php 用户注册并且设置为己登录状态实现方法

php教程 用户注册并且设置为己登录状态实现方法,下面实例讲述了如何把表单提交的数据保存到mysql教程数据库教程，而没有实现用户注册后自动登录的功能，而实例二就实现了这种做法。

$self = $_SERVER['PHP_SELF'];
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$username = $_POST['username'];
$password = $_POST['password'];

if( (!$firstname) or (!$lastname) or (!$username) or (!$password) )
{
  $form ="Please enter all new user details...";
  $form.="

  $form.=" method="post">First Name: ";
  $form.="   $form.=" value="$firstname">
Last Name: ";
  $form.="   $form.=" value="$lastname">
User Name: ";
  $form.="   $form.=" value="$username">
Password:   ";
  $form.="   $form.=" value="$password">
";
  $form.="";
  $form.="

";
  echo($form);
}
else
{
  $conn = @mysql_connect("localhost","root", "") or die("Could not connect to MySQL");
  $db = @mysql_select_db("my_database",$conn) or die("Could not select database");
  $sql = "insert into users (first_name,last_name,user_name,password)values ("$firstname","$lastname","$username",password("$password") )";
  $result = @mysql_query($sql,$conn)or die("Could not execute query");
  if($result){
    echo("New user $username added");
  }
}
?>

下面个实例更详细，用户注册后并且设置用户的为登录状态，本实现利用了setcookie来保存用户登录信息

create table user_info (
user_id char(18),
fname char(15),
email char(35));

//File: index.php

$form = "

Your first name?:



Your email?:

";
if ((! isset ($seenform)) && (! isset ($userid))) :
     print $form;
elseif (isset ($seenform) && (! isset ($userid))) :
     $uniq_id = uniqid(rand());
     @mysql_pconnect("localhost", "root", "") or die("Could not connect to MySQL server!");
     @mysql_select_db("user") or die("Could not select user database!");
     $query = "INSERT INTO user_info VALUES('$uniq_id', '$fname', '$email')";
     $result = mysql_query($query) or die("Could not insert user information!");
     setcookie ("userid", $uniq_id, time()+2592000);

     print "Congratulations $fname! You are now registered!.";
elseif (isset($userid)) :
     @mysql_pconnect("localhost", "root", "") or die("Could not connect to MySQL server!");
     @mysql_select_db("user") or die("Could not select user database!");
     $query = "SELECT * FROM user_info WHERE user_id = '$userid'";
     $result = mysql_query($query) or die("Could not extract user information!");

     $row = mysql_fetch_array($result);
     print "Hi ".$row["fname"].",
";
     print "Your email address is ".$row["email"];

endif;

?>