whats a good way to write this function

sarmenhb · 01-20-2008, 02:58 AM

what im trying to do is, if i were to type the function as br(); without any arguments i want the output to display my own message instead of php's error being

Warning: Missing argument 1 for br()

<?php

function br($num)
{
$num = 0;
if ($num == null) { die('br function must contain a value');
return;
}

else { echo "<br>"; }
return;
}

br();
?>

sarmenhb · 01-20-2008, 03:45 AM

i think im getting somewhere :p but the die function keeps showing and the br wont work.

Code:

<?php

function br() { 
$num = func_num_args;
if ($num > 0 ) 
{

 echo "<br>";
 
  }

 
else { die('too many arguments specified or no arguments entered'); } 

}

br(5); 
?>

xenon · 01-20-2008, 04:34 AM

PHP Code:


			
function br($num = null)
{
    if ($num == null) {
        echo 'br function must contain a value';
    }
    else {
        echo "<br>";
    }
}

Do you have any idea at all what you are doing? If yes, what are you TRYING to do?

Alan @ CIT · 01-20-2008, 10:39 AM

Hi Sarmenhb,

There are a couple of ways to achieve what you want:

The first is to provide your function with a default value in the event that one isn't provided - see Xenon's example. In this case, he has set the default value to NULL, then checks against that in the function.

The second way is to use func_num_args() as you tried:

PHP Code:


			
function br()

{

    $numargs = func_num_args();

    

    if ($numargs == 0)

    {

        die('You must provide a number!');

    }

    else

    {

        $num = func_get_args(0);    // Gets the first argument

        echo "You provided $num !";

    }

}

This method also allows you to accept an unknown number of arguments if your function needs it.

Alan

ETbyrne · 01-20-2008, 03:49 PM

Change <br> to <br />.