Is there a way in PHP to use a function which has optional parameters in its declaration where I do not have to pass an optional arguments which already have values declared and just pass the next argument(s) which have different values that are further down the parameter list.
Assuming I have a function that has 4 arguments, 2 mandatory, 2 optional. I don't want to use null values for the optional arguments. In usage, there are cases where I want to use the function and the value of the 3rd argument is the same as the default value but the value of the 4th argument is different.
I am looking for a not so verbose solution that allows me to just pass the argument that differs from the default value without considering the order in the function declaration.
createUrl($host, $path, $protocol='http', $port = 80) {
//doSomething
return $protocol.'://'.$host.':'.$port.'/'.$path;
}
I find myself repeating declaring variables so that I could use a function i.e to use $port, I redeclare $protocol with the default value outside the function scope i.e
$protocol = "http";
$port = 8080;
Is there any way to pass the 2nd optional parameter($port) without passing $protocol and it would "automatically" fill in the default value of $protocol i.e
getHttpUrl($server, $path, $port);
This is possible in some languages like Dart in the form of Named Optional parameters.See usage in this SO thread. Is their a similar solution in PHP
You could potentially
use a variadic function for this.
Example:
<?php
function myFunc(...$args){
$sum = 0;
foreach ($args as $arg) {
$sum += $arg;
}
return $sum;
}
Documentation: http://php.net/manual/en/functions.arguments.php#functions.variable-arg-list
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With