php - 将 MySQL 查询结果传递到 PHP session 登录

Question

我正在尝试修复 friend 的旧项目的登录系​​统，但我似乎无法让它接受合法的登录，而不是从数据库传递用户名和 ID，它只是即使它可以从数据库获取正确的用户名和 ID，也会引发错误。知道是否是这部分代码导致了错误以及解决方案是什么？

MySQL 查询和 PHP session

<?php
$row = mysqli_fetch_assoc(mysqli_query($con,'SELECT id,usr FROM tz_members WHERE usr="{.$_POST[username]}" AND pass="{.$_POST[password]}"'));
if($row['usr'])
{
    $_SESSION['usr'] = $row['usr'];
    $_SESSION['id'] = $row['id'];
    $_SESSION['rememberMe'] = $_POST['rememberMe'];
}
else {
    //(display error message)
}
?>

登录表单

<form>
    <label class="grey" for="username">Username:</label>
    <input class="field" type="text" name="username" id="username" value="" size="23" />
    <label class="grey" for="password">Password:</label>
    <input class="field" type="password" name="password" id="password" size="23" />
    <label><input name="rememberMe" id="rememberMe" type="checkbox" checked="checked" value="1" /> &nbsp;Remember me</label>
    <div class="clear">
    </div>
    <input type="submit" name="submit" value="Login" class="bt_login" />
</form>

完整代码

http://pastebin.com/kj6i9Y8h

Answer 1

我希望这有帮助:)

$row = mysqli_fetch_assoc(mysqli_query($con,'SELECT "id", "usr" FROM "tz_members" WHERE usr="'.$_POST['username'].'" AND pass="'.$_POST['password'].'"'));