Invalid argument supplied for foreach()

Question

It often happens to me to handle data that can be either an array or a null variable and to feed some foreach with these data.

$values = get_val         


        

Answer 1

I usually use a construct similar to this:

/**
 * Determine if a variable is iterable. i.e. can be used to loop over.
 *
 * @return bool
 */
function is_iterable($var)
{
    return $var !== null 
        && (is_array($var) 
            || $var instanceof Traversable 
            || $var instanceof Iterator 
            || $var instanceof IteratorAggregate
            );
}

$values = get_values();

if (is_iterable($values))
{
    foreach ($values as $value)
    {
        // do stuff...
    }
}

Note that this particular version is not tested, its typed directly into SO from memory.

Edit: added Traversable check

Answer 2

There seems also to be a relation to the environment:

I had that "invalid argument supplied foreach()" error only in the dev environment, but not in prod (I am working on the server, not localhost).

Despite the error a var_dump indicated that the array was well there (in both cases app and dev).

The if (is_array($array)) around the foreach ($array as $subarray) solved the problem.

Sorry, that I cannot explain the cause, but as it took me a while to figure a solution I thought of better sharing this as an observation.

Answer 3

$values = get_values();

foreach ((array) $values as $value){
  ...
}

Problem is always null and Casting is in fact the cleaning solution.

Answer 4

First of all, every variable must be initialized. Always.
Casting is not an option.
if get_values(); can return different type variable, this value must be checked, of course.

Answer 5

More concise extension of @Kris's code

function secure_iterable($var)
{
    return is_iterable($var) ? $var : array();
}

foreach (secure_iterable($values) as $value)
{
     //do stuff...
}

especially for using inside template code

<?php foreach (secure_iterable($values) as $value): ?>
    ...
<?php endforeach; ?>

Answer 6

Warning invalid argument supplied for foreach() display tweets. go to /wp-content/plugins/display-tweets-php. Then insert this code on line number 591, It will run perfectly.

if (is_array($tweets)) {
    foreach ($tweets as $tweet) 
    {
        ...
    }
}