我必须制作一个简单的网站，你可以登录和退出，如果用户登录了，他们会看到一些功能，否则他们不会看到。我不太擅长网络开发，但我已经设法找到了一些似乎奏效的东西。我决定在登录和注销时不想将用户重定向到另一个页面，所以这让我有点难以理解。
我只是想知道我是否会以正确的方式开始和结束这场比赛，如果可能的话，是否有人能给我一些建议，让比赛变得更好。
<?php
if(isset($_POST['logout'])) {
session_destroy();
}
}
session_start();
if(!isset($_SESSION['username'])) {
if (!empty($_POST['username']) && !empty($_POST['password'])) {
$result = mysql_query("SELECT * FROM users WHERE username ='$_POST['username']' AND password = '$_POST['password']'");
if(mysql_num_rows($result))
$_SESSION['username'] = $_POST['username'];
}
else {
echo "";
}
}
}
?>
<?php if(!isset($_SESSION['username'])) {
echo '<div id = "account">
<form name="input" action="index.php" method="post">
Username:<input type="text" name="username" /> Password:<input type="password" name="password" />
<input type="submit" value="GO!" />
</form>
}
else {
echo "Signed in"
<form name='logout' action='index.php' method='post'>
<input type='submit'value='Reset' name='logout'/>
";
} ?>
<?php
$test = mysql_query("SELECT * FROM posts ORDER BY post_id DESC");
if($test) {
while($row = mysql_fetch_array($test)) {
echo '<div class="posts">';
echo "$row[post]";
echo '</div>';
}
}
我对您的代码进行了修改。我试着添加了很多评论，让它更容易理解。希望没有语法错误，但我实际上无法测试，因为我没有MySQL数据库之类的。
这是你的主要代码：
<?php
//When you are developing and testing, set the error level as high as possible.
//This will help you find problems early. A well written program will have no errors and warnings, ever.
error_reporting(E_ALL | E_STRICT);
//Starting the session should be one of the first things your code does, and should only be done once.
session_start();
require 'config.php';
if(isset($_POST['logout']))
{
//I don't think there is any reason to check if username is set. If you are logging out, just destroy.
session_destroy();
//Also unset the session username since session_destroy() does not affect existing globals.
unset($_SESSION['username']);
}
//I changed this to elseif, because there should not be a condition where you are logging out and checking for a login.
elseif(!isset($_SESSION['username']))
{
//You should not assume that variables are set, because accessing them if they are not set
//will cause a warning. I've added isset().
if(isset($_POST['username']) && !empty($_POST['username']) && isset($_POST['password']) && !empty($_POST['password']))
{
//You absolutely MUST escape your strings or you are at risk of SQL injection.
//Use mysql_real_escape_string() for this.
$username = mysql_real_escape_string($_POST['username']);
$password = mysql_real_escape_string($_POST['password']);
$result = mysql_query("SELECT * FROM members WHERE username ='$username' AND password = '$password'");
//You should probably check that the value === 1 here.
//I'm assuming it should always be 1 or 0.
if(0 === mysql_num_rows($result))
{
$_SESSION['username'] = $username;
}
else {
echo "Fail :(";
}
}
//If you put an else statement here, you could print an error for if the username was not specified.
}
//You should not have SQL queries in your template, so I moved this here.
//Notice that I'm just setting $posts to the data. It's best to just pass
//the data, and format it in the template.
$result = mysql_query("SELECT * FROM posts ORDER BY post_id DESC");
if($result)
{
$posts = array();
while($row = mysql_fetch_array($result))
{
$posts[] = $row['post'];
}
}
else
{
$posts = false;
}
//Try to separate code logic from templates.
//Your program is small, so it's not that important, but I would do it anyway.
require 'template.php';
?>
这是您的模板代码，应该放在一个名为template.php的新文件中：
<div id = "container">
<h1>#HookyGear Bay</h1>
<div id = "login">
<?php if(!isset($_SESSION['username'])) {
echo '<div id = "accountBox">
<form name="input" action="index.php" method="post">
Username:<input type="text" name="username" /> Password:<input type="password" name="password" />
<input type="submit" value="Sign In" />
</form>
</div>';
}
else {
echo "<div id='accountBox'>You Are logged in as ".$_SESSION['username']."
<form name='logout' action='index.php' method='post'>
<input type='submit'value='Reset' name='logout'/>
</div> ";
} ?>
</div>
<div id = "content">
<?php
if(false !== $posts)
{
foreach($posts as $post)
{
echo '<div class="blogPosts">'.$post.'</div>';
}
}
else { ?>
<div class="blogPosts"><?php echo "no blog posts"; ?></div>
<?php
}
?>
<div style="clear:both;"></div>
</div>
</div>
您可以将用户重定向到不同的注销特定页面，而不是每次加载页面时都检查$_POST['logout']。
您可以尝试调用session中的方法。就像我的
在这里，它会检查你是否连接到你的数据库，我将其命名为connect.inc.php
<?php
if(!mysql_connect('localhost', 'root', '')|| !mysql_select_db('byp_db'))
{
die(mysql_error());
}
?>
接下来，我创建了我的core.inc.php，它将检查你是否已经在session中，你将在中使用loggedin()方法
<?php
error_reporting(E_ALL ^ E_NOTICE);
ob_start();
session_start();
$current_file = $_SERVER['SCRIPT_NAME'];
$http_referer = $_SERVER['HTTP_REFERER'];
function loggedin() {
if(isset($_SESSION['user_p_info_id'])&&!empty($_SESSION['user_p_info_id'])) {
return true;
}else {
return false;
}
}
function getuserfield($field){
$query = "SELECT `$field` FROM `user_p_info` where `user_p_info_id`='".$_SESSION['user_p_info_id']."'";
if($query_run = mysql_query($query)){
if($query_result = mysql_result($query_run, 0, $field)){
return $query_result;
}
}
}
?>
下一步，你将创建你的日志形式
<?php
require 'connections/connect.inc.php';
require 'connections/core.inc.php';
if(isset($_POST['uname']) && isset($_POST['password'])){
$uname = $_POST['uname'];
$pword = $_POST['password'];
//echo $uname;
//echo $pword;
if(!empty($uname)&&!empty($pword)){
$query_login = "SELECT * FROM user_a_info where username = '$uname' and password = '$pword'";
//echo $query_login;
$query_result = mysql_query($query_login);
$num_rows = mysql_num_rows($query_result);
if($num_rows == 0){
?>
<script type="text/javascript">
alert("Invalid Data !");
</script>
<?php
}else{
//echo "validated";
$user_p_info_id = mysql_result($query_result, 0, 'user_p_info_id');
$_SESSION['user_p_info_id']=$user_p_info_id;
header('Location: index.php');
}
}
}
?>
<form action="login.php" method="POST">
<p> USERNAME : <input type="text" name="uname" /> </p>
<p> PASSWORD : <input type="password" name="password" /> </p>
<p> <input type="submit" value="LOGIN" /> </p>
</form>
然后你的注销功能会像这个
<?php
require 'core.inc.php';
session_destroy();
header('Location: ../index.php');
?>
只需注意，如果你想检查你是否在session中，只需将此条件设置为
<?php
require 'connections/connect.inc.php';
require 'connections/core.inc.php';
if(loggedin()) {
// Do something
}
?>
希望这能帮助