Hi Guys,

I need to create a random image to be generated on a home page I am doing.

Basically I need to put three random images in three seperate boxes on the homepage.

These images need to be based on a category id.

1. Bi-Folding doors
2. Windows and Doors
3. Conservatories.

The image (image1) is in a table called portfolio, the cat_id is in the portfolio table and linked with portfolio_cat table.

I need the code for the following

get image1 from portfolio table where its cat_id is = 1,2 or 3 and limit that to just one image to view.

I have this piece of code which generates random images I just don't know how to change it to do what I need.

Here is the code


<?PHP
mysql_select_db("$database", $dbh);
$sql = "SELECT * FROM portfolio ORDER BY ????????? RAND() LIMIT 1";
$query = mysql_query($sql, $dbh) or die ("Error in query: $sql" . mysql_error());


//Now we loop through each record returned.
while ($row = mysql_fetch_array($query))
{

//get values from database
$id = $row['??????']; //usually database record id

//now generate using above values - note html <option>
echo $row['??????'];

}//end of loop



?>


Can anybody help with this?

I have a feeling the echo statement might need to change also so if anybody could do the code for that then that would be great.


Cheers

Dan

$sql = "select * from portfolio where cat_id = 1 order by RAND() limit 0,1";

Thanks for your reply.


But that query does not include where i am pulling the image from


would it not be



$sql = "SELECT image1 FROM portfolio WHERE cat_id=1 ORDER BY RAND() limit 1";



So if the above query is correct...how would I echo it into my html???

I'm assuming your holding the IMAGE PATH, not the actual image in the database.


$sql = "SELECT image1 FROM portfolio WHERE cat_id=1 ORDER BY RAND() limit 1";
$result = mysql_query($sql) or die('Could not look up user data; '.mysql_error());
$db_info = mysql_fetch_assoc($result);
echo "<img src=\"".$db_info['image1']."\">";

Yeah

There is an image folder on the server.


The database just links the path.

Thanks for the help.

Hi guys


thanks for all your help.

Here is the final code that works great.



<?php

mysql_select_db("$database", $dbh);

$sql = "SELECT image1 FROM portfolio WHERE cat_id=1 ORDER BY RAND() Limit 1";

$query = mysql_query($sql, $dbh) or die ("Error in query: $sql" . mysql_error());

while($row = mysql_fetch_array($query))

{

echo "<img src=\"images/".$row['image1']."\">";

} // End of while loop
?>