I know i was having troubles with this a while ago but as i found a way in a php book on how to solve it i thought i will share it.
lets say your db has a coloumn for websites and another for images and you want to make a list that displays all the info in the db including the website as a link and an image. well here is how to do it:
PHP Code:
<?php
$db = mysql_connect("localhost", "root", "");
mysql_select_db("db_name");
// -----------------------------------------------
$result = mysql_query("SELECT * FROM table order by 'id'")or die(mysql_error());
$num_links = mysql_num_rows($result);
for ($i=0; $i<$num_links; $i++)
{
//part 1
$row= mysql_fetch_array($result);
$id = $row["id"];
$name = $row["name"];
$email = $row["email"];
$website = $row["website"];
$description = $row["brief_description"];
$image_dir = $row["image_dir"];
//part 2
print "User Id : $id <br />";
print "Name : $name <br />";
print "E-mail : $email <br />";
print "Website : <a href=http://$website>$website</a><br />";
print "brief description : $description <br />";
print "image is <img src='$image_dir'> <br /><br />";
}
?>
As you see in part 1 we got the data from the database. this is all what we need for now.
In Part 2 we print the data out that we stored before as variables. so it will view a block containing id,name,email,website,brief discription and an image.
For the link i used:
PHP Code:
print "Website : <a href=http://$website>$website</a><br />";
what this does is it writes website then makes a link command tellink it to hyperlink to the website that we got from the variable.
the same with the image image tag with a variable for the acctual image. so it looks like this:
PHP Code:
print "image is <img src='$image_dir'> <br /><br />";
I hope that helps someone.
