MySQL php create individual named page - php

Question

I am learning to create my own database and so far I have "thanks to tutorials and stackoverflow" to create a register page.
Once registered, they enter username and password and it opens the upload page.
Its to upload house details
On the upload page they enter City: Price: Decription: Bedrooms: Bathrooms: Photo:
So far, once they input the data it displays on the index page, I have done pagination, re-sized the uploaded image, and all ok.
My next step is on the index page each entry for it to open to a new window displaying their data.
The thing is that makes it hard for me to find the code, is that I want the user to input a name for their details page when uploading the house details.
And for it not to end in .php or .html just www.mysite.com/the-desired-name
my code so far might be using bad code but it works fine, and its helping me achieve what I want till I have the experience to perfect it all.
my code is upload page
<form enctype="multipart/form-data" action="add.php" method="POST">
City: <input type="text" name="city"><br>
Price: <input type="text" name = "price"><br>
Decription: <input type="text" name ="description"><br>
Bedrooms: <input type="text" name="bed"><br>
Bathrooms: <input type="text" name="bath"><br>
Your desired link name MYSITE.COM/ <input type="text" name="link"><br>
Photo: <input type="file" name="photo"><br>
<input type="submit" value="Add">
</form>
And then it uploads via
<?php
//This is the directory where images will be saved
$target = "upload/";
$target = $target . basename( $_FILES['photo']['name']);
//This gets all the other information from the form
$city=$_POST['city'];
$price=$_POST['price'];
$description=$_POST['description'];
$bed=$_POST['bed'];
$bath=$_POST['bath'];
$link=$_POST['link'];
$pic=($_FILES['photo']['name']);
// Connects to your Database
mysql_connect("host", "username", "password") or die(mysql_error()) ;
mysql_select_db("mydatabase") or die(mysql_error()) ;
//Writes the information to the database
mysql_query("INSERT INTO `employees` VALUES ('$city', '$price', '$description', '$bed', '$bath', '$link', '$pic')") ;
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
require_once 'SimpleImage.php';
$image = new SimpleImage();
$image->load($target);
$image->resize(50,50);
$image->save($target);
//Tells you if its all ok
echo "<script>window.location = 'http://www.mysite.com'</script>";
}
else {
//Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
?>
and then it displays on the index page
<?php
// Connects to your Database
mysql_connect("host", "username", "password") or die(mysql_error()) ;
mysql_select_db("mydatabase") or die(mysql_error()) ;
if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; };
$start_from = ($page-1) * 2;
$data = mysql_query("SELECT * FROM employees ORDER BY bath ASC LIMIT $start_from, 2") or die(mysql_error());
//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
?>
<?php
//Outputs the image and other data
echo "<img src=http://www.mysite.com/upload/".$info['photo'] . " /><br />";
echo "<b>City:</b> ".$info['city'] . " ";
echo "<b>Price:</b> ".$info['price'] . " ";
echo "<b>Bed:</b> ".$info['bed'] . " ";
echo "<b>Bath:</b> ".$info['bath'] . " ";
echo "<b>Extra:</b> ".$info['description'] . " ";
echo "<b>Link:</b> <u>www.mysite.com/</u> ".$info['link'] . " <br /><br /></a>";
}?>
<?php
$data = mysql_query("SELECT COUNT(photo) FROM employees") or die(mysql_error());
$info = mysql_fetch_row($data);
$total_records = $info[0];
$total_pages = ceil($total_records / 2);
for ($i=1; $i<=$total_pages; $i++) {
echo "<a href='index.php?page=".$i."'>".$i."</a> ";
};
?>

Answer 1

for it not to end in .php or .html just www.mysite.com/the-desired-name
Assuming you are on Apache, have a look at mod_rewrite. You can use that to turn requests like the one you stated into internal requests to /showpage.php/the-desired-name. Inside that showpage.php script, you can access the-desired-name via the $_SERVER['PATH_INFO'] variable. Or, if that does not work, via some other variable which phpinfo() will tell you.