Display DB of sites

[shyish]

Ok I have a MySQL database of all the websites and templates I make to put on a portfolio site.

I was wondering how to make a database so that each website design has a seperate link, without me having to make a page for each one... so it would display as portfolio.php?0001 or whatever?

 

Currently the database just contains ID, thumbnail, large preview, demo link, colours, information and what it's made in (flash, HTML CSS etc.)

 

My PHP is really rusty, so Im not sure what to do to achieve this...

 

Currently my intention first is to have a list of all the sites displaying their thumbnail, ID, what used, colours in a grid-style-list and to try and add a search feature (not sure how to do that either yet)

 

 

Thanks in advance

~Shyish.

[DEVILofDARKNESS]

Is this what you mean?

 

$id = $_GET['id'];
switch($id){
case 1:
    //Show detailed info about site with id = 1;
break;
case 2:
    //Show detailed info about site with id = 2;
break;
...
default:
  $query = "SELECT * FROM sites";
  $result = mysql_query($query);
  while($sites = mysql_fetch_array($result)){
     echo "<a href='./sites.php?id='" . $sites['id'] . "'>" . $sites['thumbnail'] . "</a>";
  }
break;
}

[shyish]

...

cough...

 

Umm possibly :L

 

Pretty much I just want it so I dont need to make a new page for every website I add, because with that I might aswell not use PHP..

[DEVILofDARKNESS]

Okay I see,

 

Do you know what this(the code I wrote above) does?

 

(if not, you really should learn the basics of php,)

[shyish]

I learn by example.

And I kind of do, I know what parts of it mean/do etc. but other parts Im yet to use.

[shyish]

damn the lack of an edit button I can see...

 

The main issue is I have no idea where I should be putting that code to use it...

 

 

[shyish]

Bumping so it doesn't fall onto the next page...

 

Ok well I've got the list done and it works great with all the JS etc, I just need the pages done now really, so if someone can help me I would love them for ever and give them 50% of the first profit I make

 

thanks

[peter_anderson]

You could always use:

 

<?php

//load config
require_once("config.php");

//get ID
$site = $_GET['siteid'];

//connect to DB
mysql_connect($host, $user, $pass) or die(mysql_error());
mysql_select_db($db) or die(mysql_error());
//load
$result = mysql_query("SELECT * FROM portfolio WHERE id=$site");  

if(mysql_num_rows($query) ==0) 
{
header("Location: /page/404"); 
exit; 
}

//code to display content
?>

[shyish]

Thanks but I've tried simply using these scripts but they aren't really helping me =(

 

Please if someone can actually explain to me what I need to do, I will be so happy.

[shyish]

Ok I've tried flicking through numerous PHP guides and googling and asking on other sites still nothing...

 

Can someone help

 

Also

bumping before this falls into the pit of PAGE 2!

[DEVILofDARKNESS]

I would suggest you learn

the W3Schools Tutorial

[peter_anderson]

Thanks but I've tried simply using these scripts but they aren't really helping me =(

 

Please if someone can actually explain to me what I need to do, I will be so happy.

 

Here's a little more for mine.

 

For  the DB, you need:

CREATE TABLE sites(
id INT NOT NULL AUTO_INCREMENT, 
PRIMARY KEY(id),
`title` VARCHAR(30), 
`link` VARCHAR(30),
`description` VARCHAR(3000))

 

For the display page, you need:

<?php

//DB info
$host = 'localhost';
$user = 'user';
$pass = 'pass';
$db = 'db';

//get ID
$site = $_GET['siteid'];

//connect to DB
mysql_connect($host, $user, $pass) or die(mysql_error());
mysql_select_db($db) or die(mysql_error());
//load
$query = mysql_query("SELECT * FROM portfolio WHERE id=$site");  

//if there is no such site, 
if(mysql_num_rows($query) ==0) 
{
//go to 404
header("Location: /page/404"); 
exit; 
}

//for content
$row = mysql_fetch_array($query);
//display the site name
echo '<p>Site Name: '.$row['title'].' </p>';
//display the link
echo '<a href="'.$row['link'].'" target="_blank">Click here to visit the site</a></p>';
//display the description
echo '<p>'.$row['description'].'</p>';
?>