我想知道是否有人可以帮助我 - 我已经成功创建了一个登录系统,允许用户(学生)登录。我的系统还需要管理员登录,管理员有权查看学生没有的页面。管理员和学生信息都来自两个不同的表格。以下是我用于学生登录的代码(有两个不同的页面 - 用户和登录)。我坚持如何实施管理登录。任何帮助表示赞赏! (管理员将登录使用 'adminnum' 和 'ADMINPASSWORD'。管理员和用户从两个不同的表登录
的login.php
<?php
include "core/init.php";
include "includes/content.php";
if (empty($_POST) === false) {
$studentemail = $_POST ['studentemail'];
$studentpassword = $_POST ['studentpassword'];
if (empty($studentemail) === true || empty($studentpassword) === true) {
$errors[] = "You need to enter an email address and password";
} else if (user_exists($studentemail) === false) {
$errors[] = "We can't find that email address. Have you registered?";
} else {
if (strlen($studentpassword) > 32) {
$errors[] = 'Password too long';
}
$login = login($studentemail, $studentpassword);
if ($login === false) {
$errors[] = 'That email/password combination is incorrect';
} else {
$_SESSION['studentid'] = $login;
header('Location: index.php');
exit();
}
}
} else {
$errors[] = 'No data received';
}
include "includes/overall/overall_header.php";
if (empty($errors) === false) {
?>
<h2> We tried to log you in, but...</h2>
<?php
echo output_errors($errors);
}
?>
<center><input id="submit" type="submit" value="Back" onclick="location.href='Login2.php'"></center>
<?php
include "includes/overall/overall_footerloggedout.php";
?>
users.php
<?php
function logged_in() {
return (isset($_SESSION['studentid'])) ? true : false;
}
function user_exists($studentemail) {
$studentemail = sanitize($studentemail);
$query = mysql_query("SELECT COUNT(`studentid`) FROM `student` WHERE `studentemail`
= '$studentemail'");
return (mysql_result($query, 0) == 1) ? true : false;
}
function studentid_from_student ($studentemail) {
$studentemail = sanitize($studentemail);
return mysql_result(mysql_query("SELECT `studentid` FROM `student` WHERE `studentemail` = '$studentemail'"), 0, 'studentid');
}
`function login($studentemail, $studentpassword) {
$studentid = studentid_from_student($studentemail);
$studentemail = sanitize($studentemail);
$studentpassword = md5($studentpassword);
return (mysql_result(mysql_query("SELECT COUNT(`studentid`) FROM `student` WHERE `studentemail` = '$studentemail' AND `studentpassword` = '$studentpassword'"), 0) == 1) ? $studentid : false;
}
?>
为什么将列标记为“flag”而不是像“is_admin”那样更有用?名字不好的领域和写得不好的代码一样糟糕。 – TheOx 2013-04-04 13:45:28
is_admin名称过于严格,对于更多类型的用户不灵活。我认为'flag'具有普遍性,但user_type可能更好。因为在进一步使用中,您可能有10种类型的用户,并且只有其中一个用户是admin。为什么列必须是is_admin呢? – 2013-04-04 13:51:11
我想我需要两张桌子?一个给学生,一个给管理员。我更喜欢有两个独立的表格,因为它们都有不同的列并存储不同的数据。当添加新学生时,他们被添加到学生表中,并给出类型'0',并且当添加新管理员时,他们被添加到管理表并且给出类型'1'。我只是想知道如何使用存储在指定表中的值作为学生或管理员登录。我需要一个连接子句的SQL查询还是有一个更简单的方法? – user2244961 2013-04-04 14:07:50