PHP warning: in_array() wrong datatype for second argument in

Question

I recently got this error on my error log:

   PHP Warning:  in_array() [<a href='function.in-array'>function.in-array</a>]: Wrong datatype for second argument in... on line 423

and refers to the following piece of code:

    <?php foreach($services_a as $key=>$service) { ?>
    <div class="oneservice">
    <input type="checkbox" name="services[]" value="<?php echo $key; ?>" id="service<?php echo $key; ?>"<?php if( in_array($key, $services) ) { echo ' checked="checked"'; } ?> />
    <label for="service<?php echo $key; ?>"><?php echo $service; ?></label>
    </div>

Any views are very welcome, Thanks

Is $services an array? You can use var_dump() to find out — knittl
– knittl, Commented Jan 22, 2012 at 19:26
It is defined as follows: $services = $_POST['services']; if ($services && is_array($services)) { foreach ($services as $i => $service) { $service = (int)$service; $array = array( 'key' => 'services', 'value' => '%"'.$service.'"%', 'compare' => 'LIKE' ); $meta_query[] = $array; unset($array); } } — user1163797
– user1163797, Commented Jan 22, 2012 at 19:28
$services_a = array( "1" => _d('Green)',263), "2" => _d('Blue',264), "3" => _d('Yellow',265), "4" => _d('Red',266), "5" => _d('Purple',267), "6" => _d('Red',268), ); — user1163797
– user1163797, Commented Jan 22, 2012 at 19:52

konsolenfreddy · Accepted Answer · 2012-01-22 19:46:26Z

4

in_array() checks for the value, not the key.

Use array_key_exists() if xou want to check for the key:

<input type="checkbox" name="services[]" value="<?php echo $key; ?>" id="service<?php echo $key; ?>"<?php if( array_key_exists($key, $services) ) { echo ' checked="checked"'; } ?> />

When you open the form the first time, your $_POST['services'] will be empty. to overcome the error, initialize an empty array if nothing is coming from post:

$services = is_array($_POST['services') && count($_POST['services']) > 0 ? $_POST['services'] : array();

edited Jan 22, 2012 at 19:46

answered Jan 22, 2012 at 19:26

konsolenfreddy

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3 Comments

user1163797 Over a year ago

PHP Warning: array_key_exists() expects parameter 2 to be array, null given in I get this error

konsolenfreddy Over a year ago

I've added some code to initialize the $_POST['services'] if it's empty.

user1163797 Over a year ago

Will this be of any help

$services_a = array( 	"1" => _d('Green)',263),  	"2" => _d('Blue',264),  	"3" => _d('Yellow',265),  	"4" => _d('Red',266),  	"5" => _d('Purple',267),  	"6" => _d('Red',268),  );

Oyeme · Accepted Answer · 2012-01-22 19:28:14Z

0

You should check "an array" Is it an array?

<?php if(is_array($services)): ?>
 <?php if( in_array($key, $services) ) { echo ' checked="checked"'; } ?>
<? endif; ?>

answered Jan 22, 2012 at 19:28

Oyeme

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1 Comment

user1163797 Over a year ago

$services_a = array( 	"1" => _d('Green)',263),  	"2" => _d('Blue',264),  	"3" => _d('Yellow',265),  	"4" => _d('Red',266),  	"5" => _d('Purple',267),  	"6" => _d('Red',268),  );

Repox · Accepted Answer · 2012-01-22 19:30:25Z

0

Probably because you called the array you are iterating for $services_a but the array you use for second argument in in_array() is called plain $services. Problem is probably that the second argument has a NULL value, which gives you the warning.

answered Jan 22, 2012 at 19:30

Repox

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