simple random image .PHP

wakerider017

Hi, guys.

I am trying to make a simple .PHP code that will create a random banner in 1 location

I put all the picture/link info in a table in my MYSQL DB..

And now I just ned a simple .PHP code that will create 1 random image at single location.

Hope this makes sense!

I did something like this before except it involved using several images and was also for my vbulletin forums. I am just trying to create something for a regular .PHP website...

Well here was the old code I used:

Maybe this will help in the creation of the new code?

$sql = "SELECT * FROM `ads` ORDER BY RAND()";  

$result = mysql_query($sql) or die(mysql_error()); 
$ads = array();  

while ($row = mysql_fetch_object($result))  

{  

    $ad =& $ads[];  

    $ad['HREF'] = $row->HREF;  

    $ad['SRC'] = $row->SRC;  

    $ad['ALT'] = $row->ALT;  

}

 <tr><td> 
<a href='{$ads[0]['HREF']}'><img src='{$ads[0]['SRC']}' alt='{$ads[0]['ALT']}' border=0></a></td></tr>

n_wattam

if all your images are within your database, what about taking the number of records within the database, then by using the "rand" with value range 0 - X, X being your number of records, you then retrive the randomly selected record.

not much code needed, and straight forward enough

wakerider017

n_wattam wrote:
if all your images are within your database, what about taking the number of records within the database, then by using the "rand" with value range 0 - X, X being your number of records, you then retrive the randomly selected record.

not much code needed, and straight forward enough

sounds good... but what would the code look like... I am a big time newbie!

wakerider017

Can anyone help me create this code?

Thanks!

wakerider017

anyone?

Kardall

well you aren't that big of a noob, you just don't know the syntax's for mysql_ in php 🙂 🙂

Get to know them, at least the more common ones 🙂 Read more at

mysql_* on the docpage

$sql = "SELECT * FROM table WHERE 1";
$my_query = mysql_query($sql);
echo "This returned: " . mysql_num_rows($my_query) . " results.";