MySQL and PHP display error

Question

Hello i m new to MySQL and PHP under training please tell me where i m wrong

My problem occurs in ad01 and ad03 they are echo Advertise Here but ad02 echo properly

<?php require_once('Connections/localhost.php'); ?>
<?php
mysql_select_db($database_localhost, $localhost);
$query_advtDisplay = "SELECT * FROM advt";
$advtDisplay = mysql_query($query_advtDisplay, $localhost) or die(mysql_error());
$row_advtDisplay = mysql_fetch_assoc($advtDisplay);
$totalRows_advtDisplay = mysql_num_rows($advtDisplay);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Download Links</title>
<link href="css/style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div class="wrapper">
    <div class="ad01">
        <?php if ($row_advtDisplay['advt-no']=='ad01')
            {
            echo $row_advtDisplay['advt-content'];
            }
            else{ echo "Advertise Here";}
            ?>
    </div> 
    <div class="middlebox">  
        <div class="ad02">
            <?php if ($row_advtDisplay['advt-no']=='ad02')
            {
            echo $row_advtDisplay['advt-content'];
            }
            else{ echo "Advertise Here";}
            ?>
        </div>
        <div class="linkbox">
            <p>Download Links</p>
            <ul>
                <li><a href="#">Link 01</a></li>
                <li><a href="#">Link 02</a></li>
                <li><a href="#">Link 03</a></li>
                <li><a href="#">Link 04</a></li>
                <li><a href="#">Link 05</a></li>
            </ul>
            <ul>
                <li><a href="#">Link 06</a></li>
                <li><a href="#">Link 07</a></li>
                <li><a href="#">Link 08</a></li>
                <li><a href="#">Link 09</a></li>
                <li><a href="#">Link 10</a></li>
            </ul>
            <ul>
                <li><a href="#" target="_blank">Link 11</a></li>
                <li><a href="#" target="_blank">Link 12</a></li>
                <li><a href="#" target="_blank">Link 13</a></li>
                <li><a href="#" target="_blank">Link 14</a></li>
                <li><a href="#" target="_blank">Link 15</a></li>
            </ul>
        </div>
        <div class="passwordbox">
            <p>RAR Password</p>
        </div> 
    </div>
    <div class="ad03"><?php if ($row_advtDisplay['advt-no']=='ad03')
            {
            echo $row_advtDisplay['advt-content'];
            }
            else{ echo "Advertise Here";}
            ?></div>  
    <div class="clear"></div> 
</div>
</body>
</html>
<?php
mysql_free_result($advtDisplay);
?>

only ad02 shows properly not ad01 and ad03 No Problem with Table only Coding Problem sorry for bad english

I would have to assume, then... that $row_advtDisplay['advt-no']='ad02' so ad01 and ad03 hit the else statement. — gloomy.penguin
– gloomy.penguin, Commented Oct 10, 2014 at 23:15
You are fetching only the first row from the result set. Please, read about looping throw a result set... here for example: php.net/manual/en/mysqli-result.fetch-assoc.php — Alejandro Arbiza
– Alejandro Arbiza, Commented Oct 10, 2014 at 23:15

Community · Accepted Answer · 2023-11-17 20:38:18Z

2

mysql_fetch_assoc will only return one row ever... so you need to loop through to see all your data.

$just_one_row = mysql_fetch_assoc($advtDisplay); 

// how to loop through all the rows 
while ($row = mysql_fetch_assoc($advtDisplay)) {
    echo $row["advt-no"];
    echo $row["advt-content"]; 
}

// maybe do something like... 
$mydata = array(); 
while ($row = mysql_fetch_assoc($advtDisplay)) {
    $mydata[ $row['advt-no'] ] = $row['advt-content'];
}

<!-- and then -->
<div class="ad01">
    <?php 
     if ( isset($mydata['ad01']) && !empty($mydata['ad01']) ) {
        echo $mydata['ad01'];    
     } else { echo "Advertise Here"; }  ?>
</div>

and I have to say it... you should really be using PDO or mysqli.

edited Nov 17, 2023 at 20:38

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answered Oct 10, 2014 at 23:22

gloomy.penguin

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