在`$_SESSION`中正确存储mysql DB中的userlvl以调用它

Question

我正在检查mysql数据库中的用户级别，以便在匹配的情况下显示链接。

这是我当前的代码：

<?php

$mysqli = new mysqli('host', 'name', 'psw', 'db');
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}
if ($result = $mysqli->query("SELECT user_level FROM users WHERE user_level = 1 ")) {

    $row_cnt = $result->num_rows;

    printf("Result set has %d rows.\n", $row_cnt);
    }
    /* until here everything works fine - the code below seems to be not working */

if ($result == $_SESSION['userlevel']) {

    $link = 'AdminLayout.php';
    printf ('<a href="' .$link. '">Adminpanel </a>');
    /* close result set */
    $result->close();


}

/* close connection */
$mysqli->close();
?>

printf("Result set has %d rows.\n", $row_cnt);只是检查连接是否工作，以及它是否检测到user_level = 1。

我使用以下命令将$_SESSION['userlevel']存储在我的登录表单中：

if(isset($_POST['submitted']))
{ 
session_start();

$mysqli = new mysqli('host', 'name', 'psw', 'db');


 if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}
 else ($result = $mysqli->query("SELECT user_level FROM users WHERE user_level = 1 ")) {
 if (mysqli_num_rows($result == 1) {
$ulevel = $mysqli_fetch_array($result);
  $ulevel = $_SESSION['userlevel'];
  }
}
}
?>

不幸的是，添加了这些行之后，我的登录表单就不再有效了。

有没有人能帮我把它修好？如何正确存储已登录帐户的用户级别，以便在另一个站点上重新调用它？

长话短说:我想检查我的MySQL数据库中的用户级别，如果登录的用户与之匹配，就让链接出现。

Answer 1

请按如下所示更正您的代码。登录页面：

if(isset($_POST['submitted']))
{ 
session_start();

$mysqli = new mysqli('host', 'name', 'psw', 'db');


 if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}
 elseif ($result = $mysqli->query("SELECT user_level FROM users WHERE user_level = 1 ")) {
 if (mysqli_num_rows($result) == 1) {
$_SESSION['userlevel'] = $mysqli_fetch_array($result);
  }
}
}
?>

另一个代码是：

<?php

$mysqli = new mysqli('host', 'name', 'psw', 'db');
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}
if ($result = $mysqli->query("SELECT user_level FROM users WHERE user_level = 1 ")) {

    $row_cnt = mysqli_num_rows($result);
    $result = mysqli_fetch_array($result);

    printf("Result set has %d rows.\n", $row_cnt);
    }
    /* until here everything works fine - the code below seems to be not working */

if ($result == $_SESSION['userlevel']) {

    $link = 'AdminLayout.php';
    printf ('<a href="' .$link. '">Adminpanel </a>');
    /* close result set */
    $result->close();
}