I am having trouble getting this to work correctly (obviously) – I am ALMOST there, and I have a good idea of WHY it is not working – just not sure how to make it work.
This is suppose to attempt to read a file into memory, if it fails it goes to the “except” clause of the block of code (that part is ‘duh’). The error file prints: “<main.DebugOutput instance at 0x04021EB8>”. What I want it to do is print the actual Error. Like a FileIOError or TraceBackError or whatever it is to that error file. This is just the beginning stages and I plan to add things like date stamps and to have it append, not write/create – I just need the actual error printed to the file. Advice?
import os, sys
try:
myPidFile = "Zeznadata.txt"
myOpenPID_File = open(myPidFile, "r") #Attempts to open the file
print "Sucessfully opened the file: \"" + myPidFile + "\"."
except:
print "This file, \"" + myPidFile + "\", does not exist. Please check the file name and try again. "
myFileErr = open("PIDErrorlog.txt", "w")
myStdError = str(sys.stderr)
myFileErr.write(myStdError)
myFileErr.close()
print "\nThis error was logged in the file (and stored in the directory): "
First, you should use a logging library. It will help you deal with different logging levels (info/warn/error), timestamps and more.
http://docs.python.org/library/logging.html
Second, you need to catch the error, and then you can log details about it. This example comes from the Python documentation.
See how it catches an IOError and assigns the error number and error message to variables? You can have an
exceptblock for each type of error you want to deal with. In the last (generic error) block, it usessys.exc_info()[0]to get the error details.http://docs.python.org/tutorial/errors.html