PHP Syntax Error

From the following code I am trying to grab the URL in my mysql database. Whats wrong with my code? It grabs the "logo" but how do I turn that into an iamge so it appears on my page?

PHP Code:

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<?php
$sql = mysql_query("SELECT * FROM `ALEast` ORDER BY `id` ASC");

while( $row = mysql_fetch_assoc($sql) ) { 
     echo '<tr class="A"> 
    <td>' . $row['logo'] . '</td>
    <td>' . $row['W'] . '</td>
    <td>' . $row['L'] . '</td> 
    <td>' . $row['PCT'] . '</td> 
    <td>' . $row['GB'] . '</td> 
  </tr>'; 
}
?>

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You need to use a separate page which queries only the logo form the database and outputs it with the appropriate Content-Type, i.e image/gif. Also, there is no need to use echo to produce HTML:

PHP Code:

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<?php while( $row = mysql_fetch_assoc($sql)): ?>
    <tr class="A">
        <td><img src="get_logo.php?id=<?php echo($row['id']) ?>" alt="Logo" ></td>
        <td><?php echo($row['W']) ?></td>
        <td><?php echo($row['L']) ?></td>
        <td><?php echo($row['PCT']) ?></td>
        <td><?php echo($row['GB']) ?></td>
    </tr>
<?php endwhile; ?>

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Quote:

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Originally Posted by rgyankees23

From the following code I am trying to grab the URL in my mysql database. Whats wrong with my code? It grabs the "logo" but how do I turn that into an iamge so it appears on my page?

PHP Code:

---

<?php
$sql = mysql_query("SELECT * FROM `ALEast` ORDER BY `id` ASC");

while( $row = mysql_fetch_assoc($sql) ) { 
     echo '<tr class="A"> 
    <td>' . $row['logo'] . '</td>
    <td>' . $row['W'] . '</td>
    <td>' . $row['L'] . '</td> 
    <td>' . $row['PCT'] . '</td> 
    <td>' . $row['GB'] . '</td> 
  </tr>'; 
}
?>

---

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Your syntax error is cause by the presence of array variables in your string. you need to enclose these in curly brackets:

PHP Code:

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echo "Test {$array['index']}"; 


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http://robbyzworld.com/personal/Yank...tstandings.php

Where you see the link, its suppose to be that image.