php Insert Query - Why Are There Two Connections to a Database? -


in script link page connects database :

include "connect.php";

connect.php

<?php error_reporting(e_error); /* allows php connect database */ // database variables $host = "myhost"; $user = "username"; $password = "password"; $dbname = "database"; // connect database $connect = mysql_connect($host, $user, $password)  or die ("could not connect server ... \n" . mysql_error ()); mysql_select_db($dbname)  or die ("could not connect database ... \n" . mysql_error ()); ? 

then in script have insert query:

include "connect.php";  $link = mysql_connect($host, $user, $password); $query = "insert mytable values ('0','".mysql_escape_string($forename)."','".mysql_escape_string($surname)."', '".mysql_escape_string($username)."', '".mysql_escape_string($password)."', '".mysql_escape_string($email)."')";          if(mysql_db_query ($dbname, $query, $link)) {             $message = "you have registered";             header("location: register.php?message=".urlencode($message));         } else {              die("query was: $query. error: ".mysql_error($link));         }     } } 

why necessary :

$link = mysql_connect($host, $user, $password); 

hasn't connection been established?

there no point in doing this, mysql_* functions assume last opened connection if none given.

however, 2 calls mysql_connect, only 1 connection made. docs:

if second call made mysql_connect() same arguments, no new link established, instead, link identifier of opened link returned. new_link parameter modifies behavior , makes mysql_connect() open new link, if mysql_connect() called before same parameters.

so default, existing connection returned.


Comments

Popular posts from this blog

PHPMotion implementation - URL based videos (Hosted on separate location) -

javascript - Using Windows Media Player as video fallback for video tag -

c# - Unity IoC Lifetime per HttpRequest for UserStore -