Develop Reference

Is there a simple way to comment out or diable a python try statement without re-indenting

Problem:
Sometimes it is nice to be able to remove or apply a try statement temporarily. Is there a convenient way to disable the try statement without re-indenting?
For example, if there was a python block statement equivalent called "goForIt:" one could edit the word "try:" to "goForIt:" and it would just execute the block as though it were not wrapped in a "try" and ignore the "except" line too.
The problem I'm trying to solve is that while I want the try statement in production I want to be able to remove it temporarily while debugging to see the error traceback rather than have it trap and process the exception.
Currently I work around this by commenting out the "try" then re-indent the code in the block. Then comment out the entire "except" block. This seems clumsy.

Instead of removing the try, you could make the except re-raise the exception:
try:
raise ValueError('whoops')
except ValueError as e:
raise # <-- just put this here
print('caught')
This will raise the error, just as if it were not caught:
ValueError Traceback (most recent call last)
<ipython-input-146-a6be6779c161> in <module>
1 try:
----> 2 raise ValueError('whoops')
3 except ValueError as e:
4 raise
5 print('caught')
ValueError: whoops

I do not believe there is a way to fix this.
Is it adequate to catch the exception, print it out and, if wanted, end the program?
try:
# code
except Exception as e:
print(e)
# end program if wanted

If you want a none code solution you need to use an IDE or a good text editor.
I am using Visual Studio code where I can indent with a keyboard shortcut (Ctrl + ` in my case)

(This is not my answer, but the answer from Alani's comment. If credits, find the original comment under the question.)
The solution is good because it allows global replacement:
try: --> if 1: # try:
except --> if 0: # except
The 2nd one is a little unsure. Exact and full word match should be used, or replace twice (for except<space> and except:). Or you can fix error fast by hand if there is any.
The replacement back is sure.
I need now such solution to debug error in strawberry which is only printed. So I think I need deactivate all try/except structures in 3 libraries (strawberry + 2 libs for django).

Pycharm variable assignment not recognized inside try block

try:
driver = launch_browser()
except:
print "Browser launch failed"
driver.get("http://www.example.com/")
The last line above is flagged by PyCharm with the following issue:
Local variable "driver" might be referenced before assignment
However, something like this makes the error go away:
driver = None
try:
driver = launch_browser()
except:
print "Browser launch failed"
driver.get("http://www.example.com/")
Is there a way to setup PyCharm so that it will see the assignements inside try blocks?
Secondarily, can PyCharm figure out the type based on the return value of the function (in this case launch_browser()) if it has docstrings?
BTW, code works just fine in both cases. It's just a matter of getting PyCharm to understand the assignment inside the try block without having to resort to a band-aid.
EDIT 1:
A return in the except: block fixes the problem as far as PyCharm is concerned. I was working on something else and inadvertently commented it out. Proof that coding for 16 hours straight is a really bad idea...

If launch_browser() fails, your code will error at the driver.get("http://www.example.com/") line. PyCharm is letting you know this.
The only way to avoid this is by not executing anything below the except, e.g. throwing an exception inside it, or putting everything that relies on driver inside an else block, which will only run if no exception is caught. E.g.
try:
driver = launch_browser()
except:
print "Browser launch failed"
else:
driver.get("http://www.example.com/")

Possible motif behind using pass after raise in Python exception handling

try:
[code]
except Exception:
[something]
raise
pass
Above code snippet looks a bit weird, doesn't it? Am I missing something? Why is there a raise and a pass following it?

Its a NOP. It does nothing. Nothing at all. No clue why they left it there, because it does nothing.

seems like initially it was just
except Exception:
pass
somebody just forgot to remove it after adding exception handling

Any way to save a traceback object in Python

I was looking to possibly try and save a traceback object and somehow pickle it to a file that I can access. An example of a use case for this is if I am submitting some python code to a farm computer to run and it fails, it would be nice to be able to open a session and access that traceback to debug the problem rather than just seeing a log of the traceback. I do not know if there is any sort of way to do this but thought it would be worth asking why it couldn't if so.

okay so you can use traceback.print_exception(type, value, traceback[, limit[, file]]) and save it in a text or json or you can refer to docs
if you find it helpful please mark it correct or upvote thanx..:)

Depending on how you've written your code, the try statement is probably your best answer. Since any error is just a class that inherits Python's builtin Exception, you can raise custom errors everywhere you need more information about a thrown error. You just need to rename your errors or pass in an appropriate string as the first argument. If you then try your code and use the except statement except CustomError as e, you can pull all the information you want out of e in the except statement as a regular instance. Example:
Your code would be:
def script():
try: codeblock
except Exception as e: raise Error1('You hit %s error in the first block'% e)
try: codeblock 2
except Exception as e: raise Error2('You hit %s error in the second block' % e)
try: script()
except Exception as e:
with open('path\to\file.txt','w') as outFile:
outFile.write(e)
The last part is really nothing more than creating your own log file, but you have to write it down somewhere, right?
As for using the traceback module mentioned above, you can get error information out of that. Any of the commands here can get you a list of tracebacks:
http://docs.python.org/2/library/traceback.html
On the otherhand, if you're trying to avoid looking at log files, the traceback module is only going to give you the same thing a log file would, in a different format. Adding your own error statements in your code gives you more information than a cryptic ValueError about what actually happened. If you print the traceback to your special error, it might give you still more information on your issue.

Failed to write to file but generates no Error

I'm trying to write to a file but it's not working. I've gone through step-by-step with the debugger (it goes to the write command but when I open the file it's empty).
My question is either: "How do I see what the error is so I can debug?", "What can go wrong when trying to write to a file that would cause it to behave this way?".
sqlScript = open('script-file.sql', 'a')
try:
sqlScript.write(sqlLine)
except IOError as (errno, strerror):
print "I/O error({0}): {1}".format(errno, strerror)
This should be simple but I can't seem to find an answer. Also, I apologize in advance as English is a second language.
Edit: I put a print statement just before and the string is not empty.
Edit2: I'm using python 2.6 if that factors in somehow.
Edit 3: I've found a "solution" to my error. I decided to try and run my script using IDLE instead of PyCharm and it works like a charm (pun intended). I have no clue why, but there it is. Any reason why that would happen?!

Building on Chris Morris' answer, maybe do something like this:
try:
sqlScript.write(sqlLine)
except Exception as e:
print type(e)
print str(e)
This will catch any Exception thrown (provided it's a subclass of Exception, of course) and tell you the type and the error message.
Also, it's possible to define multiple except: cases for different possible exceptions, so maybe try doing that for each exception that might be potentially thrown/raised.

The following code allows you to see what exception it is that is being thrown, and see a trace of where it originated from.
try:
sqlScript.write(sqlLine)
except:
print "Unexpected error:", sys.exc_info()[0]
raise
See http://docs.python.org/tutorial/errors.html for more info.

You have to put your cursor at the beginning of the file. You can do that with seek method:
myScript.seek(0)
See this answer: https://stackoverflow.com/a/2949648/2119117

If no exception is being tossed, I'd suspect the string variable 'sqlLine' is empty.
Did you print it before the write statement?

Are the keywords in the script in lowercase? The same thing happened to me in another db and I solved it by changing words to UPPERCASE.

Your current working directory isn't what you expect it to be and it's successfully writing to some script-file.sql in another directory. Try printing os.getcwd() and make sure it's what you expect, and look in that directory.

It happened on my linux environment but work on windows try
sqlScript = open('script-file.sql', 'a', buffering=False)
or
sqlScript = open('script-file.sql', 'a', 0)