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Python script called from PHP can't write a file

I have a problem with converting docx to pdf files in my script.
At first I tried to use a pure php-based solution, described here:
https://stackoverflow.com/a/20035739/12812601
Unfortunatelly this does not work (it creates an empty com object, then throws a fatal error).
So I've tried to use a python script to do this.
I use a great script from here:
https://stackoverflow.com/a/20035739/12812601
So here is a problem.
The Python script standalone (run via a command line) works just fine, and saves the converted pdf. Unfortunatelly when I try to call it via PHP it can't save a converted file.
PHP scripts can create and write files oin the same directory without any problem
This supposed to be a local configuration, so I do not care about any portability
Scripts:
*******PHP*******
<?php
//Script only for testing Python calls, tried different methods
error_reporting(E_ALL);
echo '<h1>Begin</h1>';
echo '<h2>Before call</h2>';
exec ('python dp.py');
echo '<h2>After exec call</h2>';
system('python dp.py');
echo '<h2>After Sys Call</h2>';
passthru('python dp.py');
echo '<h2>After Pass Call</h2>';
$w = get_current_user();
var_dump($w);
?>
*****Python*****
import sys
import os
import comtypes.client
import win32com.client
wdFormatPDF = 17
#static file names for testing
in_file = 'C:\\Users\\fake_user\\OneDrive\\Stuff\\f1.docx'
out_file = 'C:\\Users\\fake_user\\OneDrive\\Stuff\\f3.pdf'
print('BEGIN<br>\n')
word = win32com.client.Dispatch('Word.Application')
word.Visible = False
doc = word.Documents.Open(in_file)
print('\nOpened Docx\n<br>')
print(in_file);
doc.SaveAs(out_file, FileFormat=wdFormatPDF)
print('\nSaved\n<br>')
doc.Close()
word.Quit()
print('DONE\n')
*****Output from the browser*****
Begin
Before call
After exec call
BEGIN
Opened Docx
C:\Users\fake_user\OneDrive\Stuff\f1.docx
After Sys Call
BEGIN
Opened Docx
C:\Users\fake_user\OneDrive\Stuff\f1.docx
After Pass Call
string(5) "fake_user"
System configuration
Windows 7 Professional Edition Service Pack 1
Apache/2.4.26 (Win32)
OpenSSL/1.0.2l
PHP/7.1.7
Python 3.8.1
I tried to run Apache both as a system service and as a user who owns the OneDrive (name changed to "fake_user" here), so it shouldn't be a permissions issue (I think)
Any help appreciated

accessing files via python using a service account

I am playing with windows server 2012 r2.
I have some files on the server. I have a separate service account which has the read access to the files. What i want to do is using python access the files by network share(any other suggestions welcomed) but only through the service account.
PS: i cannot use RDP.

The underlying WINAPIs for this task are part of [MS.Docs]: WNetAddConnection2W function family.
The [GitHub]: mhammond/pywin32 - Python for Windows (pywin32) Extensions wrapper is [ActiveState]: Module win32wnet (it's not the official doc (I couldn't find any at this point) - I don't know for how long the URL will be valid, but it's the best I could find).
I've prepared a trivial example.
code00.py:
#!/usr/bin/env python3
import sys
import os
import pywintypes
import win32wnet
CONNECT_INTERACTIVE = 0x00000008
HOST_NAME = "192.168.1.3"
SHARE_NAME = "Work"
SHARE_FULL_NAME = os.path.sep * 2 + os.path.sep.join((HOST_NAME, SHARE_NAME))
SHARE_USER = "cfati"
SHARE_PWD = "********"
def main():
net_resource = win32wnet.NETRESOURCE()
net_resource.lpRemoteName = SHARE_FULL_NAME
flags = 0
#flags |= CONNECT_INTERACTIVE
print("Trying to create connection to: {:s}".format(SHARE_FULL_NAME))
try:
win32wnet.WNetAddConnection2(net_resource, SHARE_PWD, SHARE_USER, flags)
except pywintypes.error as e:
print(e)
else:
print("Success!")
if __name__ == "__main__":
print("Python {:s} on {:s}\n".format(sys.version, sys.platform))
main()
Notes:
The password is obfuscated (obviously)
This is the simplest functionality (the equivalent of your command), however the function can do much more:
One thing that I want to point out. If you:
Input some invalid credentials, and
Decomment the flags |= CONNECT_INTERACTIVE line
A credentials dialog box will then pop up
Output:
(py35x64_test) e:\Work\Dev\StackOverflow\q050602112>net use
New connections will be remembered.
There are no entries in the list.
(py35x64_test) e:\Work\Dev\StackOverflow\q050602112>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" code00.py
Python 3.5.4 (v3.5.4:3f56838, Aug 8 2017, 02:17:05) [MSC v.1900 64 bit (AMD64)] on win32
Trying to create connection to: \\192.168.1.3\Work
Success!
(py35x64_test) e:\Work\Dev\StackOverflow\q050602112>net use
New connections will be remembered.
Status Local Remote Network
-------------------------------------------------------------------------------
OK \\192.168.1.3\Work Microsoft Windows Network
The command completed successfully.
(py35x64_test) e:\Work\Dev\StackOverflow\q050602112>net use * /delete /y
You have these remote connections:
\\192.168.1.3\Work
Continuing will cancel the connections.
The command completed successfully.
(py35x64_test) e:\Work\Dev\StackOverflow\q050602112>net use
New connections will be remembered.
There are no entries in the list.

Python os.environ throws key error?

I'm accessing an environment variable in a script with os.environ.get and it's throwing a KeyError. It doesn't throw the error from the Python prompt. This is running on OS X 10.11.6, and is Python 2.7.10.
What is going on?
$ python score.py
Traceback (most recent call last):
File "score.py", line 4, in <module>
setup_logging()
File "/score/log.py", line 29, in setup_logging
config = get_config()
File "/score/log.py", line 11, in get_config
environment = os.environ.get('NODE_ENV')
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/UserDict.py", line 23, in __getitem__
raise KeyError(key)
KeyError: 'NODE_ENV'
$ python -c "import os; os.environ.get('NODE_ENV')"
$
As requested, here's the source code for score.py
from __future__ import print_function
from log import get_logger, setup_logging
setup_logging()
log = get_logger('score')
And here's log.py
import json
import os
import sys
from iron_worker import IronWorker
from logbook import Logger, Processor, NestedSetup, StderrHandler, SyslogHandler
IRON_IO_TASK_ID = IronWorker.task_id()
def get_config():
environment = os.environ.get('NODE_ENV')
if environment == 'production':
filename = '../config/config-production.json'
elif environment == 'integration':
filename = '../config/config-integration.json'
else:
filename = '../config/config-dev.json'
with open(filename) as f:
return json.load(f)
def setup_logging():
# This defines a remote Syslog handler
# This will include the TASK ID, if defined
app_name = 'scoreworker'
if IRON_IO_TASK_ID:
app_name += '-' + IRON_IO_TASK_ID
config = get_config()
default_log_handler = NestedSetup([
StderrHandler(),
SyslogHandler(
app_name,
address = (config['host'], config['port']),
level = 'ERROR',
bubble = True
)
])
default_log_handler.push_application()
def get_logger(name):
return Logger(name)

Try running:
find . -name \*.pyc -delete
To delete your .pyc files.
Researching your problem I came across this question, where a user was experiencing the same thing: .get() seemingly raising a KeyError. In that case, it was caused, according to this accepted answer, by a .pyc file which contained code where a dict value was being accessed by key (i.e., mydict['potentially_nonexistent_key']), while the traceback was showing the code from the updated .py file where .get() was used. I have never heard of this happening, where the traceback references current code from a .py file, but shows an error raised by an outdated .pyc file, but it seems to have happened at least once in the history of Python...
It is a long shot, but worth a try I thought.

I encountered a similar error when I set the environment variable without exporting it. So if you do this:
me#host:/# NODE_ENV=foo
You will get this:
me#host:/# python3
Python 3.8.2 (default, Apr 27 2020, 15:53:34)
[GCC 9.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> node_env = os.environ['NODE_ENV']
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python3.8/os.py", line 675, in __getitem__
raise KeyError(key) from None
KeyError: 'NODE_ENV'
>>>
But if you do this:
me#host:/# NODE_ENV=foo
me#host:/# export NODE_ENV
It works:
me#host:/# python3
Python 3.8.2 (default, Apr 27 2020, 15:53:34)
[GCC 9.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> node_env = os.environ['NODE_ENV']
>>> print(node_env)
foo
>>>

Command for windows to delete the .pyc files:
del /S *.pyc

I had the same problem. I solved that by making some corrections on the .env file:
Before:
Key = Value
After my correction:
Key=Value
without blank spaces and worked!

I was getting this error while trying to source from a .env file.
I didn't explicitly export the env vars so I had to change this.
ENVIRONMENT=DEV
to this
export ENVIRONMENT=DEV

Use export a=10 instead of a=10 while setting env variable. Add the same in ~./bashrc to reload the env var wherever you login.
Doing this resolved the issue

I'd recommend you start debugging os.py, for instance, on windows it's being used this implementation:
def get(self, key, failobj=None):
print self.data.__class__
print key
return self.data.get(key.upper(), failobj)
And if I test it with this:
import os
try:
os.environ.get('NODE_ENV')
except Exception as e:
print("-->{0}".format(e.__class__))
os.environ['NODE_ENV'] = "foobar"
try:
os.environ.get('NODE_ENV')
except Exception as e:
print("{0}".format(e.__class__))
The output will be:
<type 'dict'>
PYTHONUSERBASE
<type 'dict'>
APPDATA
<type 'dict'>
NODE_ENV
<type 'dict'>
NODE_ENV
So it makes sense the exception is not spawned reading dict.get docs.
In any case, if you don't want to mess up or debugging the python modules, try cleaning up the *.pyc files, try to set up properly NODE_ENV. And if all that don't work, restart your terminal to clear up.

Python - How do you run a .py file?

I've looked all around Google and its archives. There are several good articles, but none seem to help me out. So I thought I'd come here for a more specific answer.
The Objective: I want to run this code on a website to get all the picture files at once. It'll save a lot of pointing and clicking.
I've got Python 2.3.5 on a Windows 7 x64 machine. It's installed in C:\Python23.
How do I get this script to "go", so to speak?
=====================================
WOW. 35k views. Seeing as how this is top result on Google, here's a useful link I found over the years:
http://learnpythonthehardway.org/book/ex1.html
For setup, see exercise 0.
=====================================
FYI: I've got zero experience with Python. Any advice would be appreciated.
As requested, here's the code I'm using:
"""
dumpimages.py
Downloads all the images on the supplied URL, and saves them to the
specified output file ("/test/" by default)
Usage:
python dumpimages.py http://example.com/ [output]
"""
from BeautifulSoup import BeautifulSoup as bs
import urlparse
from urllib2 import urlopen
from urllib import urlretrieve
import os
import sys
def main(url, out_folder="C:\asdf\"):
"""Downloads all the images at 'url' to /test/"""
soup = bs(urlopen(url))
parsed = list(urlparse.urlparse(url))
for image in soup.findAll("img"):
print "Image: %(src)s" % image
filename = image["src"].split("/")[-1]
parsed[2] = image["src"]
outpath = os.path.join(out_folder, filename)
if image["src"].lower().startswith("http"):
urlretrieve(image["src"], outpath)
else:
urlretrieve(urlparse.urlunparse(parsed), outpath)
def _usage():
print "usage: python dumpimages.py http://example.com [outpath]"
if __name__ == "__main__":
url = sys.argv[-1]
out_folder = "/test/"
if not url.lower().startswith("http"):
out_folder = sys.argv[-1]
url = sys.argv[-2]
if not url.lower().startswith("http"):
_usage()
sys.exit(-1)
main(url, out_folder)

On windows platform, you have 2 choices:
In a command line terminal, type
c:\python23\python xxxx.py
Open the python editor IDLE from the menu, and open xxxx.py, then press F5 to run it.
For your posted code, the error is at this line:
def main(url, out_folder="C:\asdf\"):
It should be:
def main(url, out_folder="C:\\asdf\\"):

Usually you can double click the .py file in Windows explorer to run it. If this doesn't work, you can create a batch file in the same directory with the following contents:
C:\python23\python YOURSCRIPTNAME.py
Then double click that batch file. Or, you can simply run that line in the command prompt while your working directory is the location of your script.

Since you seem to be on windows you can do this so python <filename.py>. Check that python's bin folder is in your PATH, or you can do c:\python23\bin\python <filename.py>. Python is an interpretive language and so you need the interpretor to run your file, much like you need java runtime to run a jar file.

use IDLE Editor {You may already have it} it has interactive shell for python and it will show you execution and result.

Your command should include the url parameter as stated in the script usage comments.
The main function has 2 parameters, url and out (which is set to a default value)
C:\python23\python "C:\PathToYourScript\SCRIPT.py" http://yoururl.com "C:\OptionalOutput\"

If you want to run .py files in Windows, Try installing Git bash
Then download python(Required Version) from python.org and install in the main c drive folder
For me, its :
"C:\Python38"
then open Git Bash and go to the respective folder where your .py file is stored :
For me, its :
File Location : "Downloads"
File Name : Train.py
So i changed my Current working Directory From "C:/User/(username)/" to "C:/User/(username)/Downloads"
then i will run the below command
" /c/Python38/python Train.py "
and it will run successfully.
But if it give the below error :
from sklearn.model_selection import train_test_split
ModuleNotFoundError: No module named 'sklearn'
Then Do not panic :
and use this command :
" /c/Python38/Scripts/pip install sklearn "
and after it has installed sklearn go back and run the previous command :
" /c/Python38/python Train.py "
and it will run successfully.
!!!!HAPPY LEARNING !!!!

Internal Server Error 500 - Python, CGI

My .py file executes ok in terminal, but gives this error in the browser
<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN">
<html><head>
<title>500 Internal Server Error</title>
</head><body>
<h1>Internal Server Error</h1>
...
...
Here is the .py file:
#!/usr/bin/python
import cgi
import cgitb; cgitb.enable()
print "Content-Type: text/html\n\n" # HTML is following
print # blank line, end of headers
print "<TITLE>CGI script output</TITLE>"
print "<H1>This is my first CGI script</H1>"
print "Hello, world!"
Should i be saving this as a .cgi file? I have tried with the same errors, i have tried many files like this and none work, i am sure the apache server is working as there are other .cgi scripts running from the same directory without issues.
I have also tried:
#!/usr/local/bin/python &
#!/usr/bin/local/python
Any help appreciated.
EDIT
error log output:
(2) No such file or directory: exec of '.../.../.../test.py' failed
Premature end of script headers: test.py

Here is something I wrote up a while ago. These are some good things to look for when troubleshooting Python CGI.
There are some tips to getting Python working in CGI.
Apache setup: This may be old
Add python as a CGI by modifying the following in the configuration:
Options Indexes FollowSymLinks ExecCGI
AddHandler cgi-script .cgi .py
Always browse the pages through Apache.
Note that viewing files in the filesystem through a browser works for most things on an html page but will not work for CGI. For scripts to work they must be opened through the htdocs file system. The address line of your browser should look like:
\\127.0.0.1\index.html or
\\localhost\index.html
If you open a file up through the file system the CGI will not work. Such as if this is in the location bar of your browser:
c:\Apache\htdocs\index.html (or some other example location)
Convert end of lines of scripts to Unix format:
Most editors have options to "show end of lines" and then a tool to convert from Unix to PC format. You must have the end of lines set to Unix format.
State the path to the Python interpreter on the first line of the CGI script:
You must have one of the following lines as the first line of your Python CGI script:
#!C:\Python25\Python.exe
#!/usr/bin/python
The top line is used when you are debugging on a PC and the bottom is for a server such as 1and1. I leave the lines as shown and then edit them once they are up on the server by deleting the first line.
Print a content type specifying HTML before printing any other output:
This can be done simply by adding the following line somewhere very early in your script:
print "Content-Type: text/html\n\n"
Note that 2 end of lines are required.
Setup Python scripts to give debugging information:
Import the following to get detailed debugging information.
import cgitb; cgitb.enable()
An alternative if cgitb is not available is to do the following:
import sys
sys.stderr = sys.stdout
On the server the python script permissions must be set to execute.
After uploading your files be sure to edit the first line and set the permissions for the file to execute.
Check to see if you can hit the python script directly. If you can't, fix with the above steps (2-6). Then when the Python script is working, debug the shtml.