PHP function "if a second parameter is specified"

Question

I've got a simple function to conditionally convert units into feet or inches for display. I'd also like to use it to conditionally add the actual text ft or in to the end. I can't just add that text in every case, because I'd end up with 36 in x 24 in x 10 in when what you want is 36x24x10 in.

So basically I'm trying to do this:

function ex($x, $y) {
    if only $x is specified {
        if ($x < 50) {
            return $x/25.4
        }
        else {
            return $x/304.8
        }
    }
    elseif $x and $y are both specified {
        if ($x < 50) {
            return ' in';
        }
        else {
            return ' ft';
        }
    }
}

echo ex($NumberToConvert).'x'.ex($OtherNumberToConvert).example($NumberIWantToConvert,1);

But of course that results in "missing argument 2" warnings. The obvious hack is to use if ($y > 0) and always have to specify two arguments (making the latter 0 when I'm looking for numerical output and 1 when I'm looking for text). But is there something more elegant?

(yes I know this only makes sense if both numbers are <50 or neither are)

function ex($x, $y='') {

user6763587
– user6763587

2017-01-15 01:54:45 +00:00
Commented Jan 15, 2017 at 1:54 — user6763587
– user6763587, Commented Jan 15, 2017 at 1:54

Rajdeep Paul · Accepted Answer · 2017-01-15 02:12:54Z

1

Make use of default argument values of a funciton. So your ex function should be like this:

function ex($x = null, $y = null) {
    if($x != null && $y == null){
        if ($x < 50) {
            return $x/25.4
        }else{
            return $x/304.8
        }
    }elseif($x != null && $y != null){
        if ($x < 50) {
            return ' in';
        }else{
            return ' ft';
        }
    }
}

Take the default argument as null for both $x and $y in the function, and process these variables inside the function accordingly.

edited Jan 15, 2017 at 2:12

answered Jan 15, 2017 at 1:58

Rajdeep Paul

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