php - 基于 session 的登录无法正常工作

Question

我目前正在开发一个网站，并且正在制作一个登录系统。我正在使用 session 来跟踪登录状态。目前涉及到登录的页面如下所示: (我更改在 url 中定义帖子变量的页面，如下所示:site.domain.com/?home)

index.php

/* CONNECTION TO CHECK LOGIN STATUS */
<?php
if (isset($_GET["sair"])){
    session_start();
    session_destroy();
}
if (isset($_SESSION["user"]) && isset($_GET["entrar"])){
    mysql_connect("localhost","dbusr","password") or die("Can't connect to DB");
    mysql_select_db("mydb") or die("Can't select DB");
    $userinfo = $_SESSION["uinfo"];
}
?> /* START OF THE PAGE, RANDOM UNIMPORTANT HTML */
<?php if (!isset($_SESSION["user"])) { ?> /*NOT LOGGED IN...*/
<form id="loginform" method="post" action="scripts/checklogin.php">
<h2>Login</h2>
<?php if (isset($_GET["falha"])) { echo "<span class='erro'>Nome ou senha incorretos</span>"; }?> /*IF LOGIN ERROR*/
<input type="text" name="user" autofocus placeholder="Apelido"/><br />
<input type="password" name="pass" placeholder="Senha"/><br /> /*LOGIN FORM*/
<input type="submit" value="Entrar"/>
</form>
<?php
} else { /*LOGGED IN...*/
if ($userinfo["sexo"]=="0"){ /*GENDER*/
echo "Bem-vindo, ".$userinfo["nome"];
} else {
echo "Bem-vinda, ".$userinfo["nome"];
}?>
<?php }?>

脚本/checklogin.php

<?php

mysql_connect("localhost","dbusr","password") or die("Can't connect to DB");
mysql_select_db("mydb") or die("Can't select DB");
$user = mysql_real_escape_string(stripslashes($_POST["user"]));
$pass = md5(mysql_real_escape_string(stripslashes($_POST["pass"])));
$result = mysql_query("SELECT * FROM users WHERE apelido = \"".$user."\" AND senha = \"".$pass."\"");
if (count($result)==1) {
    session_start();
    $_SESSION["user"] = $user;
    $_SESSION["pass"] = $pass;
    while ($row = mysql_fetch_assoc($result)) {
        $_SESSION["uinfo"] = $row;
    }
    header("location:../?entrar");
} else {
    header("location:../?falha");
}
?>

但是当我进入页面并输入我的信息时，它仍然不显示已登录的部分。另外，当我输入错误的登录名时，它不会在登录表单上显示文本(<span>)

Answer 1

在检查$_SESSION["user"]之前，您需要在index.php中添加session_start()。