Fatal error with fetch_array(mysqli_both)

Question

I am receiving a "Fatal error" once the function fetch_array(mysqli_both) is invoked. The file below on the while instance is exactly where the error is coming from.

    <?php
include 'connect.php';
$id = $_SESSION['user_id'];
$res = mysqli_query($mysqli, "SELECT a.name,j.link,j.points, a.submit_time from journal j, article a where a.journal_id = j.id and a.professor_id = $id");
?>
<table>
    <tr>
        <th>Article</th>
        <th>Journal</th>
        <th>Points</th>
        <th>Time</th>
    </tr>
<?php
while ($row = $res->fetch_array(MYSQLI_BOTH)) {
?>
    <tr>
        <td><?php echo $row[0] ?></td>
        <td><?php echo $row[1] ?></td>
        <td><?php echo $row[2] ?></td>
        <td><?php $date = date_create($row[3]);
echo date_format($date, 'Y-m-d H:i:s'); ?></td>
    <tr>
<?php   
}
?>
</table>

When i try it on my wampserver i don't have this problem as i activated couple of PHP extensions. Online, i don't think it's easy to activate such extensions. Do you think this error is related to the fact that i didn't invoked the $query as a variable with $mysqli in line 4?

What $_SESSION['user_id'] contains? ... You should post the error info. — Cyrus
– Cyrus, Commented Feb 7, 2015 at 13:05

LUXS · Accepted Answer · 2015-02-07 12:58:55Z

1

First of all you should use require_once('connect.php'); because it is a needed file for this code.

Also you have to check your $mysqli variable if it's a correct connection to your database.

For more help you might have to post the full Error-Message

answered Feb 7, 2015 at 12:58

LUXS

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3 Comments

Leb_Broth Over a year ago

Fatal error: Call to a member function fetch_array() on a non-object in /homepages/journaltable.php on line 16

LUXS Over a year ago

then you might have an error in your SQL-Syntax. Try to run it on phpmyadmin.

LUXS Over a year ago

I think it has to be journal AS j and article AS a

Alex · Accepted Answer · 2015-02-07 13:16:18Z

0

you should check result before fetch:

if ($res) {
while ($row = $res->fetch_array(MYSQLI_BOTH)) {
...
...
}
} else {
echo printf("Errormessage: %s\n", mysqli_error($mysqli));
}

answered Feb 7, 2015 at 13:16

Alex

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