Execute php code in Python

Question

For some reason, I have to run a php script to get an image from Python. Because the php script is very big and it is not mine, it will takes me days to find out the right a

Answer 1

You can use php.py. This would allow you to execute php code in python, like in this example (taken from here):

php = PHP("require '../code/private/common.php';")
code = """for ($i = 1; $i <= 10; $i++) { echo "$i\n"; }"""
print php.get_raw(code)

Answer 2

You can simply execute the php executable from Python.

Edit: example for Python 3.5 and higher using subprocess.run:

import subprocess

result = subprocess.run(
    ['php', 'image.php'],    # program and arguments
    stdout=subprocess.PIPE,  # capture stdout
    check=True               # raise exception if program fails
)
print(result.stdout)         # result.stdout contains a byte-string

Answer 3

Example code:

import subprocess

# if the script don't need output.
subprocess.call("php /path/to/your/script.php")

# if you want output
proc = subprocess.Popen("php /path/to/your/script.php", shell=True, stdout=subprocess.PIPE)
script_response = proc.stdout.read()

Answer 4

Make a wrapper around the PHP script, which:

• performs the stuff (If I understand well, it's an image creation),
• then redirects (301 Moved Permanently) to the result image,
• (also, the image should be purged some day).

So you can refer to this service (the PHP script) with a simple HTTP request, from anywhere, you can test it with browser, use from Python prg, you need just download the image the usual way.

Also, if you have a such standalone sub-system, don't feel bad about implement it with different language/technique. It has several advantages, e.g. you can install that service on a different host.

Recommended reading: Service-Oriented Architecture on Wikipedia.

Answer 5

If you can run the PHP script locally from the command-line, subprocess.check_output() will let you can PHP and will capture the return value.

If you are accessing PHP via a socket, then you can use urllib.urlopen() or urllib.urlretrieve() to pull down the resource.