Php Links + images form Mysql db.

asgsoft · 26-04-2005, 18:45

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.

26-04-2005, 18:45	#1 (permalink)
asgsoft WMT Addict Join Date: Apr 2005 Location: Brighton,UK Posts: 162 iTrader: 0 / 0%	Php Links + images form Mysql db. 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.

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Php Links + images form Mysql db.

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