Question

我有一张桌子，它有更多列：

id | username | picture

图片是varchar类型，它保留用户图片的路径用户名是varchar并且是唯一的

 <?php
session_start();

if (isset($_POST["logIn"])) {
    $connection = new mysqli(...);

    $email = $connection->real_escape_string($_POST["email"]);
    $password = sha1($connection->real_escape_string($_POST["password"]));
    $data = $connection->query("SELECT * FROM users WHERE email='$email' AND password='$password'");

    if ($data->num_rows > 0) {
        $_SESSION["email"] = $email;
        $_SESSION["loggedIn"] = 1;
        $picture = $connection->query("SELECT picture FROM users WHERE email='$email' AND password='$password'");
        $_SESSION["picture"]=$picture;
        header("Location:../index.php");
        exit();

    }

所以我想做的是，创建一个$_SESSION["picture"]变量，该变量将保留用户名图片的值，我的意思是试图立即登录的用户的图片

Answer 1

您必须调用提取函数才能从表中获取数据。

$data = $connection->query("SELECT picture FROM users WHERE email='$email' AND password='$password'");
if ($data->num_rows > 0) {
    $row = $data->fetch_assoc();
    $_SESSION["email"] = $email;
    $_SESSION["loggedIn"] = 1;
    $_SESSION["picture"]=$row['picture'];
    header("Location:../index.php");
    exit();
}

创建当前会话变量不起作用

1 个答案: