showing images

[stuworx]

below is the code i'm using to view records in my database. when i enter the info into the db i put the address of the image, what do i have to do to this code to make the image show rather then the address of it?

<!-- output the column headers -->
<FORM METHOD="POST" ACTION="<?php echo("addToBasket.php3"); ?>">
<TABLE border = 1> <TR>
<TD ALIGN="center"> <em>Product</TD>
<TD ALIGN="center"> <em>Description</TD>
<TD ALIGN="center"> <em>Price</TD>
<TD ALIGN="center"> <em>URL</TD>
<TD ALIGN="center"> <em>Image</TD>
<TD ALIGN="center"> <em>Qty ordered</TD>
</em></TR>
<?php
/ now work through the products file, creating a table with a
data entry field to contain the quantity ordered /
$dbcnx = myConnect();
$query = "SELECT PROD_ID, PROD_NAME, PROD_PRICE, PROD_URL, PROD_IMAGE FROM $table_products";
$result=mysql_query($query);
if (! $result) printf ("Error: %s<br>", mysql_error ());
for ($i = 0; $i < mysql_num_rows ($result); $i++) {
$row = mysql_fetch_row($result);
echo("<tr>");
for($j = 0 ; $j<mysql_num_fields($result); $j++) {
echo("<td>");
if($j==3) echo "<a href=\"".$row[$j]."\" target=\"_blank\">".$row[$j]."</a>";
else echo($row[$j]);

		echo("</td>");
	};
	echo("<TD><INPUT TYPE=TEXT NAME=QUANTITY[");
	echo($row[0]);
	echo("] SIZE=3></tr>");
};

?>
</TABLE>
<INPUT TYPE=SUBMIT VALUE="Submit Form">
<INPUT TYPE=RESET VALUE="Reset Form">

</FORM>
<?php };

[richfm]

<IMG alt="something in your basket" src="the address goes here">

[stuworx]

yes but i need to take the information in the image field which will be the address of the image.

above only makes sence to me, i'll try explain again.

in the table i have a field that stores the address of the image i want. they will be many rows in the table and each image will be different. in the code above i need to add something that takes the relevant image address and display the relevant image when i view the table. it works in a similar way as the url field, i just cant get it to work.

think this is a better explaination.

[der_metzger]

if you wand to know where the image is...
ask the one who has uploaded the image...

if you want to show an "PROD" use something like:

<?php 
$query = "SELECT PROD_ID, PROD_NAME, PROD_PRICE, PROD_URL, PROD_IMAGE FROM $table_products WHERE PROD_ID='$id'"; 
$result=mysql_query($query); 
$row = mysql_fetch_array($result); 
?>
<table>
 <tr>
  <td>
   NUMBER!
  </td>
  <td>
   LINK 2 PRODUCT
  </td>
  <td>
   IT COSTS
  </td>
  <td>
   IMAGE!
  </td>
 </tr>
 <tr>
  <td>
   <?php echo $row['PROD_ID']; ?>
  </td>
  <td>
   <a href="<?php echo $row['PROD_URL']; ?>">
    <?php echo $row['PROD_NAME']; ?>
   </a>
  </td>
  <td>
   <?php echo $row['PROD_PRICE']; ?>
  </td>
  <td>
   <a href="<?php echo $row['PROD_URL']; ?>">
    <img src="<?php echo $row['PROD_PRICE']; ?>" alt="<?php echo $row['PROD_NAME']; ?>">
   </a>
  </td>
 </tr>
</table>