从一个表中检索用户 ID 并将其插入到另一个表中

Question

我正在做一个有两张桌子的项目 1.用户表| （id 为主键） 2.电影桌| （movie_id为主键，id为外键）

我想从 users 表中获取 user_id 并想在该用户登录时插入电影相关信息 我正在为此使用 session_variables 但我不知道我的代码是否正确我尝试了以下代码但没有得到任何结果。

这是我的进程.php

$errors = [];
$fav = "";
$rank = "";
$rewatched = "";
$status = "";
$recommend = "";

if (isset($_POST['submit'])) {
    if (empty($_POST['fav'])) {
        $errors['fav'] = 'Favourite required';
    }
    if (empty($_POST['rank'])) {
        $errors['rank'] = 'Rank required';
    }
    if (empty($_POST['rewatched'])) {
        $errors['rewatched'] = 'Rewatched required';
    }
    if (isset($_POST['status'])) {
        $errors['status'] = 'Status required';
    }
    if (isset($_POST['recommend'])) {
        $errors['recommend'] = 'Recommend required';
    }

    $fav = $_POST['fav'];
    $rank = $_POST['rank'];
    $rewatched = $_POST['rewatched'];
    $status = $_POST['status'];
    $recommend = $_POST['recommend'];

    // Select id from users and insert into movie
    if (count($errors) === 0) {
        $sql = "SELECT * FROM movie WHERE id = (SELECT id FROM users WHERE id = '".$_SESSION['id']."')";
        $query = "INSERT INTO movie SET fav=?, rank=?, rewatched=?, status=?, recommend=? WHERE id = '".$_SESSION['id']."'";
        $stmt = $conn->prepare($query);
        $stmt->bind_param('siiss', $fav, $rank, $rewatched, $status, $recommend);
        $result = $stmt->execute();

        if ($result) {
            $anime_id = $stmt->insert_id;
            $stmt->close();

            $_SESSION['id'] = $movie_id;
            $_SESSION['fav'] = $fav;
            $_SESSION['rank'] = $rank;
            $_SESSION['rewatched'] = $rewatched;
            $_SESSION['status'] = $status;
            $_SESSION['recommend'] = $recommend;
            $_SESSION['message'] = 'Details have been submitted successfully!';
            $_SESSION['type'] = 'alert-success';
        } else {
            $_SESSION['error_msg'] = "Database error: Could not update details";
        }
    }
}

简单来说，我想从用户表中获取 id，当该特定用户登录并填写与电影相关的表单并提交结果时，它应该与他/她的用户 id 一起存储在电影表中。

谢谢

Answer 1

<?php

// Assuming $_SESSION['id'] is the User ID of the User Who Currently Logged in.

$errors = [];
$fav = "";
$rank = "";
$rewatched = "";
$status = "";
$recommend = "";

if (isset($_POST['submit'])) {
    if (empty($_POST['fav'])) {
        $errors['fav'] = 'Favourite required';
    }
    if (empty($_POST['rank'])) {
        $errors['rank'] = 'Rank required';
    }
    if (empty($_POST['rewatched'])) {
        $errors['rewatched'] = 'Rewatched required';
    }
    if (isset($_POST['status'])) {
        $errors['status'] = 'Status required';
    }
    if (isset($_POST['recommend'])) {
        $errors['recommend'] = 'Recommend required';
    }

    $fav = $_POST['fav'];
    $rank = $_POST['rank'];
    $rewatched = $_POST['rewatched'];
    $status = $_POST['status'];
    $recommend = $_POST['recommend'];
    $id = $_SESSION['id'];

    // Insert into Movie Table
    if (count($errors) === 0) {
        $query = "INSERT INTO movie SET fav=?, rank=?, rewatched=?, status=?, recommend=?, id =?";
        $stmt = $conn->prepare($query);
        $stmt->bind_param('siissi', $fav, $rank, $rewatched, $status, $recommend, $id);
        $result = $stmt->execute();

        if ($result) {
            // If Insertion Successful
        } else {
            // If Something Went Wrong
        }
    }
}

在上面的代码中，当用户提交表单时，它将捕获表单详细信息并将其存储在电影表中，外键作为用户 ID。我假设$_SESSION['id'] 是当前登录用户的用户ID。