Question

I have a problem running another file from php. I want my php params to be the output of running a python file that calls another file itself.

Here is my php file:

<?php
    if (isset($_POST['submit'])) {
    $params = solve();
}

function solve() {
  exec("python array.py", $output);
  return $output;
}
?>

If array.py is simply:

if __name__ == "__main__":
    print 1
    print 2
    print 3
    print 4

I will get 1,2,3,4 for my output, but I as soon as I change array.py to the following file that calls os.system, I don’t get anything. So the new array.py is:

import os

def main():
    os.system("python test.py") #test.py creates tmp.txt with 4 lines w/ values 1,2,3,4


def output():
    f = open("tmp.txt", "r")
    myReturn = []
    currentline = f.readline()

    while currentline:
          val = currentline[:-1]  #Getting rid of '\n'
          val = int(val)
          myReturn = myReturn + [val]
          currentline = f.readline()
    f.close()
    return myReturn


if __name__ == "__main__":
     main()
     o = output()
     print o[0]
     print o[1]
     print o[2]
     print o[3]

Also if I just run test.py, the output is the file tmp.txt:

 1
 2
 3
 4

So now, when I run my php file, the output tmp.txt is not even created in the directory and as a result I don’t get any output from my php either.
I am not sure why this is happening because when I just run array.py myself, I get the desired output, and the tmp file is created.

EDIT:
I forgot to include: import os above.

Answer 1

Change exec to:

exec("python array.py 2>&1", $output)

Or check the web server or php error log. This will return the error output from the python script to your php script (not normally what you want in production).