Question

——-EDIT——-
hi guys, seeing that you solved this problem for me, i thought it would be a good idea to solve the same problem again but on a different page. i cannot get the image to show up.

    <?php
$id = $_GET['product_id'];
$query = mysql_query("SELECT * FROM products WHERE serial = '$id'")
or die(mysql_error());  

while($info = mysql_fetch_array($query)) {
echo "";

$name = $info['name'];
$description = $info['description'];
$price = $info['price'];
$picture = $info['picture'];

}
?>

<form action="editsuccess.php?product_id=<?php echo $id; ?>" method="post">

Product ID:<br/>
<input type="text" value="<?php echo $id;?>" name="product_id" disabled/>

<br/>

Name:<br/>
<input type="text" value="<?php echo $name;?>" name="name"/>

<br/>

Description:<br/>
<input type="text" value="<?php echo $description;?>" name="description"/>

<br/>

Price:<br/>
<input type="text" value="<?php echo $price;?>" name="price"/>

<br/>

Picture:<br/>
<? echo'<img src="../getImage.php?id=' . $info['serial'] .'"/>'?> 

</br>

<input type="submit" value="Update Product"/>

</form>

This is a page where an admin can edit a product from a row in a table.
The image is not showing up for some reason.

Thanks for any suggestions.

——EDIT ENDS HERE——–

I still cannot get my PHP image to show up even after following the right method in uploading an image to the database. the following code is for displaying the image:

<form name="form1">
    <input type="hidden" name="productid" />
    <input type="hidden" name="command" />
</form>

    <table border="0" cellpadding="2px" width="600px">
        <?
            $result=mysql_query("select * from products");
            while($row=mysql_fetch_array($result)){

        ?>
        <tr>
            <td><?php '<img src="getImage.php?id=' . $row['serial'] .'"/>'
?>

            </td>
            <td>    <b><a href="products.php?product_id=<?=$row['serial']?>"><?=$row['name']?></a></b><br />
                    <?=$row['description']?><br />
                    Price:<big style="color:green">
                        £<?=$row['price']?></big><br /><br />
                    <input type="button" value="Add to Cart" onclick="addtocart(<?=$row['serial']?>)" />
            </td>
        </tr>
        <tr><td colspan="2"><hr size="1" /></td>
        <? } ?>
    </table>

the getImage.php looks like this:

...
$link = mysql_connect($host, $user, $passwd);
mysql_select_db($dbName);
$query = 'SELECT picture FROM products WHERE serial="' . $_GET['id'] . '"';
$result = mysql_query($query,$link);
$row = mysql_fetch_assoc($result);
echo $row['picture'];
?>

only the name, description and price is showing up on the webpage. my MySQL table looks like this:

• serial
• name
• description
• price
• picture (blob)

Answer 1

You are not setting the correct Content-type header before echoing out the image data.

You MUST also escape the $_GET['id'] parameter.

// Escape $id
$id = mysql_real_escape_string($_GET['id']);   

$link = mysql_connect($host, $user, $passwd);
mysql_select_db($dbName);

// Use the escaped $id
$query = "SELECT picture FROM products WHERE serial='$id'";
$result = mysql_query($query,$link);

if ($result) {
  $row = mysql_fetch_assoc($result);

  // Set the Content-type
  // This assumes image/jpeg. If you have different image types,
  // you'll need logic to supply the correct MIME type
  // image/jpeg image/png image/gif, etc
  header("Content-type: image/jpeg");
  echo $row['picture'];
}
?>

In your main script, it looks like you are merely missing an echo

        <td><?php '<img src="getImage.php?id=' . $row['serial'] .'"/>'
        // Should be
        <td><?php echo '<img src="getImage.php?id=' . $row['serial'] .'"/>'
        // ------^^^^^^