Develop Reference

How to open and print the contents of a file line by line using subprocess?

I am trying to write a python script which SSHes into a specific address and dumps a text file. I am currently having some issues. Right now, I am doing this:
temp = "cat file.txt"
need = subprocess.Popen("ssh {host} {cmd}".format(host='155.0.1.1', cmd=temp),shell=True,stdout=subprocess.PIPE, stderr=subprocess.PIPE).communicate()
print(need)
This is the naive approach where I am basically opening the file, saving its output to a variable and printing it. However, this really messes up the format when I print "need". Is there any way to simply use subprocess and read the file line by line? I have to be SSHed into the address in order to dump the file otherwise the file will not be detected, that is why I am not simply doing
f = open(temp, "r")
file_contents = f.read()
print (file_contents)
f.close()
Any help would be appreciated :)

You don't need to use the subprocess module to print the entire file line by line. You can use pure python.
f = open(temp, "r")
file_contents = f.read()
f.close()
# turn the file contents into a list
file_lines = file_contents.split("\n")
# print all the items in the list
for file_line in file_lines:
print(file_line)

Directing print output to a .txt file

Is there a way to save all of the print output to a txt file in python? Lets say I have the these two lines in my code and I want to save the print output to a file named output.txt.
print ("Hello stackoverflow!")
print ("I have a question.")
I want the output.txt file to to contain
Hello stackoverflow!
I have a question.

Give print a file keyword argument, where the value of the argument is a file stream. The best practice is to open the file with the open function using a with block, which will ensure that the file gets closed for you at the end of the block:
with open("output.txt", "a") as f:
print("Hello stackoverflow!", file=f)
print("I have a question.", file=f)
From the Python documentation about print:
The file argument must be an object with a write(string) method; if it is not present or None, sys.stdout will be used.
And the documentation for open:
Open file and return a corresponding file object. If the file cannot be opened, an OSError is raised.
The "a" as the second argument of open means "append" - in other words, the existing contents of the file won't be overwritten. If you want the file to be overwritten instead at the beginning of the with block, use "w".
The with block is useful because, otherwise, you'd need to remember to close the file yourself like this:
f = open("output.txt", "a")
print("Hello stackoverflow!", file=f)
print("I have a question.", file=f)
f.close()

You can redirect stdout into a file "output.txt":
import sys
sys.stdout = open('output.txt','wt')
print ("Hello stackoverflow!")
print ("I have a question.")

Another method without having to update your Python code at all, would be to redirect via the console.
Basically, have your Python script print() as usual, then call the script from the command line and use command line redirection. Like this:
$ python ./myscript.py > output.txt
Your output.txt file will now contain all output from your Python script.
Edit:
To address the comment; for Windows, change the forward-slash to a backslash.
(i.e. .\myscript.py)

Use the logging module
def init_logging():
rootLogger = logging.getLogger('my_logger')
LOG_DIR = os.getcwd() + '/' + 'logs'
if not os.path.exists(LOG_DIR):
os.makedirs(LOG_DIR)
fileHandler = logging.FileHandler("{0}/{1}.log".format(LOG_DIR, "g2"))
rootLogger.addHandler(fileHandler)
rootLogger.setLevel(logging.DEBUG)
consoleHandler = logging.StreamHandler()
rootLogger.addHandler(consoleHandler)
return rootLogger
Get the logger:
logger = init_logging()
And start logging/output(ing):
logger.debug('Hi! :)')

Another Variation can be... Be sure to close the file afterwards
import sys
file = open('output.txt', 'a')
sys.stdout = file
print("Hello stackoverflow!")
print("I have a question.")
file.close()

Suppose my input file is "input.txt" and output file is "output.txt".
Let's consider the input file has details to read:
5
1 2 3 4 5
Code:
import sys
sys.stdin = open("input", "r")
sys.stdout = open("output", "w")
print("Reading from input File : ")
n = int(input())
print("Value of n is :", n)
arr = list(map(int, input().split()))
print(arr)
So this will read from input file and output will be displayed in output file.
For more details please see https://www.geeksforgeeks.org/inputoutput-external-file-cc-java-python-competitive-programming/

Be sure to import sys module. print whatever you want to write and want to save. In the sys module, we have stdout, which takes the output and stores it. Then close the sys.stdout . This will save the output.
import sys
print("Hello stackoverflow!" \
"I have a question.")
sys.stdout = open("/home/scilab/Desktop/test.txt", "a")
sys.stdout.close()

One can directly append the returned output of a function to a file.
print(output statement, file=open("filename", "a"))

How to execute a python script and write output to txt file?

I'm executing a .py file, which spits out a give string. This command works fine
execfile ('file.py')
But I want the output (in addition to it being shown in the shell) written into a text file.
I tried this, but it's not working :(
execfile ('file.py') > ('output.txt')
All I get is this:
tugsjs6555
False
I guess "False" is referring to the output file not being successfully written :(
Thanks for your help

what your doing is checking the output of execfile('file.py') against the string 'output.txt'
you can do what you want to do with subprocess
#!/usr/bin/env python
import subprocess
with open("output.txt", "w+") as output:
subprocess.call(["python", "./script.py"], stdout=output);

This'll also work, due to directing standard out to the file output.txt before executing "file.py":
import sys
orig = sys.stdout
with open("output.txt", "wb") as f:
sys.stdout = f
try:
execfile("file.py", {})
finally:
sys.stdout = orig
Alternatively, execute the script in a subprocess:
import subprocess
with open("output.txt", "wb") as f:
subprocess.check_call(["python", "file.py"], stdout=f)
If you want to write to a directory, assuming you wish to hardcode the directory path:
import sys
import os.path
orig = sys.stdout
with open(os.path.join("dir", "output.txt"), "wb") as f:
sys.stdout = f
try:
execfile("file.py", {})
finally:
sys.stdout = orig

If you are running the file on Windows command prompt:
python filename.py >> textfile.txt
The output would be redirected to the textfile.txt in the same folder where the filename.py file is stored.
The above is only if you have the results showing on cmd and you want to see the entire result without it being truncated.

The simplest way to run a script and get the output to a text file is by typing the below in the terminal:
PCname:~/Path/WorkFolderName$ python scriptname.py>output.txt
*Make sure you have created output.txt in the work folder before executing the command.

Use this instead:
text_file = open('output.txt', 'w')
text_file.write('my string i want to put in file')
text_file.close()
Put it into your main file and go ahead and run it. Replace the string in the 2nd line with your string or a variable containing the string you want to output. If you have further questions post below.

file_open = open("test1.txt", "r")
file_output = open("output.txt", "w")
for line in file_open:
print ("%s"%(line), file=file_output)
file_open.close()
file_output.close()

using some hints from Remolten in the above posts and some other links I have written the following:
from os import listdir
from os.path import isfile, join
folderpath = "/Users/nupadhy/Downloads"
filenames = [A for A in listdir(folderpath) if isfile(join(folderpath,A))]
newlistfiles = ("\n".join(filenames))
OuttxtFile = open('listallfiles.txt', 'w')
OuttxtFile.write(newlistfiles)
OuttxtFile.close()
The code above is to list all files in my download folder. It saves the output to the output to listallfiles.txt. If the file is not there it will create and replace it with a new every time to run this code. Only thing you need to be mindful of is that it will create the output file in the folder where your py script is saved. See how you go, hope it helps.

You could also do this by going to the path of the folder you have the python script saved at with cmd, then do the name.py > filename.txt
It worked for me on windows 10

Python non-specific write exception

Currently teaching myself Python, and learning file I/O by writing a script to both read from and add text to an existing file. The script runs up until I call the write() method, at which point it throws out a non-specific exception - this is the traceback:
File "test.py", line 13, in <module>
f.write(txt)
IOError: [Errno 0] Error
My code:
from sys import argv
script, filename = argv
f = open(filename, 'a+')
print("The contents of %s are:") % filename
print f.read()
txt = raw_input("What would you like to add? ")
f.write(txt)
print("The new contents are:")
print f.read()
f.close()
My environment is Python 2.7.3 in Win7, PowerShell, and Notepad++.
What is causing this? How would I fix it? In my understanding, the a+ access mode should allow me to both read and append to the file. Changing the access mode to r+ yields the same exception.
Clarifications:
I have an existing text file (a.txt) with a single word in it that I pass as an argument to the script, like so:
python test.py a.txt
I am under an admin account in Windows.
Results:
At the minimum, adding two seek() commands fixes the issue - detailed in the answer post.

A problem when one tries to add a text of little size: it remains in the buffer, that keeps the text before the real writing is done after receiving more data.
So, to be sure to write really, do as it is described in the doc concerning os.fsync() and flush()
By the way, it is better to use the with statement.
And it's still more better to use binary mode. In your case, there shouldn't be a problem because you just add text after the reading and just use seek(o,o) . But when one wants to move correctly the file's pointer into the bytes of the file, it is absolutely necessary to use binary mode [ the 'b' in open(filename, 'rb+') ]
I personnaly never use 'a+', I've never understood what are its effects.
from sys import argv
from os import fsync
script, filename = argv
with open(filename, 'rb+') as f:
print("The contents of %s are:") % filename
print f.read()
f.seek(0,2)
txt = raw_input("What would you like to add? ")
f.write(txt)
f.flush()
fsync(f.fileno())
f.seek(0,0)
print("The new contents are:")
print f.read()

For some reason print f.read() doesn't work for me on OS X when you have opened the file in a+ mode.
On Max OS X, changing the open mode to r+ and then adding a f.seek(0) line before the second read makes it work. Sadly, this doesn't help windows.
This is the working code on Mac OS:
from sys import argv
script, filename = argv
f = open(filename, 'r+')
print("The contents of %s are:") % filename
print f.read()
txt = raw_input("What would you like to add? ")
f.write(txt)
print("The new contents are:")
f.seek(0)
print f.read()
f.close()
This is the only way I could get it to work on windows 7:
from sys import argv
script, filename = argv
f = open(filename, 'r')
print("The contents of %s are:") % filename
print f.read()
f.close()
txt = raw_input("What would you like to add? ")
f = open(filename, 'a')
f.write(txt)
f.close()
f = open(filename, 'r')
print("The new contents are:")
print f.read()
f.close()
Which seems super hacky. This should also work on Mac OS X too.

Python file.read() function output?

I am trying to output a file's contents to terminal using the File.read() function in Python, but keep receiving the following output which doesn't match my ".txt" file contents.
Python Code
from sys import argv
script, input_file = argv
def print_all(f):
print f.read
current_file = open(input_file)
print "Print File contents:\n"
print_all(current_file)
current_file.close()
Output:
Print File contents:
<built-in method read of file object at 0x1004bd470>

If you want to call a function, you will need () after the name of the function (along with any required arguments)
Therefore, in your function print_all replace:
print f.read # this prints out the object reference
with:
print f.read() # this calls the function

You just need to change
print f.read
to say
print f.read()

You should do a read()
current_file = open(input_file)
print "Print File contents:\n"
print_all(current_file.read())

You need to actually call the function in your print_all definition:
def print_all(f):
print f.read()

You haven't called the read method, you've just got it from the file class. In order to call it you have to put the braces. f.read()