CODEWITHSUNDEEP
phparrays
zhenming
zhenming

Reputation: 1089

return empty array in php

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();

Upvotes: 13

Views: 25996

Answers (4)

Oran Fry
Oran Fry

Reputation: 31

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

Upvotes: 0

rikhzan amir
rikhzan amir

Reputation: 1

You can simplify this code to

return (isset($arr)) ? $arr : [] ;

if it is true, return $arr else return empty array

Upvotes: 0

Jasbir Rana
Jasbir Rana

Reputation: 277

To avoid complexity, Simply

return [];

Upvotes: 1

cegfault
cegfault

Reputation: 6632

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.

Upvotes: 16

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