hgfdr8324, it must be something in your environment if you test it [here](http://www.compileonline.com/execute_python_online.php) you won't see the comma that you are seeing for some reason in your machine. – Mansueli Jul 14 '14 at 21:32

score 1 · Answer 3 · answered Jul 14 '14 at 19:55

1

if (condition):    
    raise Exception("test")

will accomplish what you want. Give it a try.

answered Jul 14 '14 at 19:55

Paul Becotte

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I tried this but the text doesn't display. Any ideas? – Kyle Bridenstine Jul 14 '14 at 19:56
It did in my case- look at the end of the stack trace. If you want no stack trace, you'll have to catch it or define an exception that prints the error message without the trace. – Paul Becotte Jul 14 '14 at 19:58

Mike Driscoll · Answer 4 · 2014-07-14T20:04:21.127

1

If you don't want to catch the exception, then this is probably what you want:

if condition:
    raise Exception("something bad happened!")

If you DO want to catch the exception, then using Python's try/except is the way to go.

edited Jul 14 '14 at 20:04

answered Jul 14 '14 at 19:56

Mike Driscoll

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score 1 · Answer 5 · answered Jul 14 '14 at 19:59

1

raise StandardError("message")

is perfectly valid code anywhere. In fact, raising an exception within a try/except block is usually only done to show case error handling. It doesn't actually make sense to raise and handle an exception within the same function.

answered Jul 14 '14 at 19:59

Midnighter

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`StandardError` doesn't exist in Python 3, is pretty obscure in Python 2, and is intended as superclass for most exceptions (to allow catching "errors" without catching "control flow exceptions") rather than being instantiated. `Exception` is a closer equivalent to Java. – Jul 14 '14 at 20:01

Does Python have a Java equivalent of throw new Exception("text here")

5 Answers5