Display SQL table content in HTML page using PHP

Question

I have a SQL table and I would like to display it in a HTML page. My code is displaying a table but the content is empty although my table is filled with data.

    <?php
        ini_set('display_errors', 'On');
        error_reporting(E_ALL);

        $database = 'karin';
        $connect = mysql_connect('localhost','root', 'karin');
        if (!$connect) {
            die(mysql_error());
        }
        if (!mysql_select_db($database))
            die("Can't select database");

        $results = mysql_query("SELECT * FROM service_messages");
        if (!$results) {
            die("Query to show fields from table failed");
        }


        while($row = mysql_fetch_array($results)) {
        ?>
            <tr>
                <td><?php echo $row['partner']?></td>
                <td><?php echo $row['GUID']?></td>
                <td><?php echo $row['serviceName']?></td>
                <td><?php echo $row['serviceID']?></td>
                <td><?php echo $row['countries']?></td>
                <td><?php echo $row['rate']?></td>
                <td><?php echo $row['ips']?></td>
                <td><?php echo $row['callback']?></td>
            </tr>

        <?php
        }
        ?>
        </tbody>
        </table>
</body>

Why doesn't it show the table content? Thanks!

var_dump() your $row to see if there is something in the result — Gopal Sharma
– Gopal Sharma, Commented Nov 8, 2015 at 13:40
mysql_select_db expect two(2) parameters right ? mysql_select_db('YourDB', $connection) — Jrey
– Jrey, Commented Nov 8, 2015 at 14:22
I don't think $results receives any value for some reason... $result = mysqli_query("SELECT * FROM service_messages"); — Freckles
– Freckles, Commented Nov 8, 2015 at 14:27

Akshay · Accepted Answer · 2015-11-08 13:41:30Z

1). Don't use mysql_*. Please use PDO or MySQLi.

2). I don't find any <table> (table starting tag) anywhere in your code.

3). Use mysql_fetch_assoc instead of mysql_fetch_array.

Here's the updated code.

    $database = 'karin';
    $connect = mysql_connect('localhost','root', 'karin');
    if (!$connect) {
        die(mysql_error());
    }
    if (!mysql_select_db($database))
        die("Can't select database");

    $results = mysql_query("SELECT * FROM service_messages");
    if (!$results) {
        die("Query to show fields from table failed");
    }
     ?>
    <table>
    <?php
    while($row = mysql_fetch_assoc($results)) {
    ?>
        <tr>
            <td><?php echo $row['partner']?></td>
            <td><?php echo $row['GUID']?></td>
            <td><?php echo $row['serviceName']?></td>
            <td><?php echo $row['serviceID']?></td>
            <td><?php echo $row['countries']?></td>
            <td><?php echo $row['rate']?></td>
            <td><?php echo $row['ips']?></td>
            <td><?php echo $row['callback']?></td>
        </tr>

    <?php
    }
    ?>
    </tbody>
    </table>

Manthan Dave · Accepted Answer · 2015-11-08 13:41:30Z

0

Try this

 while($row = mysql_fetch_row($results)) {
        ?>
            <tr>
                <td><?php echo $row['partner']?></td>
                <td><?php echo $row['GUID']?></td>
                <td><?php echo $row['serviceName']?></td>
                <td><?php echo $row['serviceID']?></td>
                <td><?php echo $row['countries']?></td>
                <td><?php echo $row['rate']?></td>
                <td><?php echo $row['ips']?></td>
                <td><?php echo $row['callback']?></td>
            </tr>

        <?php
        }
        ?>

For more refrence : http://php.net/manual/en/function.mysql-fetch-row.php

answered Nov 8, 2015 at 13:41

Manthan Dave

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Comments

Freckles · Accepted Answer · 2017-01-08 10:39:55Z

0

Well, for some reason it works like this:

    <?php
    //ini_set('display_errors', 'On');
    //error_reporting(E_ALL);

    $mysqli = new mysqli("localhost", "root", "2014", "cd");

    $q = $mysqli->query("select * from service_messages");

    while($row = $q->fetch_assoc()){
        $arr[] = $row;
    }
    echo json_encode($arr);



?>

edited Jan 8, 2017 at 10:39

answered Nov 8, 2015 at 14:48

Freckles

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