PHP exec function passing a value to command line

You need to use the $argv variable built in to PHP. Exec() method emulate the command line, meaning exec("example.php") works the same as php -f example.php on the command line. This means you can pass arguments to by putting a space between the file name and all the arguements (exactly like you would do in the command line). So, exec("example.php $var1 $var2") is the same as php -f example.php $var1 $var2 on the command line. Now the $argv variable come into pay in the executed script. Inside of example.php if you var_dump $argv you will see an array, where the zero index is the name of the script being executed and all the other indexes are the variables passed to the script. Check out the PHP Manual for more information about $argv.

An important thing to remember is the $argv breaks up arguments by spaces. So if you pass a string into the exec() function, and it has spaces in it, $argv will break the string into separate arguments. For example:

$string = "this is a string";
exec("example.php $string");

//inside example.php
var_dump($argv);

this will output an array that looks like this:

array[4]{
    [0]=>"this",
    [1] =>"is",
    [2]=>"a",
    [3]=>"string".
}

If you want to pass in a string as an argument you need to escape all the spaces in the string.