Question

I am programming a game in PHP and have the following code to connect to a database

//$sqldb=mysql_connect('godaddy.hostedresource.com', 'godaddyUserName', 'godaddyPassword') OR die ('Unable to connect to database! Please try again later.');
$sqldb=mysql_connect('localhost', 'root', 'mypassword') OR die ('Unable to connect to database! Please try again later.');

The trick here is that if I am on the production server I comment out the godaddy database; when I upload the code to the server I then comment out the localhost code instead.

Unfortunately the ineveitable has happened and I uploaded the code with the wrong connection commented out; this led to 24 hours of locked out customers! 🙁

Is there a way to have the code to tell if it is on the localhost server, and if it isn’t it then looks for the godaddy connection?

Answer 1

you can try this to identify if its on live or localhost

if($_SERVER["SERVER_NAME"] == "localhost" 
    && 
   $_SERVER["SERVER_ADDR"] == "127.0.0.1"){
  // in localhost
  $hostname = "localhost";
  $username = "localuser";
  $password = "localpassword";
}else{
  // not in localhost
 $hostname = "livehost";
 $username = "liveuser";
 $password = "livepassword";
}

and fail if couldn’t connect to database but save the error into a file.

if(!mysql_connect($hostname,$username,$password)){
 file_put_contents("mysql_connect.error",mysql_error(),FILE_APPEND);
 die("Couldn't connect to database");
}

a suggestion, try not to use mysql_* anymore, switch to PDO or mysqli ..