Originally Posted by ryan_vietnow

Try to make this code as a function.
I'll give you an example:

Help with Code Tags

php Syntax (Toggle Plain Text)

<?php
include('dbinfo.inc.php');
include('../functions.php');
mysql_connect("localhost","username","password")
or die("Failure to communicate");
 
mysql_select_db("db")
or die("Could not connect to Database");
 
function displayimage($id)
{
$query = "SELECT content,size,type FROM upload WHERE id='$id';";//$id is the id of your pic or whatever...
$result=mysql_query($query);
mysql_close();
 
$content=mysql_result($result,0,"content");
$size=mysql_result($result,0,"size");
$type=mysql_result($result,0,"type");
 
header('Content-Type: '.$type);
imagejpeg($content); //I have changed "print $content;" to this one...
}
$one=1;
$two=2;
$three=3;
 
//just enter in the parameter of your function whatever id you want..
$one1=displayimage($one);
$two2=displayimage($two);
$three3=displayimage($three);
 
echo '<img src="'.$one1.'">'; 
echo '<img src="'.$two2.'">';
echo '<img src="'.$three3.'">';
 
?>

<?php include('dbinfo.inc.php'); include('../functions.php'); mysql_connect("localhost","username","password") or die("Failure to communicate"); mysql_select_db("db") or die("Could not connect to Database"); function displayimage($id) { $query = "SELECT content,size,type FROM upload WHERE id='$id';";//$id is the id of your pic or whatever... $result=mysql_query($query); mysql_close(); $content=mysql_result($result,0,"content"); $size=mysql_result($result,0,"size"); $type=mysql_result($result,0,"type"); header('Content-Type: '.$type); imagejpeg($content); //I have changed "print $content;" to this one... } $one=1; $two=2; $three=3; //just enter in the parameter of your function whatever id you want.. $one1=displayimage($one); $two2=displayimage($two); $three3=displayimage($three); echo '<img src="'.$one1.'">'; echo '<img src="'.$two2.'">'; echo '<img src="'.$three3.'">'; ?>

I think that's it.
3,2,1...ZERO.