PHP 数据库从 MySQL 检索，其中 user_name 保存在 session 中

Question

我正在尝试检索附加到特定用户的所有计划的计划信息。他们已登录，并且他们的用户名保存在 session 中，我知道我需要将其与 MySQL WHERE 语句结合使用。这是我的代码:

<?php
$servername = "localhost";
$username = "root";
$password = "XXX";
$dbname = "name";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "SELECT plan_id, plan_name, plan_type, plan_active FROM plans WHERE user_name ='$_SESSION['user_name']'";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo "id: " . $row["plan_id"]. "<br>";
        echo "id: " . $row["plan_name"]. "<br>";
        echo "id: " . $row["plan_type"]. "<br>";
        echo "id: " . $row["plan_active"]. "<br>";
    }
} else {
    echo "0 results";
}

mysqli_close($conn);
?>

请注意，我在 WHERE 语句中使用了 $_SESSION['user_name']。我做错了什么？根本没有显示任何内容。

Answer 1

您没有在查询中正确传递 $_SESSION['user_name'] 变量。

试试这个:-

替换:-

$sql = "SELECT plan_id, plan_name, plan_type, plan_active FROM plans WHERE user_name ='$_SESSION['user_name']'";

带有:-

$sql = "SELECT plan_id, plan_name, plan_type, plan_active FROM plans WHERE user_name ='" . $_SESSION['user_name'] . "'";

我建议您使用准备好的语句。那么你就不会犯这些类型的错误。