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Check whether an array is empty

Asked 2011-11-30T16:03:00.577
249

I have the following code

<?php

$error = array();
$error['something'] = false;
$error['somethingelse'] = false;

if (!empty($error))
{
    echo 'Error';
}
else
{
    echo 'No errors';
}

?>

However, empty($error) still returns true, even though nothing is set.

What's not right?

php arrays
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3 Answers

72

You can also check it by doing.

if(count($array) > 0)
{
    echo 'Error';
}
else
{
    echo 'No Error';
}
answered 2011-11-30T16:06:06.953
16

PHP's built-in empty() function checks to see whether the variable is empty, null, false, or a representation of zero. It doesn't return true just because the value associated with an array entry is false, in this case the array has actual elements in it and that's all that's evaluated.

If you'd like to check whether a particular error condition is set to true in an associative array, you can use the array_keys() function to filter the keys that have their value set to true.

$set_errors = array_keys( $errors, true );

You can then use the empty() function to check whether this array is empty, simultaneously telling you whether there are errors and also which errors have occurred.

answered 2011-11-30T16:12:35.837
1

I can't replicate that (php 5.3.6):

php > $error = array();
php > $error['something'] = false;
php > $error['somethingelse'] = false;
php > var_dump(empty($error));
bool(false)

php > $error = array();
php > var_dump(empty($error));
bool(true)
php >

exactly where are you doing the empty() call that returns true?

answered 2011-11-30T16:06:29.677

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