I am writing code where i want to redirect output two different files in python
one for log purpose and another for fabric code creation.
Below is the code which redirect output to two different files:
import sys
print('###################################Creating storage commands Variables##########################')
storage_file = open("Storage-output.txt", 'w')
sys.stdout = storage_file
print ('network port show')
print ('Storage Commands are completed')
storage_file.close()
sys.stdout = sys.__stdout__
# write fabric code
fabric_file = open("Fabric_code.py", 'w')
print"from fabric.api import run"
print"def host_type():"
print" run('rows 0', shell=False)"
print" run('networ port show', shell=false)"
fabric_file.close()
sys.stdout = sys.__stdout__
storage_file = open("Storage-output.txt", 'a')
sys.stdout = storage_file
print('###############Genarating aggregate command#############')
print ('storage aggregate create -aggregate aggr_fab -diskcount 6 -raidtype raid_dp", shell=False')
storage_file.close()
sys.stdout = sys.__stdout__
# write fabric code
fabric_file = open("Fabric_code.py", 'a')
print" run('storage aggregate create -aggregate aggr_fab -diskcount 6 - raidtype raid_dp', shell=False) "
fabric_file.close()
sys.stdout = sys.__stdout__
In Above code creating one for log file Storage_file to store this file for records and Fabric_code file to genarate fabric code.
My code generates 1000's of commands I do not want to open and close multiple times for two different files again and again in the python code.
Instead of this is there any solution where i can redirect print output to two direct files without opening and closing
You should refactor your code by opening the files once in the beginning, then writing your files and closing the files in the end. From the above example we can do the following:
import sys
# Open files
storage_file = open("Storage-output.txt", 'w')
fabric_file = open("Fabric_code.py", 'w')
# ===================
# Write Block
# ===================
print('###################################Creating storage commands Variables##########################')
sys.stdout = storage_file
print ('network port show')
print ('Storage Commands are completed')
sys.stdout = sys.__stdout__
# write fabric code
print"from fabric.api import run"
print"def host_type():"
print" run('rows 0', shell=False)"
print" run('networ port show', shell=false)"
sys.stdout = sys.__stdout__
sys.stdout = storage_file
print('###############Genarating aggregate command#############')
print ('storage aggregate create -aggregate aggr_fab -diskcount 6 -raidtype raid_dp", shell=False')
sys.stdout = sys.__stdout__
# write fabric code
print" run('storage aggregate create -aggregate aggr_fab -diskcount 6 - raidtype raid_dp', shell=False) "
sys.stdout = sys.__stdout__
# closing files
storage_file.close()
fabric_file.close()
Related
I am having trouble figuring out what the issue with my code snippet for writing print messages in my console to multiple log-files is doing.
The code snippet I have posted below is supposed to create a new directory test, then write 11 log-files, 1 global log file, and 10 loop log files to this directory. However, the 1st 2 print messages to my global log file is missing when I run this and I cannot figure out what the issue is?
import sys
import os
# Create a test folder to store these global and loop log files.
path = os.getcwd()
test_dir_name = 'test'
test_dir_path = os.path.join(path, test_dir_name)
os.mkdir(test_dir_path)
# Keep a reference to the original stdout.
orig_stdout = sys.stdout
# Define global logfile path.
global_log_name = "global-log.txt"
global_log_path = os.path.join(test_dir_path, global_log_name)
# Problematic code-snippet
sys.stdout = open(global_log_path, 'w')
print("This is a global log file.") # Why is my code omitting this line?
print("The loop is now creating 10 individual log files.") # And this one?
sys.stdout.close()
for i in range(10):
sys.stdout = open(global_log_path, 'w')
print("Creating loop log file {}...".format(i))
sys.stdout.close()
loop_log_name = "local-log-{}.txt".format(i)
loop_log_path = os.path.join(test_dir_path, loop_log_name)
sys.stdout = open(loop_log_path, 'w')
print("This is loop log file {}".format(i))
print("Closing this loop log file...")
sys.stdout.close()
sys.stdout = open(global_log_path, 'w')
print("Loops have concluded.") # But then it includes this line.
print("Now closing global log file.") # And this line in the global log file.
sys.stdout.close()
sys.stdout = orig_stdout
print("Back to original console.")
Some assistance would be greatly appreciated.
The principal issue with this code snippet is the inappropriate use of open(global_log_path, 'w') to append further print messages to global-log.txt. After you have initially executed:
sys.stdout = open(global_log_path, 'w')
print("This is a global log file.") # Why is my code omitting this line?
print("The loop is now creating 10 individual log files.") # And this one?
Subsequent redirections of stdout to global-log.txt instead require passing the argument a, standing for append to open() like so:
sys.stdout = open(global_log_path, 'a')
print("Creating loop log file {}...".format(i))
This prevents previously redirected text from being overwritten, which was happening with your code snippet.
I want to get the console output in a .txt file.
This is what i have:
import sys
print('some text')
a='moretext.1'.split('.')
sys.stdout = open('output.txt', 'w')
print(a)
sys.stdout.close()
here it works but in my program don't.
Does someone know what it could be?
It says that that its on line 2 or something
And I already searched on Stackoverflow and in the internet but i cant find anything
Do not mess with sys.stdout, instead open the file and print to it like so:
print('some text')
a='moretext.1'.split('.')
with open('output.txt', 'w') as out:
print(a, file=out)
Multiple ways to do so
1--
python3 myprogram.py > output.txt
2--
import sys
print('some text')
a='moretext.1'.split('.')
output = open('output.txt', 'w')
print(a, file=output)
output.close()
3--
import sys
print('some text')
a='moretext.1'.split('.')
stdout = sys.stdout
sys.stdout = open('output.txt', 'w')
print(a)
sys.stdout.close()
sys.stdout = sys.__stdout__
4--
As #Timur Shtatland suggested you can use a with statement
Suppose I am writing stdout to a file, like this:
sys.stdout = open("file.txt", "w")
# print stuff here
Doing this doesn't work:
sys.stdout.close()
How can I close a file after writing stdout to it?
I took your question to mean: "How can I redirect sys.stdout to a file?"
import sys
# we need this to restore our sys.stdout later on
org_stdout = sys.stdout
# we open a file
f = open("test.txt", "w")
# we redirect standard out to the file
sys.stdout = f
# now everything that would normally go to stdout
# now will be written to "test.txt"
print "Hello world!\n"
# we have no output because our print statement is redirected to "test.txt"!
# now we redirect the original stdout to sys.stdout
# to make our program behave normal again
sys.stdout = org_stdout
# we close the file
f.close()
print "Now this prints to the screen again!"
# output "Now this prints to the screen again!"
# we check our file
with open("test.txt") as f:
print f.read()
# output: Hello World!
Is this an answer to your question?
You can also do this if you want to redirect all print() to a file, which is a fast way and also usefull by my opinion but it could have other effects. If I'm wrong please correct me.
import sys
stdoutold = sys.stdout
sys.stdout = fd = open('/path/to/file.txt','w')
# From here every print will be redirected to the file
sys.stdout = stdoutold
fd.close()
# From here every print will be redirected to console
You can do this:
import sys
class writer(object):
""" Writes to a file """
def __init__(self, file_name):
self.output_file = file_name
def write(self, something):
with open(self.output_file, "a") as f:
f.write(something)
if __name__ == "__main__":
stdout_to_file = writer("out.txt")
sys.stdout = stdout_to_file
print "noel rocks"
The file is only open when you write to it like this.
Looking for some help logging/saving the prints to two file locations as seen below, does anyone know a way to do this?
### Create output file/open it for editing
output_file = open('FILE.txt','w')
output_file1 = open('FILE_APPENDING.txt','a')
## Create a backup of current setting
old_stdout = sys.stdout
sys.stdout = output_file
sys.stdout = output_file1
print "stuff here"
## loop here printing stuff
## Revert python to show prints as normal
sys.stdout=old_stdout
## Close the file we are writing too
output_file.close()
output_file1.close()
Thanks in advance
- Hyflex
You can reassign sys.stdout with some class that writes to multiple files:
class MultiWrite(object):
def __init__(self, *files):
self.files = files
def write(self, text):
for file in self.files:
file.write(text)
def close(self):
for file in self.files:
file.close()
import sys
# no need to save stdout. There's already a copy in sys.__stdout__.
sys.stdout = MultiWrite(open('file-1', 'w'), open('file-2', 'w'))
print("Hello, World!")
sys.stdout.close()
sys.stdout = sys.__stdout__ #reassign old stdout.
Anyway, I agree with Ashwini. It seems that you are searching a hack to obtain something, when you should really use a different approach.
Simply use file.write:
with open('FILE.txt','w') as output_file:
#do something here
output_file.write(somedata) # add '\n' for a new line
with open('FILE_APPENDING.txt','a') as output_file1:
#do something here
output_file1.write(somedata)
help on file.write:
>>> print file.write.__doc__
write(str) -> None. Write string str to file.
Note that due to buffering, flush() or close() may be needed before
the file on disk reflects the data written.
from sys import stdout
stdout = open('file', 'w')
print 'test'
stdout.close()
does create the file, but it contains nothing.
I had to use
import sys
sys.stdout = open('file', 'w')
print 'test'
sys.stdout.close()
But wouldn't the from ... import... automatically make the name available? Why do I still have to use sys.stdout instead of stdout?
The problem is this: print is equivalent to sys.stdout.write().
So when you do from sys import stdout, the variable stdout won't be used by print.
But when you do
import sys
print 'test'
it actually writes to sys.stdout which is pointing to the file you opened.
Analysis
from sys import stdout
stdout = open('file', 'w')
print 'test' # calls sys.stdout.write('test'), which print to the terminal
stdout.close()
import sys
sys.stdout = open('file', 'w')
print 'test' # calls sys.stdout.write('test'), which print to the file
sys.stdout.close()
Conclusion
This works...
from sys import stdout
stdout = open('file', 'w')
stdout.write('test')
stdout.close()