php 用户注册并且设置为己登录状态实现方法_PHP教程

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;

?>

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