Conditional Display of Images

genista

Hi all,

I have some images stored on my server, the details of which are stored in a database. What I need to do is to display the image in a table on two conditions:

a - If there is actually an image uploaded, and...
b - It is approved by the moderator

Here is the php, and the problem I am having is displaying an image in a table only when the two criteria above are met:

echo "<br><b>Here is the Photo portfolio for this supplier:</b></br>"; 
//Images to go under here 
$query3 = "select * from supplierimages where supplierid='" . mysql_real_escape_string($id) . "'"; 

  $result = mysql_query($query3) or die("Couldn't execute query"); 
//echo $query3; 

//Set up rows of result set   

while ($row= mysql_fetch_array ($result)) { 

$image1 = $row["image1"]; 
$image1approved = $row["image1approved"]; 
$image2 = $row["image2"]; 
$image2approved = $row["image2approved"]; 
$image3 = $row["image3"]; 
$image3approved = $row["image3approved"]; 
$image4 = $row["image4"]; 
$image4approved = $row["image4approved"]; 
$image5 = $row["image5"]; 
$image5approved = $row["image5approved"]; 
} 


echo "<br><table border=\"1\" width=\"100%\" bordercolordark=\"#000000\" bordercolorlight=\"#000000\" name=\"hello\"> 
    <tr> 

<td>&nbsp;<img src=\"../Supplier/files/'.$image1.'\"</td> 
<td>&nbsp;</td> 
<tr> 
    <td>&nbsp;</td> 
    <td>&nbsp;</td> 
</tr> 
</table></br>";

Any ideas what I will need?

Thanks,

leatherback

hm.. Why do you not just store the images in a separate table? Makes future changes a lot easier (Say, you want to store other things for the images too)..

Please condier this first, as this decision will change your code here considerably, and make the puzzle you have not much easier to solve.

bgallagher

you could always try to place condtions in your query .. so that the query only returns values that exist.
ex:

in your database you would need fields for condition1 and condition2
ex:

condition 1 perhaps being the image file name
condition 2 perhaps being the approval true or false
then you could query your database like such ..to return only the values you want.
this query also assumes that if there is no image there will be no approval. so the only images you will get are ones that are approved when the approved field is set to 'true'.

$query3 = "select images from supplierimages where supplierid= $varialbe for the supplier id' AND approval = 'true' ";

this query will look through the db find images for the defined supplierid AND images that are approved.

bgallagher

ps.
that query is rough you may need to mind some of the syntax.

~ hope it helps