MySQL and php connection error - php

Question

I created a form using php, mysql and xampp server. The problem is that whatever I write in this form it shows that the "message failed to send" and even when I check my db using http://localhost/phpmyadmin/ the message is not there. Here is the code.
P.S I followed a video tutorial and it is exactly the same that made me totally lost. Please help.
The Connection code is:
<?php
$db_host = 'localhost';
$db_user= 'root';
$db_pass= 'the password';
$db_name= 'chat';
if ($connection= mysql_connect($db_host, $db_user, $db_pass)) {
echo "Connected to Database Server...<br />";
if ($database= mysql_select_db($db_name, $connection)) {
echo "Database has been selected... <br />";
} else {
echo "Database was not found. <br />";
}
} else {
echo "Unable to connect to MYSQL server.<br />";
}
?>
And the function code is:
<?php
function get_msg() {
$query = "SELECT 'Sender', 'Message' FROM 'chat' . 'chat'";
$run = mysql_query($query);
$messages= array();
while($message = mysql_fetch_assoc($run)) {
$messages[]= array ('sender' =>$message['Sender'],
'message'=>$message['Message']);
}
return $messages;
}
function send_msg($sender, $message) {
if(!empty($sender) && !empty($message)){
$sender = mysql_real_escape_string($sender);
$message = mysql_real_escape_string($message);
$query = "INSERT INTO 'chat'.'chat' VALUES (null, '{$sender}', '$message')";
if($run = mysql_query($query)) {
return true;
} else {
return false;
}
} else {
return false;
}
}
?>

Answer 1

I have found the problem in your SQL query:
$query = "INSERT INTO 'chat'.'chat' VALUES (null, '{$sender}', '$message')";
You have to specify the fields there, the SQL query is invalid. Also you have used the wrong escape characters. This works:
$query = "INSERT INTO `chat`.`chat` (`ID`, `sender`, `message`) VALUES (null, '{$sender}', '$message')";
You do not have to specify a null value if you use AUTO_INCREMENT:
$query = "INSERT INTO `chat`.`chat` (`sender`, `message`) VALUES ('{$sender}', '$message')";
And please use MySQLi instead of MySQL because it is deprecated. Furthermore, the database should not be specified twice, simple use:
$query = "INSERT INTO `chat` (`sender`, `message`) VALUES ('{$sender}', '$message')";

Answer 2

$query = "INSERT INTO `chat`.`chat` VALUES (null, '{$sender}', '$message')";
Note the ` and it's not '

Answer 3

I am a novice in PHP and MySQL so I tend to always explicitly exit or die on MySQL on errors.
<?php
$db = new mysqli('host', 'username', 'password', 'db');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
// no reason to continue, no db connection
}
$statement="SELECT * from `bufferlines` WHERE `beloeb` >'-400' AND `tekst` LIKE 'Dankort-nota SuperB%'";
if(!$res=$db->query($statement)){
printf("Error: %s\n", $db->error); //show MySQL error
echo "<br />".$statement; // show the statement that caused that error
exit("Error 4");//no reason to continue, show where in code
}
?>
This way I get it thrown in my face and cannot get any further until I have pinned down and corrected the error.
The number in exit("Error 4") is only to find the place in the code where it went wrong and thus is unique.
I know this is counter productive when you know your stuff, but for me it's an invaluable learning tool together with php.net dev.mysql.com and stackoverflow.com