Context : Implement prod Django web project in Centos env.
I am trying to solve a function on file storage action that didn't work but i cant see why?... maybe a permission problem...
I have a pretty same function that work...
The function
import os
def manager_file_tasks(file, archive_path):
print(file, archive_path)
# i got this : file.csv /MYHOME/Site_Django/media/archive_file/uploadedFiles
tmp_archive_path = os.path.join(BASE_DIR, 'visualisation', 'UploadFiles', 'file')
# Test if the folder archive exist
if not os.path.exists(tmp_archive_path):
os.makedirs(tmp_archive_path)
print(tmp_archive_path)
# i got this : /MYHOME/Site_Django/visualisation/UploadFiles/file
fs = FileSystemStorage(location=tmp_archive_path)
print(fs)
# i got this : <django.core.files.storage.FileSystemStorage object at 0x7f57f622bda0>
print(armdb_file.name, armdb_file)
# i got this : file.csv file.csv
print(fs.save(armdb_file.name, armdb_file))
# NO RESULT .... :(
filename = fs.save(armdb_file.name, armdb_file)
print(filename)
armdb_file = fs.path(filename)
print(armdb_file)
# some other stuffs are made after....
Here the rights of the folder:
[me#web01]$ ll /MYHOME/Site_Django/visualisation/UploadFiles/
total 0
drw-r--r--. 2 apache apache 6 10 oct. 11:22 file
EDIT 1
I think that i resolve a part of the problem (that encounter with SELinux)
with this command line :
sudo chcon -t httpd_sys_content_t /MYHOME/Site_Django -R
EDIT 2
The file dont exist in my tmp folder :/
Even the python code :
print(file.file.name)
return
/tmp/tmp1t_ahpg4.upload.csv
I cant read this file with that :
with open(file,"r") as f:
print(f.read())
Thanks in advance for the help.
Related
I've been having trouble deploying my Flask app on Heroku. My current problem is Heroku can't locate the files in my static folder. After trying suggestions on SO, I am now getting this error:
with open(file_dir, encoding="utf-8") as f:
IsADirectoryError: [Errno 21] Is a directory: '/app'
Here's one of the suggestions I found to help Heroku find the encouragement.txt file:
# Absolute path to the file
file_dir = os.path.dirname(os.path.abspath('encouragement.txt'))
# Absolute path to this file's root directory
parent_dir = os.path.join(file_dir, os.pardir)
app_dir = os.path.join(parent_dir, '/app')
The '/app' was added because one of the posts mentioned that Heroku had an app directory as the parent directory, so instead of 'static/encouragement.txt', it would be 'app/static/encouragement.txt'. The same error was given with and without joining '/app'.
Here's where the error came from:
#app.route("/")
def index():
# Read data from the file as str
with open(file_dir, encoding="utf-8") as f:
messages = f.read()
I tried using heroku bash command to check the file system structure and confirm the app directory but the command executed and returned blank.
My files are organized as follow:
MyProject
> static
> encouragement.txt
app.py
Not sure if it's needed, but I'm using windows. Thank you.
Update:
I tried adding index.php and composer.json as a couple tutorials suggested. They didn't work for the txt files; not sure if they helped with the other static files though. Not the best solution, but I ended up just copying and pasting all the txt content to my app.py.
I ran into this issue today after doing a bit of "clean-up" to my environment variables/configs. The issue is you're trying to open a directory itself as if it were a file, which causes the error. I recommend the answer in this article (as shown).
import os
# path to your file
file_name = r'/app/static/encouragement.txt'
print(os.path.isfile(file_name))
with open(file_name, 'r', encoding='utf-8') as f:
lines = f.readlines()
print(lines)
I am new to GitPython and I am trying to get the content of a file within a commit. I am able to get each file from a specific commit, but I am getting an error each time I run the command. Now, I know that the file exist in GitPython, but each time I run my program, I am getting the following error:
returned non-zero exit status 1
I am using Python 2.7.6 and Ubuntu Linux 14.04.
I know that the file exist, since I also go directly into Git from the command line, check out the respective commit, search for the file, and find it. I also run the cat command on it, and the file contents are displayed. Many times when the error shows up, it says that the file in question does not exist. I am trying to go through each commit with GitPython, get every blob or file from each individual commit, and run an external Java program on the content of that file. The Java program is designed to return a string to Python. To capture the string returned from my Java code, I am also using subprocess.check_output. Any help will be greatly appreciated.
I tried passing in the command as a list:
cmd = ['java', '-classpath', '/home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*:', 'java_gram.mainJava','absolute/path/to/file']
subprocess.check_output(cmd, stderr=subprocess.STDOUT, shell=False)
And I have also tried passing the command as a string:
subprocess.check_output('java -classpath /home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*: java_gram.mainJava {file}'.format(file=entry.abspath.strip()), shell=True)
Is it possible to access the contents of a file from GitPython?
For example, say there is a commit and it has one file foo.java
In that file is the following lines of code:
foo.java
import java.io.FileInputStream;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.List;
public class foo{
public static void main(String[] args) throws Exception{}
}
I want to access everything in the file and run an external program on it.
Any help would be greatly appreciated. Below is a piece of the code I am using to do so
#! usr/bin/env python
__author__ = 'rahkeemg'
from git import *
import git, json, subprocess, re
git_dir = '/home/rahkeemg/Documents/GitRepositories/WhereHows'
# make an instance of the repository from specified path
repo = Repo(path=git_dir)
heads = repo.heads # obtain the different repositories
master = heads.master # get the master repository
print master
# get all of the commits on the master branch
commits = list(repo.iter_commits(master))
cmd = ['java', '-classpath', '/home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*:', 'java_gram.mainJava']
# start at the very 1st commit, or start at commit 0
for i in range(len(commits) - 1, 0, -1):
commit = commits[i]
commit_num = len(commits) - 1 - i
print commit_num, ": ", commit.hexsha, '\n', commit.message, '\n'
for entry in commit.tree.traverse():
if re.search(r'\.java', entry.path):
current_file = str(entry.abspath.strip())
# add the current file or blob to the list for the command to run
cmd.append(current_file)
print entry.abspath
try:
# This is the scenario where I pass arguments into command as a string
print subprocess.check_output('java -classpath /home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*: java_gram.mainJava {file}'.format(file=entry.abspath.strip()), shell=True)
# scenario where I pass arguments into command as a list
j_response = subprocess.check_output(cmd, stderr=subprocess.STDOUT, shell=False)
except subprocess.CalledProcessError as e:
print "Error on file: ", current_file
# Use pop on list to remove the last string, which is the selected file at the moment, to make place for the next file.
cmd.pop()
First of all, when you traverse the commit history like this, the file will not be checked out. All you get is the filename, maybe leading to the file or maybe not, but certainly it will not lead to the file from different revision than currently checked-out.
However, there is a solution to this. Remember that in principle, anything you could do with some git command, you can do with GitPython.
To get file contents from specific revision, you can do the following, which I've taken from that page:
git show <treeish>:<file>
therefore, in GitPython:
file_contents = repo.git.show('{}:{}'.format(commit.hexsha, entry.path))
However, that still wouldn't make the file appear on disk. If you need some real path for the file, you can use tempfile:
f = tempfile.NamedTemporaryFile(delete=False)
f.write(file_contents)
f.close()
# at this point file with name f.name contains contents of
# the file from path entry.path at revision commit.hexsha
# your program launch goes here, use f.name as filename to be read
os.unlink(f.name) # delete the temp file
I have a process that scans a tape library and looks for media that has expired, so they can be removed and reused before sending the tapes to an offsite vault. (We have some 7 day policies that never make it offsite.) This process takes around 20 minutes to run, so I didn't want it to run on-demand when loading/refreshing the page. Rather, I set up a django-cron job (I know I could have done this in Linux cron, but wanted the project to be as self-contained as possible) to run the scan, and creates a file in /tmp. I've verified that this works -- the file exists in /tmp from this morning's execution. The problem I'm having is that now I want to display a list of those expired (scratch) media on my web page, but the script is saying that it can't find the file. When the file was created, I use the absolute filename "/tmp/scratch.2015-11-13.out" (for example), but here's the error I get in the browser:
IOError at /
[Errno 2] No such file or directory: '/tmp/corpscratch.2015-11-13.out'
My assumption is that this is a "web root" issue, but I just can't figure it out. I tried copying the file to the /static/ and /media/ directories configured in django, and even in the django root directory, and the project root directory, but nothing seems to work. When it says it cant' find /tmp/file, where is it really looking?
def sample():
""" Just testing """
today = datetime.date.today() #format 2015-11-31
inputfile = "/tmp/corpscratch.%s.out" % str(today)
with open(inputfile) as fh: # This is the line reporting the error
lines = [line.strip('\n') for line in fh]
print(lines)
The print statement was used for testing in the shell (which works, I might add), but the browser gives an error.
And the file does exist:
$ ls /tmp/corpscratch.2015-11-13.out
/tmp/corpscratch.2015-11-13.out
Thanks.
Edit: was mistaken, doesn't work in python shell either. Was thinking of a previous issue.
Use this instead:
today = datetime.datetime.today().date()
inputfile = "/tmp/corpscratch.%s.out" % str(today)
Or:
today = datetime.datetime.today().strftime('%Y-%m-%d')
inputfile = "/tmp/corpscratch.%s.out" % today # No need to use str()
See the difference:
>>> str(datetime.datetime.today().date())
'2015-11-13'
>>> str(datetime.datetime.today())
'2015-11-13 15:56:19.578569'
I ended up finding this elsewhere:
today = datetime.date.today() #format 2015-11-31
inputfilename = "tmp/corpscratch.%s.out" % str(today)
inputfile = os.path.join(settings.PROJECT_ROOT, inputfilename)
With settings.py containing the following:
PROJECT_ROOT = os.path.abspath(os.path.dirname(__file__))
Completely resolved my issues.
Im trying to execute the following code in Python 2.7 on Windows7. The purpose of the code is to take back up from the specified folder to a specified folder as per the naming pattern given.
However, Im not able to get it work. The output has always been 'Backup Failed'.
Please advise on how I get resolve this to get the code working.
Thanks.
Code :
backup_ver1.py
import os
import time
import sys
sys.path.append('C:\Python27\GnuWin32\bin')
source = 'C:\New'
target_dir = 'E:\Backup'
target = target_dir + os.sep + time.strftime('%Y%m%d%H%M%S') + '.zip'
zip_command = "zip -qr {0} {1}".format(target,''.join(source))
print('This is a program for backing up files')
print(zip_command)
if os.system(zip_command)==0:
print('Successful backup to', target)
else:
print('Backup FAILED')
See if escaping the \'s helps :-
source = 'C:\\New'
target_dir = 'E:\\Backup'
So I have the following problem:
On an ivent a javascript sends some text to the django server and there are two functions that should work:
views.py:
def log(request):
f = open('media/log.txt', 'r')
return HttpResponse(f, mimetype='text/plain')
def modelers(request):
mod_stat = request.POST['id']
time = datetime.datetime.now().strftime("%b %d %Y %H:%M:%S")
file=open('media/log.txt', 'a')
file.write(time)
file.write(' ')
file.write(mod_stat)
file.write('\n')
file.close()
return ErrorResponse()
so the user clicks on a button and the "modelers" function is getting the info and is trying to add a line to the log file. But it doesn't work!
The apache error.log says that
IOError: [Errno 13] Permission denied: 'media/log.txt', referer: ...
chmod 777 media doesn't help.. I know that I must config the apache somehow to let the django write files, but didn't find how :(
If not under apache it works great(so the url.py is OK), but I need to make it work with apache. The other part of the application also works fine but there are no operations with files.. until now..
Have you tried chmod 777 media/log.txt? Does ls -l media/ say rwxrwxrwx log.txt? If yes then try to specify absolute path to log.txt in f = open('...log.txt')