Any way to specify optional parameter values in PHP?

Question

Let\'s say I\'ve got a PHP function foo:

function foo($firstName = \'john\', $lastName = \'doe\') {
    echo $firstName . \" \" . $lastName;
}
// foo(); --&g         


        

Answer 1

Sadly what you're trying to do has no "syntactic sugar" way of doing it. They're all varying forms of WTF.

If you need a function that takes an undefined number of arbitrary parameters,

function foo () { 
     $args = func_get_args(); 
     # $args = arguments in order 
}

Will do the trick. Try avoid using it too much, because for Php this is a bit on the magical side.

You could then optionally apply defaults and do strange things based on parameter count.

function foo(){ 
   $args = func_get_args();
   if( count($args) < 1 ){ 
       return "John Smith"; 
   }
   if( count($args) < 2 ) { 
       return "John " .$args[0];
   }
   return $args[0] . " " . $args[1];
}

Also, you could optionally emulate Perl style parameters,

function params_collect( $arglist ){ 
    $config = array();
    for( $i = 0; $i < count($arglist); $i+=2 ){ 
       $config[$i] = $config[$i+1];
    }
    return $config; 
}
function param_default( $config, $param, $default ){ 
    if( !isset( $config[$param] ) ){ 
           $config[$param] = $default;
    }
    return $config;
}

function foo(){ 
   $args = func_get_args();
   $config = params_collect( $args ); 
   $config = param_default( $config, 'firstname' , 'John' ); 
   $config = param_default( $config, 'lastname' ,  'Smith' ); 
   return $config['firstname'] . '  ' . $config['lastname'];   
}

foo( 'firstname' , 'john', 'lastname' , 'bob' ); 
foo( 'lastname' , 'bob', 'firstname', 'bob' ); 
foo( 'firstname' , 'smith'); 
foo( 'lastname', 'john' );

Of course, it would be easier to use an array argument here, but you're permitted to have choice ( even bad ways ) of doing things.

_{notedly, this is nicer in Perl because you can do just foo( firstname => 'john' );}

• func_get_args at php.net