如何将参数从 python 传递给 php 脚本？

Question

I have this php script:

<?php
$argOne = $argv[1];
print "$argOne";
print "you made it to php";

And I have this python function calling it:

def phpCall(argOne):
    script = 'php ./phpScript'
    proc = subprocess.run([script, argOne], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    print(proc)
    script_response = str(proc.stdout,'utf-8')
    return script_response

this is what happens when I run the python function:

root@testSrv:~# python3 webAccess.py 
CompletedProcess(args=['php ./phpScript', 'testArgument'], returncode=0, stdout=b'you made it to php', stderr=b'')
you made it to php
root@testSrv:~# 

Calling the php script itself:

root@testSrv:~# php phpScript argumentOne
argumentOneyou made it to phproot@testSrv~# 

So my argument is making its way into the python function, but is not being passed correctly to be assinged to $argv[1] in the php script but not sure why. I yanked this syntax out of a similar thing I was doing running a bash script from python, and this syntax worked to pass to the bash script's $1

Answer 1

use sys argv:

import subprocess
from sys import argv

def phpCall(argOne):
    proc = subprocess.Popen("php phpScript.php "+argOne, shell=True, stdout=subprocess.PIPE)
    script_response = proc.stdout.read()
    print(script_response)
    return script_response

phpCall(*argv[1:])

And

python3 webAccess.py testing!

Returns

testing! you made it to php

Answer 2

dzil123 on the python discord helped me out, and while Monnomcjo's solution may work went this route:

proc = subprocess.run(['php', '-f', './phpScript', argOne], stdout=subprocess.PIPE, stderr=subprocess.PIPE)

to format the subprocess.run() call correctly, without passing in the 'script' variable I had, and removing shell=True