docker and python using symfony process - python

I am using Laradock and want to be able to run a python script from my laravel app using Symfony Process. From inside the root on my container I can run "python3 script_name.py arg1" and it runs just fine. pip list shows all modules needed. When I run it from inside Laravel, it tells me:
"import pymysql ImportError: No module named 'pymysql'"
I have used a non-docker Laravel app to do this just fine, using:
$script = storage_path().'/app/script.py';
$process = new Process('python3 '. $script." ".session('division'));
What am I missing?

On *nix make sure that PYTHONPATH is configured correctly for all users or try to set full path to python3.
How to check
At first your php user
php -r "print shell_exec( 'whoami' );" // somebody
When run
su somebody python3 script_name.py arg1

Related

Running flask command via crontab

I have a simple script that executes a flask command called sendemail (located in the "main" blueprint).
The "task" script, located in /home/ubuntu/tasks:
cd /home/ubuntu/app
source venv/bin/activate
flask main sendemail
deactivate
When I run (from anywhere, including the home directory)
bash /home/ubuntu/tasks/task
The function runs exactly as intended. However, when I add this same script to crontab, it produces an error, emailing me this message:
/home/ubuntu/tasks/task: line 4: flask: command not found
I've made sure that I have the latest flask installed and assume this might have something to do with the PATH variables - how can I fix/debug this?
The activation doesn’t work in the cron because you don’t have the same environment variables. You can use set > /path/to/your.log to diagnose…
You can simplify your scrip by calling Flask directly:
/home/ubuntu/app/venv/bin/flask main sendemail

Problem with flask using Python 3.7.6 with vscode on mac

I have a weird problem with some code I want to run. The code itself should not be the problem since it is downloaded from a Udemy class and not modified:
# coding=utf-8
from flask import Flask, render_template, request
app = Flask(__name__)
#app.route("/")
def hello():
items = ["Apfel", "Birne", "Banane"]
return render_template("start.html", name="Max Mustermann", items=items)
#app.route("/test")
def test():
name = request.args.get("name")
return render_template("test.html", name=name)
I found online that, to start the emulated webserver(?) I have to rund the following temrinal commands before I can see the output:
(base) Christophs-MBP:13-23 chris$ export FLASK_APP=run.py run flask
(base) Christophs-MBP:13-23 chris$ export FLASK_APP=run.py run flask
(base) Christophs-MBP:13-23 chris$ export FLASK_APP=run.py
(base) Christophs-MBP:13-23 chris$ run flask
bash: run: command not found
No reaction to my terminal commands
Basically there is no reaction to the command to start the server(?).It should reply with "Running on 127.0.0.1:5000" as soon as I've run the command once.
If I go to my browser, there is no page when I address http://127.0.0.1:5000. What am I doing wrong? I am pretty new to Python and an absolute rookie regarding the terminal. Not sure if I broke something there, since trying to install pyenv to manage my Python installs better as recommended by a friend does not work either (I cannot update the SDK headers as described on RealPython
What are the export statements?
On Mac, export key=value creates a new (or updates an existing one) environment variable - the tutorial most likely simply asked you to provide one where key is FLASK_APP and value is a path to your app.
To verify it's been saved correctly, you can list the variables by just typing export in the terminal and finding out what's inside each of the environment variables on your system (if you want to only view FLASK_APP you can type export | grep FLASK_APP).
Why do you need FLASK_APP?
When you call flask run in your terminal, you will see the following message:
Error: Could not locate a Flask application. You did not provide the "FLASK_APP" environment variable, and a "wsgi.py" or "app.py"
module was not found in the current directory.
I presume your file is called run.py, therefore none of the conditions are met. You could rename run.py to app.py and simply type flask run in the terminal, but you can also type export FLASK_APP=<path-to-run.py>. It seems the tutorial author decided to do the latter. Keep in mind that if you rename your file to app.py you will need to run flask run within the directory that file lives in. You can change directory in the terminal using cd command.
Why do you get bash: run: command not found?
bash is a language running inside your terminal, and it only knows of a few commands - it is not aware of any run commands. It does however know about flask command once you have installed it on your machine. Within the command's output there is a part which includes a run command:
Commands:
routes Show the routes for the app.
run Run a development server.
shell Run a shell in the app context.
Therefore, what you want to do is type flask run instead of just run in your terminal.

Can't run PyCharm interpreter as root

I am trying to use Kubernetes python SDK.
I tried to run the following code:
from kubernetes import client, config
# Configs can be set in Configuration class directly or using helper utility
config.load_kube_config()
v1 = client.CoreV1Api()
print("Listing pods with their IPs:")
ret = v1.list_pod_for_all_namespaces(watch=False)
It failed with lots of errors.
When I run the same code with python from a shell, the same issue.
When I run the same code with sudo python from shell, it works.
I am trying to run PyCharm interperte as root.
Following the instruction from JetBrains, I created a script shell with the name pythonCustomInt.sh that contains:
sudo python
I went to PyCharm settings > Project Interpreter and changed the Base interpreter to /<path>/pythonCutomInt.sh but it writes an error:
Environment location directory is not empty
I am not sure where I need to put the script.
Any idea?
I ran sudo -s and then from the pycharm folder (pycharm-community-2018.1.4/bin) I ran sh ./pycharm.sh and it worked.

manage.py command in crontab not working

I have created a executeable script .sh which contains code to run a django managemenet command.
cron.sh
#!/bin/sh
. /path/to/env/activate
cd /path/to/project
/path/to/env/bin/python manage.py some_command
I can confirm this script and manage.py command is working by executing it directly on terminal
$ /path/to/cron.sh
When i do it same via crontab its not working as expected.
** What am i doing wrong ?? I can confirm there is nothing wrong with crontab, it executing the cron.sh file but path/to/env/bin/python manage.py some_command is not working as expected.
cron log also showing
CRON[14768]: (root) CMD /path/to/cron.sh > /dev/null 2>&1
I am using bitnami django ami (ubuntu 14.04.5 LTS)
Update
After removing /dev/null i am getting this error now
"Cannot locate wrapped file"
It seems that it is a PATH problem. I do not know if django uses specific paths that must be set but AFAIK the crontab PATH is really limited due to security reasons. Just to check if that is the problem you could do in a shell terminal the following:
echo $PATH
You will get a complete PATH for instance:
/usr/local/sbin:/usr/local/bin:/usr/bin:/usr/lib/jvm/default/bin:/usr/bin/site_perl:/usr/bin/vendor_perl:/usr/bin/core_perl
In your crontab, put it above your code:
PATH=/usr/local/sbin:/usr/local/bin:/usr/bin:/usr/lib/jvm/default/bin:/usr/bin/site_perl:/usr/bin/vendor_perl:/usr/bin/core_perl
Tell me if this works. If does, try to purge the provided PATH or even better provide absolute locations in your code.
I have to say that I don't know if you can perform a cd in the cron like this. I always used absolute paths or cd /some/dir && /path/to/script args.
P.S: I cannot make comments yet, for this reason I put it in an answer.
The problem is that your not using the script that Bitnami uses to load all the environment variables (/opt/bitnami/scritps/setenv.sh).
I would try using this script:
#!/bin/sh
. /opt/bitnami/scritps/setenv.sh
. /path/to/env/activate
cd /path/to/project
/path/to/env/bin/python manage.py some_command

run python script on vagrant up

I'm trying to run a vagranfile that will result in a running python flask app. I've tried using this as the last command to achieve this -
config.vm.provision :shell, :path => "python app.py"
But it resulted with the following error -
*pathfor shell provisioner does not exist on the host system: /Users/*****/code/app-vagrant/python app.py
I understand that the script is trying to run this command from the host machine, how do I make Vagrant run the script on the vagrant machine launched?
you have 2 options to run a script using the vagrant shell provisioner you must pass either the inline or path argument:
inline (string) - Specifies a shell command inline to execute on the remote machine.
path (string) - Path to a shell script to upload and execute. It can be a script relative to the project Vagrantfile or a remote script (like a gist).
so when you pass :path => "python app.py" the system tries to find a script named python app.py on your host.
replace using inline argument and it will achieve what you want
config.vm.provision :shell, :inline => "python app.py"
note: provisioner are run as root user by default, if you want to change and run it as vagrant user:
config.vm.provision :shell, :inline => "python app.py", :privileged => false
If you want to start app when the machine start, you can do that:
move your python app code directory to the Vagrantfile's directory
Config the Vagrantfile like that:
config.vm.provision "shell", inline: <<-SHELL
python /vagrant/<your python app code directory>/app.py
SHELL

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