Call php function from python

Question

My PHP code is:

function start($height, $width) {
    # do stuff
    return $image;
}

Here is my Python code:

import subprocess
def php(script_path):
        p = subprocess.Popen(['php', script_path], stdout=subprocess.PIPE)
        result = p.communicate()[0]
            return result

    page_html = "test entry"
    output = php("file.php") 
    print page_html + output

    imageUrl = start(h,w)

In Python, I want to use this PHP launch function. I do not know how to access the launch function from Python. Can someone help me with this?

Answer 1

This is how I do it. It works like a charm.

# shell execute PHP
def php(code):
  # open process
  p = Popen(['php'], stdout=PIPE, stdin=PIPE, stderr=STDOUT, close_fds=True)

  # read output
  o = p.communicate(code)[0]

  # kill process
  try:
    os.kill(p.pid, signal.SIGTERM)
  except:
    pass

  # return
  return o

To execute a specific file, follow these steps:

width = 100
height = 100

code = """<?php

  include('/path/to/file.php');
  echo start(""" + width + """, """ + height + """);

?>
"""
res = php(code)