in_array is'nt finding a value

Question

I got an array $data['permissions'] which looks like this: Array ( [0] => 2 [1] => 11 ) and my in_array function I wrote do not work. Here is the code:

$data['permissions'] = $auth_user->getPermissions();
$stmt = $auth_user->runQuery("SELECT * FROM words;");
$stmt->execute();
while($wordsRow=$stmt->fetch(PDO::FETCH_ASSOC)){
                        echo $wordsRow['subcategory_id'];?>
                    <tr>
                        <?php
                        if($wordsRow['user_id'] == $_SESSION['user_session'] ||
                        in_array(array($wordsRow['subcategory_id']), $data['permissions'])){ ?>
                            <td><?php echo $wordsRow['name'];?></td>
                            <td><?php echo $wordsRow['subcategory_id'];?></td>
                            <?php
                        }
                        ?>
                    </tr>
                    <?php
                    }

$wordsRow['user_id'] == $_SESSION['user_session'] in if works fine but in_array function in if is not finding the value that actually is in this array.

echo $wordsRow['subcategory_id'];

in while loop shows: 11 2 11. Where is the problem?

Jeff Lambert · Accepted Answer · 2016-04-21 15:24:26Z

2

The first parameter to in_array should be a value to find and you're passing it an array to find inside of a one-dimensional array which just isn't going to work.

Try changing this:

in_array(array($wordsRow['subcategory_id']), $data['permissions'])

To something like this:

in_array($wordsRow['subcategory_id'], $data['permissions'])

answered Apr 21, 2016 at 15:24

Jeff Lambert

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1 Comment

xxyoyoxx Over a year ago

Yeah. I guess I missed that becouse of copying parts of code. Just newby. Thank you very much

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