Get Results from mySQL in PHP - php

Question

I am trying to get result from as table in mySQL in the following code.
It does not work. I am newbie in PHP.
Maybe someone can help whatws wrong in my code? I think that my SQL sentence is not written well. Is it?
<?php
header("Content-type: text/html; charset=utf8");
// header('Content-Type:text/html;charset=utf-8');
//1. create connection
$connection=mysql_connect("localhost","user","pass");
if(!$connection){
die("database connection failed:" . mysql_error());
}
//2. select database
$db = mysql_select_db("database",$connection);
if(!db) {
die("database connection failed:" . mysql_error());
}
//if i want to work with hebrew databases
mysql_query("SET NAMES 'utf8'",$connection); // reading heberer from phpadmin database - only for hebrew sites
$MessageNo = $_POST['MsgNo']
//$MessageN0=(int)$MessageNo
$query = mysql_query("SELECT * FROM MessageContent WHERE MsgNo='$MessageNo'");;;
$messages = array();
while ($row = mysql_fetch_array($query)) {
$messages[] = array('MsgContId' => $row['MsgContId'], 'MsgNo' => $row['MsgNo'],'MsgContent' => $row['MsgContent'], 'AddedBy' => $row['AddedBy'],'AddedAt'=>$row['AddedAt']);
}
// echo json_encode(array('users' => $users));
echo json_encode($messages);
mysql_close($connection);
?>

Answer 1

I can see on this line you have formatting errors:
$query = mysql_query("SELECT * FROM MessageContent WHERE MsgNo='$MessageNo'");;;
Remove the extra two semi-colons:
$query = mysql_query("SELECT * FROM MessageContent WHERE MsgNo='$MessageNo'");
Also this:
$MessageNo = $_POST['MsgNo']
Needs a semi-colon:
$MessageNo = $_POST['MsgNo'];
Additionally you're using deprecated functions. I would suggest you look into MySQLi functions or PDO.
Consult: mysqli with prepared statements, or PDO with prepared statements.
EDIT: Also in your second query, you're not using the connection string:
$query = mysql_query("SELECT * FROM MessageContent WHERE MsgNo='$MessageNo'",$connection);
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// rest of your code
Sidenote: Error reporting should only be done in staging, and never production.
Also add or die(mysql_error()) to mysql_query().
Sidenote:
Your present code is open to SQL injection.