PHP - MYSQL - Display Pages by id from One php file - php

Question

I have a problem to do the code work...I don't know why its not working...I want to display from the database title and many other by id...The name of the file is view.php...If I want to show id 1 then I type on url view.php?id=1 or id 2 view.php?id=2...here is my code please help me out...
<?php
$getid = $_GET["id"];
$dbuser = 'user';
$userpass = 'pass';
$database = 'db';
error_reporting(0);
$con = mysql_connect("localhost",$dbuser,$userpass);
if(!$con){die('Could not connect:'.mysql_error());}
mysql_select_db($database, $con);
mysql_query("SET NAMES utf8");
date_default_timezone_set('Europe/Athens');
$query = mysql_query("SELECT id,title,description FROM player WHERE id = '$getid'");
if(mysql_num_rows($query)){
$result = mysql_fetch_assoc($query);
$id = $result['id'];
$title = $result['title'];
$description = $result['description'];
echo 'ID: '.$id.'<br />Title: '.$title.'<br />Description: '.$description;
}else{
echo 'Error';
exit;
}
?>

Answer 1

Change this:
if(mysql_num_rows($query)){
to this:
if(mysql_num_rows($query) >= 1){
keep in mind that you've got to have things in your database with the ID that you give,
or it will just go to else again.
(mysql_num_rows returns FALSE or an integer with the amount of rows, and since the if statement didnt check for an integer but for true, it always went down the ELSE path)
Source: PHP.net

Answer 2

<?php
$query = mysql_query("SELECT id,title,description FROM player WHERE id = '$getid'");
$count = mysql_num_rows($query);
if($count != 0){
while($result = mysql_fetch_array($query)){
$id = $result['id'];
$title = $result['title'];
$description = $result['description'];
echo 'ID: '.$id.'<br />Title: '.$title.'<br />Description: '.$description;
}
}else{
echo 'Error';
}
?>