I have a function that returns an array, and is passed to foreach i.e.
foreach(function() as $val)
Since the array I am returning is declared in a series of if statements, I need to return an empty array if all the if statements are evaluated to false. Is this a correct way to do it?
if (isset($arr))
return $arr;
else
return array();
I would recommend declaring $arr = array(); at the very top of the function so you don't have to worry about it.
If you are doing the check immediately before you return, I do not recommend isset. The way you are using foreach is depending on an array being returned. If $arr is set to a number, for example, then it will still validate. You should also check is_array(). For example:
if (isset($arr) && is_array($arr))
return $arr;
else
return array();
Or in one line instead:
return (isset($arr) && is_array($arr)) ? $arr : array();
But, like I said, I recommending declaring the array at the very top instead. It's easier and you won't have to worry about it.
To avoid complexity, Simply
return [];
You can simplify this code to
return (isset($arr)) ? $arr : [] ;
if it is true, return $arr else return empty array
Best to do it like this:
$arr = []; // declare your array upfront
// do your conditional adding and manipulating here
// if (...) $arr['abc'] = 'ABC';
return $arr; // only one return statement needed because the $arr definitely exists
