Question

While running one PHP script. The script will call another one somewhere in the script. Those two scripts will need to run simultaneous so I can not use include or require here. The script will be called using the terminal like this:

exec( '/usr/bin/php ../process.php >> logs/' . date( 'Y_m_d' ) . '.log 2>&1 &' );

The big problem here is I want to process 6 different things, but they will all be using quite the same code. The only difference is the directory where they will be processed in.

Now the question: Is there a simple way to run the second script while passing one or more variables?

I tried something like ../process?var=something, but this clearly doesn’t work this way.

Answer 1

You can pass some variables as arguments to the script. All the arguments passed are in the variable $argv.

See php docs $argv

Then just loop through the $argv and maybe check for some keywords you define.

exec( '/usr/bin/php ../process.php -f '/path/to/folder' >> logs/' . date( 'Y_m_d' ) . '.log 2>&1 &' );

The $argv variable would then hold -f and /path/to/folder, so just check for -f in the loop and get the value by position of -f+1.