OpenShift has these default dir's:
# $_ENV['OPENSHIFT_INTERNAL_IP'] - IP Address assigned to the application
# $_ENV['OPENSHIFT_GEAR_NAME'] - Application name
# $_ENV['OPENSHIFT_GEAR_DIR'] - Application dir
# $_ENV['OPENSHIFT_DATA_DIR'] - For persistent storage (between pushes)
# $_ENV['OPENSHIFT_TMP_DIR'] - Temp storage (unmodified files deleted after 10 days)
How do reference them in a python script?
Example script "created a log file in log directory and log in data directory?
from time import strftime
now= strftime("%Y-%m-%d %H:%M:%S")
fn = "${OPENSHIFT_LOG_DIR}/test.log"
fn2 = "${OPENSHIFT_DATA_DIR}/test.log"
#fn = "test.txt"
input = "appended text " + now + " \n"
with open(fn, "ab") as f:
f.write(input)
with open(fn2, "ab") as f:
f.write(input)
Can these script be used with cron?
EDIT the BASH File:
#! /bin/bash
#date >> ${OPENSHIFT_LOG_DIR}/new.log
source $OPENSHIFT_HOMEDIR/python-2.6/virtenv/bin/activate
python file.py
date >> ${OPENSHIFT_DATA_DIR}/new2data.log
import os
os.getenv("OPENSHIFT_INTERNAL_IP")
should work.
So with your example, modify to:-
import os
OPENSHIFT_LOG_DIR = os.getenv("OPENSHIFT_LOG_DIR")
fn = os.path.join(OPENSHIFT_LOG_DIR, "test.log")
And, yes, you can call this python script with a cron by referencing your bash script if you want... Like this for example:-
#!/bin/bash
date >> ${OPENSHIFT_LOG_DIR}/status.log
chmod +x status
cd ${OPENSHIFT_REPO_DIR}/wsgi/crawler
nohup python file.py 2>&1 &
Those variables OPENSHIFT_* are provided as environment variables on OpenShift -- so the $_ENV["OPENSHIFT_LOG_DIR"] is an example to get the value inside a php script.
In python, the equivalent would just be os.getenv("OPENSHIFT_LOG_DIR").
Made edits to Calvin's post above and submitted 'em.
Re: the question of where file.py exists -- use os.getenv("OPENSHIFT_REPO_DIR") as the base directory where all your code would be located on the gear where you app is running.
So if your file is located in .openshift/misc/file.py -- then just use:
os.path.join(os.getenv("OPENSHIFT_REPO_DIR"), ".openshift", "misc", "file.py")
to get the full path.
Or in bash, the equivalent would be:
$OPENSHIFT_REPO_DIR/.openshift/misc/file.py
HTH
Related
I've been trying to run a Java program and capture it's STDOUT output to a file from the Python script. The idea is to run test files through my program and check if it matches the answers.
Per this and this SO questions, using subprocess.call is the way to go. In the code below, I am doing subprocess.call(command, stdout=f) where f is the file I opened.
The resulted file is empty and I can't quite understand why.
import glob
test_path = '/path/to/my/testfiles/'
class_path = '/path/to/classfiles/'
jar_path = '/path/to/external_jar/'
test_pattern = 'test_case*'
temp_file = 'res'
tests = glob.glob(test_path + test_pattern) # find all test files
for i, tc in enumerate(tests):
with open(test_path+temp_file, 'w') as f:
# cd into directory where the class files are and run the program
command = 'cd {p} ; java -cp {cp} package.MyProgram {tc_p}'
.format(p=class_path,
cp=jar_path,
tc_p=test_path + tc)
# execute the command and direct all STDOUT to file
subprocess.call(command.split(), stdout=f, stderr=subprocess.STDOUT)
# diff is just a lambda func that uses os.system('diff')
exec_code = diff(answers[i], test_path + temp_file)
if exec_code == BAD:
scream(':(')
I checked the docs for subprocess and they recommended using subprocess.run (added in Python 3.5). The run method returns the instance of CompletedProcess, which has a stdout field. I inspected it and the stdout was an empty string. This explained why the file f I tried to create was empty.
Even though the exit code was 0 (success) from the subprocess.call, it didn't mean that my Java program actually got executed. I ended up fixing this bug by breaking down command into two parts.
If you notice, I initially tried to cd into correct directory and then execute the Java file -- all in one command. I ended up removing cd from command and did the os.chdir(class_path) instead. The command now contained only the string to run the Java program. This did the trick.
So, the code looked like this:
good_code = 0
# Assume the same variables defined as in the original question
os.chdir(class_path) # get into the class files directory first
for i, tc in enumerate(tests):
with open(test_path+temp_file, 'w') as f:
# run the program
command = 'java -cp {cp} package.MyProgram {tc_p}'
.format(cp=jar_path,
tc_p=test_path + tc)
# runs the command and redirects it into the file f
# stores the instance of CompletedProcess
out = subprocess.run(command.split(), stdout=f)
# you can access useful info now
assert out.returncode == good_code
I'm trying to run a python script in crontab.
5 0 * * * python /home/hadoop/import_openapp.py >> /home/hadoop/openapp.out 2>&1
The python script is something like below:
import sys
import datetime
from fabric.api import local
ystd = datetime.date.today() - datetime.timedelta(days=1)
c = ystd.strftime('%Y-%m-%d')
print(c)
print('Start to format file ...')
......
print('Start to upload on HDFS ...')
local("/home/hadoop/hadoop/bin/hadoop fs -put " + finalfile + " /user/hadoop/yunying/openapp")
print('Start to upload on MaxCompute ...')
......
When the crontab is called, the log file is like:
2016-07-01
Start to format file ...
Start to upload on HDFS ...
[localhost] local: /home/hadoop/hadoop/bin/hadoop fs -put /data/uxin/nsq_client_active_collect/hadoop/openappfinal.log /user/hadoop/yunying/openapp
And then, the process is over. I cannot find it in ps -ef|grep python
Why it comes to an end while meeting local()?
It is likely that the PYTHONPATH it not set up correctly for whatever the user that CRON is using to run the script. Print out the path to a debug file to check:
with open('/path/to/debug_file.txt', 'wt') as f:
f.write(str(sys.path))
Try adding the line:
sys.path.append("/path/to/fabric.api")
before importing local
You can also dynamically get the location of the file that is being run using
import os
os.path.realpath(__file__)
this will allow you to use relative paths if you need them
I have a shell script from which I am creating some logfile in a directory. And this shell script I am executing in python as below.
cmd = "sh xyz.sh"
try:
subprocess.call(cmd,shell=True)
except OSError:
print "Failed to run the script.:
I want to attach the timestamp on those created logfile which is getting created by xyz.sh script. For this I want to use below python code.
b=time.strftime("%x")
c=time.strftime("%X")
ts=b+"_"+c
I want to attach the result of 'ts' variable on created log file. For example : if logfile is name as abc.log and output of ts is : 6/19/2016_10:20:16 then my logfile should renamed as 6/19/2016_10:20:16_abc.log
Please help me to achieve this.
If you already know the name of the generated logfile, you can use os.rename to rename it. Something along the lines of :
import time, os
b = time.strftime("%x")
c = time.strftime("%X")
ts = b + "_" + c
os.rename("abc.log", ts + "_abc.log")
If you don't know the name of the logfile, you can find it using os.listdir(os.getcwd()), which will give you a list with all the files of the current directory (i.e. the one from where you are running your script).
I made a small script in sublime that will extract commands from a json file that is on the user's computer and then it will open the terminal and run the settings/command. This works, except that it doesn't really open up the terminal. It only runs the command (and it works, as in my case it will run gcc to compile a simple C file), and pipes to STDOUT without opening up the terminal.
import json
import subprocess
import sublime_plugin
class CompilerCommand(sublime_plugin.TextCommand):
def get_dir(self, fullpath):
path = fullpath.split("\\")
path.pop()
path = "\\".join(path)
return path
def get_settings(self, path):
_settings_path = path + "\\compiler_settings.json"
return json.loads(open(_settings_path).read())
def run(self, edit):
_path = self.get_dir(self.view.file_name())
_settings = self.get_settings(_path)
_driver = _path.split("\\")[0]
_command = _driver + " && cd " + _path + " && " + _settings["compile"] + " && " + _settings["exec"]
proc = subprocess.Popen(_command, shell=True, stdout=subprocess.PIPE, stderr=subprocess.STDOUT)
I'm not sure if using subprocess.Popen is the right way to go about it as I'm new to Python.
So to re-iterate; I want it to open up the terminal, run the command, and have the terminal stay open until the user presses ENTER or something. I'm running Windows 7 and Python 3, if that matters.
subprocess.Popen simply creates a subprocess with the given command. It is in no way related to opening a terminal window or any other windows for that matter.
You'll have to look into your platform specific UI automation solutions in order to achieve what you want. Or see if maybe the Sublime plugins mechanism can already do that.
NOTES:
Also, you should be using os.path.join/os.path.split/os.path.sep etc for your path operations—Sublime also runs on OS X for example, and OS X does not use backslashes. Also, file handles need to be closed, so use:
with open(...) as f:
return json.load(f) # also not that there is no nead to f.read()+json.loads()
# if you can just json.load() on the file handle
Furthermore, strings should usually be built using string interpolation:
_command = "{} && cd {} && {} && {}".format(_driver, _path, _settings["compile"], _settings["exec"])
...and, you should not be prefixing your local variables with _—it doesn't look nice and serves no purpose in Python either; and while we're at it, I might as well use the chance to recommend you to read PEP8: http://www.python.org/dev/peps/pep-0008/.
I've been trying to trouble this for days now, and would appreciate some help --
Basically, I wrote the following Python script
import os, sys
# =__=__=__=__=__=__=__ START MAIN =__=__=__=__=__=__=__
if __name__ == '__main__':
# initialize variables
all_files = []
# directory to download data siphon files to
dDir = '/path/to/download/directory/'
# my S3 bucket
s3bucket = "com.mybucket/"
foldername = "test"
# get a list of available feeds
feeds = <huge JSON object with URLs to feeds>
for item in range(feeds['count']):
# ...check if the directory exists, and if not, create the directory...
if not os.path.exists(folderName):
os.makedirs(folderName)
... ... ...
# Loop through all the splits
for s in dsSplits:
... ... ...
location = requestFeedLocation(name, timestamp)
... ... ...
downloadFeed(location[0], folderName, nameNotGZ)
# THIS IS WHERE I AM HAVING PROBLEMS!!!!!!!!!!!
cmd = 's3cmd sync 'dDir+folderName+'/ s3://'+s3bucket+'/'
os.system(cmd)
Everything in my code works...when I run this straight from the command line, everything runs as expected...however, when I have it executed via cron -- the following DOES NOT execute (everything else does)
# THIS IS WHERE I AM HAVING PROBLEMS!!!!!!!!!!!
cmd = 's3cmd sync 'dDir+folderName+'/ s3://'+s3bucket+'/'
os.system(cmd)
To answer a few questions, I am running the cron as root, s3cmd is configured for the root user, OS is Ubuntu 12.04, python version is 2.7, all of the necessary directories have Read / Write permissions...
What am I missing?
First check variable 'foldername' in command you have used N in variable.
In command syntax you need to add plus (+) sign in prefix like +dDir+folderName+
So I hope command would be like below..
cmd = 's3cmd sync '+dDir+foldername+'/ s3://'+s3bucket+'/'
os.system(cmd)
Here I found some more s3cmd sync help: http://tecadmin.net/s3cmd-file-sync-with-s3bucket/