How to simulate args inside a php file? - php

Question

PHP file.php arg1 arg2
Now I want to hardcode arg1 and arg2 into file.php,how to do it?

Answer 1

I never tried it, but the arguments are contained in a certain array $argv. So you have to set those entries if you want to hardcode them:
$argc = 3; // number of arguments + 1
$argv[0] = 'file.php'; // contains script name
$argv[1] = 'arg1';
$argv[2] = 'arg2';
$argv is a reserved variable.
But note that the first two parameter that you specify via command line will always be overwritten by arg1 and arg2.
Instead, if you need these values always in your script you should define them as normal variables at the top of your script:
$var1 = 'arg1';
$var2 = 'arg2';
or even as constants:
define('CONST1', 'arg1');
define('CONST2', 'arg2');
If you only want to provide arg1, arg2 as parameter via the command line, then you can just access them via $argv[1] and $argv[2], no need to mess with $argv;

Answer 2

you would use argv() inside your php script to get the arguments
foreach ($argv as $args){
print $args."\n";
}
element 0 contains the script name. the rest are the arguments.

Answer 3

You want to test how many arguments to the function, then do something accordingly.
function foo()
{
$numargs = func_num_args();
echo "Number of arguments: $numargs<br />\n";
if ($numargs >= 2) {
echo "Second argument is: " . func_get_arg(1) . "<br />\n";
}
$arg_list = func_get_args();
for ($i = 0; $i < $numargs; $i++) {
echo "Argument $i is: " . $arg_list[$i] . "<br />\n";
}
}
foo(1, 2, 3);