Question

我希望从单个表单将数据发布到两个数据库表中。我的数据库安排如下：

数据库1 - '关注列表'：

watchlist_id
USER_ID
名称
描述
类别

数据库2 - 'watchlist_films'：

watchlist_id
film_id

我当前的MySQL查询如下所示：$query = "INSERT INTO watchlist_films (watchlist_id, film_id) VALUES ('$watchlist_name['watchlist_id']', '$rt_id') WHERE watchlists ('watchlist_id') = " . $watchlist_name['watchlist_id'];，但我不确定某处是否有某种形式的INNER JOIN？

不确定要提供哪些其他信息/代码，所以我很抱歉，如果这里的细节太少，但是，如果还有其他需要的东西，请给我发表评论，我会提出其他所需要的内容。我是一个相对的PHP新手，所以如果这看起来像一个非常简单的问题，请道歉！

根据评论进行更新：

我现在有一半的查询工作，并更新了逻辑以反映它。新查询基本上执行以下操作：

INSERT新'watchlists'表

SELECT watchlist_id

'watchlists'

WHERE watchlist_name = $watchlist_name（刚刚创建的新关注列表的名称）和user_id = $user_id
INSERT watchlist_id（从上一个查询中选择）和film_id到'watchlist_films'表

根据您的评论，我的查询现在看起来像这样：

if ($submit == 'Submit') {
        require_once("db_connect.php");

        $watchlist_name = clean_string($_POST['watchlist-name']);
        $watchlist_description = clean_string($_POST['watchlist-description']);
        $watchlist_category = $_POST['watchlist-category'];

        $addWatchlist_bad_message = '';
        $addWatchlist_good_message = '';

        if ($db_server) {
            if (!empty($watchlist_name)) {
                $watchlist_name = clean_string($watchlist_name);
                $watchlist_description = clean_string($watchlist_description);
                mysql_select_db($db_database);

                // Insert new Watchlist into Watchlist index
                $insert_new_watchlist = "INSERT INTO watchlists (user_id, name, description, category) VALUES ('$user_id', '$watchlist_name', '$watchlist_description', '$watchlist_category')";
                mysql_query($insert_new_watchlist) or die("Insert failed. " . mysql_error() . "<br />" . $insert_new_watchlist);

                // Select new Watchlist ID
                $select_new_watchlist = "SELECT watchlist_id FROM watchlists WHERE name = " . $watchlist_name;
                $new_watchlist_id = mysql_query($select_new_watchlist) or die("Insert failed. " . mysql_error() . "<br />" . $select_new_watchlist);

                // Add film to new Watchlist
                $add_new_film = "INSERT INTO watchlist_films (watchlist_id, film_id) VALUES ('$new_watchlist_id', '$rt_id')";
                mysql_query($add_new_film) or die("Insert failed. " . mysql_error() . "<br />" . $add_new_film);
                $addWatchlist_good_message = '<div class="alert alert-success">Watchlist created successfully!</div>';?>
                <script>
                    $('a.add-watchlist').trigger('click');
                </script><?php 
            }
        } else {
            $addWatchlist_bad_message = '<div class="alert alert-error">Error: could not connect to the database.</div.';?>
            <script>
                $('a.add-watchlist').trigger('click');
            </script><?php
        }
        require_once("db_close.php");
    }

然而，我的查询似乎在SELECT语句中失败，在将新的Watchlist添加到Watchlist索引并将电影添加到新创建的监视列表之间。

Answer 1

try this


$query1 = "INSERT INTO watchlist_films (watchlist_id, film_id) 
VALUES ('" . $watchlist_name['watchlist_id'] . "', '$rt_id')"; 

 $query2= "INSERT INTO watchlists ('watchlist_id') 
VALUES (" . $watchlist_name['watchlist_id'] . ")";

$result = mysqli_multi_query($query1, $query2);

Answer 2

您需要为每个表编写一个INSERT。

$mysqli->query("INSERT INTO watchlist_films (watchlist_id, film_id) 
  VALUES ('" . $watchlist_name['watchlist_id'] . "', '$rt_id')");
$mysqli->query("INSERT INTO watchlists ('watchlist_id') 
  VALUES (" . $watchlist_name['watchlist_id'] . ")");

将表单数据插入两个数据库表

2 个答案: