ちょっとした問題があります。専門家の 1 人が助けてくれるはずです :-p.
データベースに次のテーブルがあります。
b_admins:
id
username
password
email
locationid
b_locations
id
name
description
ログイン スクリプト:
<?php
// Inialize session
session_start();
// Connection to MySQL Database.
include ('_includes/_dbconnection.php');
include ('_includes/_dbopen.php');
// Retrieve username and password from database according to user's input
$login = mysql_query("SELECT * FROM b_admins WHERE (ausername = '" . mysql_real_escape_string($_POST['username']) . "') and (apassword = '" . mysql_real_escape_string($_POST['password']) . "')");
// Check username and password match
if (mysql_num_rows($login) == 1) {
// Set username session variable
$_SESSION['username'] = $_POST['username'];
// Set locationid session varaible //// Modified 12 April 2012.
$_SESSION['locationid'] = $row_Recordset1["locationid"];
$locationid = $_SESSION['locationid'];
// Jump to secured page
header('Location: _home.php');
}
else {
// Jump to login page
header('Location: index.php');
}
// Script Source: http://frozenade.wordpress.com/2007/11/24/how-to-create-login-page-in-php-and-mysql-with-session/
?>
私のホームページのスクリプト:
// Check, if user is already login, if login proceed to _home.php
if (isset($_SESSION['username'])) {
header('Location: _home.php');
}
// Load Header file.
include ('_includes/_header.php');
// Template Construction.
// Create Homepage Layout after $home01.
$home01 = '<div id="imglogo"></div>
<div id="container">
<div class="top">
<div id="logo">
</div>
<ul class="socialicons">
</ul>
</div>
<div id="content" >
<div id="home">
<div class="latest">';
// MySQL Load Location Name from MySQL database after $home02.
$home02 = '<h3>';
// Create Notification Area for the Administrator after $home03.
$home03 = ' Administrator Panel</h3><p>You are logged in to make changes to your Branch, only your Branch details can be changed within this Administration Panel. You may proceed to make any changes using the menu at the bottom.</p></div>
<!-- /Latest section -->
<!-- News info section -->
<ul class="news-info">
<li><label>Users Registered</label>';
//MySQL Load User count from MySQL database after $home04.
$home04 = '<span>';
// Close Template.
$home05 = '</span></li>
</ul>
<!-- /News info section -->
</div>
<!-- /Home -->
<!-- Menu -->
<div class="menu">
<ul class="tabs">
<li><a href="_home.php" class="tab-home"></a></li>
<li><a href="_events.php" class="tab-events"></a></li>
<li><a href="_gallery.php" class="tab-portfolio"></a></li>
<li><a href="_template_edit.php" class="tab-contact"></a></li>
</ul>
</div>
</div>';
////////////////////////////////////////////////////////////////////////// Combining Template with MySQL Data.
/// Build Home Page from MySQL database.
$SQL = "SELECT * FROM b_admins, b_locations WHERE locationid ='$locationid' LIMIT 0, 1";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
echo $home01;
echo $home02;
print $db_field['lcity'];
echo $home03;
}
/// Build Home Page News Section from MySQL database.
echo $home04;
$SQL = "SELECT count( * ) as total_record FROM `b_users` WHERE locationid = '$locationid'";
echo $home05;
/// Load Footer and Database close file.
include ('_includes/_footer.php');
include ('_includes/_dbclose.php');
?>
b_admins の値「locationsid」を保存して、SQL を使用して、「locationsid」がデータベース b_locations のテーブルと同じデータベースからデータを取得できるようにしたいと考えています。
また、ログイン スクリプトを使用してログインしたときに修正方法がわからないという次のエラーが表示されます。
前もって感謝します :-)