Question

I’m writing an application that sends requests to the PHP CLI. However, as i try to run the application, php complains it can’t find the script to run.

Here’s the relevant code:

char *params[] = {
    "/usr/bin/php",
    "-f /var/www/test/php/tracker/gps-upload-sh.php",
    imei,
    rmc,
    (char *) 0
};
signal(SIGCHLD, SIG_IGN);
pid_t pID = fork();
if (pID == 0) {
    if(execv("/usr/bin/php", params) == -1) {
        perror("Failed to call php.");
        _exit(1);
    }
    _exit(0);
}

Output:

$ ./socket
Could not open input file:  /var/www/test/php/tracker/gps-upload-sh.php
^C

File:

$ ls -l /var/www/test/php/tracker/gps-upload-sh.php
-rwxr-xr-x 1 onik onik 6707 2012-03-09 16:00 /var/www/test/php/tracker/gps-upload-sh.php

Run directly (where REQUIRED_PARAMS are the same as the ones passed from the execv):

$ php -f /var/www/test/php/tracker/gps-upload-sh.php [REQUIRED_PARAMS]
OK

What to do?

Answer 1

Try separating the -f and the path into separate parameters:

char *params[] = {
    "/usr/bin/php",
    "-f",
    "/var/www/test/php/tracker/gps-upload-sh.php",
    imei,
    rmc,
    (char *) 0
};

(This is what your shell does with you put a space between the two.)

---

Alternatively, remove the space between -f and the path:

char *params[] = {
    "/usr/bin/php",
    "-f/var/www/test/php/tracker/gps-upload-sh.php",
    imei,
    rmc,
    (char *) 0
};