cleanest way to skip a foreach if array is empty [duplicate] - php

Question

This question already has answers here:
Invalid argument supplied for foreach()
(20 answers)
Closed 7 years ago.
Not a major problem but I was wondering if there is a cleaner way to do this. It would be good to avoid nesting my code with an unnecessary if statement. If $items is empty php throws an error.
$items = array('a','b','c');
if(!empty($items)) { // <-Remove this if statement
foreach($items as $item) {
print $item;
}
}
I could probably just use the '#' error suppressor, but that would be a bit hacky.

Answer 1

There are a million ways to do this.
The first one would be to go ahead and run the array through foreach anyway, assuming you do have an array.
In other cases this is what you might need:
foreach ((array) $items as $item) {
print $item;
}
Note: to all the people complaining about typecast, please note that the OP asked cleanest way to skip a foreach if array is empty (emphasis is mine). A value of true, false, numbers or strings is not considered empty.
In addition, this would work with objects implementing \Traversable, whereas is_array wouldn't work.

Answer 2

The best way is to initialize every bloody variable before use.
It will not only solve this silly "problem" but also save you a ton of real headaches.
So, introducing $items as $items = array(); is what you really wanted.

Answer 3

$items = array('a','b','c');
if(is_array($items)) {
foreach($items as $item) {
print $item;
}
}

Answer 4

If variable you need could be boolean false - eg. when no records are returned from database or array - when records are returned, you can do following:
foreach (($result ? $result : array()) as $item)
echo $item;
Approach with cast((Array)$result) produces an array of count 1 when variable is boolean false which isn't what you probably want.

Answer 5

I wouldn't recommend suppressing the warning output. I would, however, recommend using is_array instead of !empty. If $items happens to be a nonzero scalar, then the foreach will still error out if you use !empty.

Answer 6

I think the best approach here is to plan your code so that $items is always an array. The easiest solution is to initialize it at the top of your code with $items=array(). This way it will represent empty array even if you don't assign any value to it.
All other solutions are quite dirty hacks to me.

Answer 7

foreach((array)$items as $item) {}

Answer 8

i've got the following function in my "standard library"
/// Convert argument to an array.
function a($a = null) {
if(is_null($a))
return array();
if(is_array($a))
return $a;
if(is_object($a))
return (array) $a;
return $_ = func_get_args();
}
Basically, this does nothing with arrays/objects and convert other types to arrays. This is extremely handy to use with foreach statements and array functions
foreach(a($whatever) as $item)....
$foo = array_map(a($array_or_string)....
etc

Answer 9

Ternary logic gets it down to one line with no errors. This solves the issue of improperly cast variables and undefined variables.
foreach (is_array($Items) || is_object($Items) ? $Items : array() as $Item) {
It is a bit of a pain to write, but is the safest way to handle it.

Answer 10

You can check whether $items is actually an array and whether it contains any items:
if(is_array($items) && count($items) > 0)
{
foreach($items as $item) { }
}

Answer 11

Best practice is to define variable as an array at the very top of your code.
foreach((array)$myArr as $oneItem) { .. }
will also work but you will duplicate this (array) conversion everytime you need to loop through the array.
since it's important not to duplicate even a word of your code, you do better to define it as an empty array at top.