好的，我有一个数据库可以存储歌曲信息和每首歌曲的星级。
我有 table 艺术家 流派 轨道 跟踪评级
在轨道中我有 trackID(设置为 autoIncrement)、genreID、artistID、title
我可以插入一首轨道并且工作正常问题是当我插入一首轨道时 我还想插入与该轨道关联的评级
在 track_rating 我有 rating_id(AutoInc), track_id, rating_number, total_points
我遇到的问题是 track_id。我无法从 tracks 表中获取值以将其插入到 track_rating 表中。
这是我的代码
add_track_form
<?php
require('includes/database.php');
$query = 'SELECT * FROM genres ORDER BY genreID';
$statement = $db->prepare($query);
$statement->execute();
$genres = $statement->fetchAll();
$statement->closeCursor();
$queryArtist = 'SELECT * FROM artists ORDER BY artistID';
$statementArtist = $db->prepare($queryArtist);
$statementArtist->execute();
$artists = $statementArtist->fetchAll();
$statementArtist->closeCursor();
?>
<!DOCTYPE html>
<html>
<!-- the head section -->
<head>
<title>MixMatcher</title>
<link rel="stylesheet" type="text/css" href="css/main.css">
</head>
<!-- the body section -->
<body>
<header><h1>Tracks</h1></header>
<main>
<h1>Add Track</h1>
<form action="add_track.php" method="post" id="add_track_form">
<label>genre:</label>
<select name="genre_id">
<?php foreach ($genres as $genres) : ?>
<option value="<?php echo $genres['genreID']; ?>">
<?php echo $genres['genreName']; ?>
</option>
<?php endforeach; ?>
</select><br>
<label>Artist:</label>
<select name="artist_id">
<?php foreach ($artists as $artists) : ?>
<option value="<?php echo $artists['artistID']; ?>">
<?php echo $artists['artistName']; ?>
</option>
<?php endforeach; ?>
</select><br>
<label>Title:</label>
<input type="text" name="title"><br>
<label>Add Track</label>
<input type="submit" value="Add track"><br>
</form>
<p><a href="index.php">View track List</a></p>
</main>
<footer>
<p>© <?php echo date("Y"); ?> MixMatcher, Inc.</p>
</footer>
</body>
</html>
添加轨道
<?php
// Get the track data
$genre_id = filter_input(INPUT_POST, 'genre_id', FILTER_VALIDATE_INT);
$artist_id = filter_input(INPUT_POST, 'artist_id', FILTER_VALIDATE_INT);
$title = filter_input(INPUT_POST, 'title', FILTER_SANITIZE_STRING);
// Validate inputs
if ($genre_id == null || $genre_id == false ||
$artist_id == null || $title == null ) {
$error = "Invalid track data. Check all fields and try again.";
include('database_error.php');
} else {
require_once('includes/database.php');
// Add the track to the database
$query = 'INSERT INTO tracks (genreID, artistID, title)
VALUES (:genre_id, :artist_id, :title)';
$statement = $db->prepare($query);
$statement->bindValue(':genre_id', $genre_id);
$statement->bindValue(':artist_id', $artist_id);
$statement->bindValue(':title', $title);
$statement->execute();
$statement->closeCursor();
$query1 = 'SELECT * FROM tracks ORDER BY trackID';
$statement1 = $db->prepare($query1);
$statement1->execute();
$tracks = $statement->fetchAll();
$statement->closeCursor();
//Problem Here
$trackID = $tracks['trackID'];
$queryRating = 'INSERT INTO track_rating (track_id, rating_number, total_points)
VALUES (:track_id, :rating_number, :total_points)';
$statementRating = $db->prepare($queryRating);
$statementRating->bindValue(':track_id', $trackID);
$statementRating->bindValue(':rating_number', 0);
$statementRating->bindValue(':total_points', 0);
$statementRating->execute();
$statementRating->closeCursor();
// Display the track List page
include('index.php');
}
?>
最佳答案
$query = 'INSERT INTO tracks (genreID, artistID, title)
VALUES (:genre_id, :artist_id, :title)';
$statement = $db->prepare($query);
$statement->bindValue(':genre_id', $genre_id);
$statement->bindValue(':artist_id', $artist_id);
$statement->bindValue(':title', $title);
$statement->execute();
$lastestID = $statement->inser_id();
$statement->closeCursor();
我从你的代码中添加了 $latestID = $statement->inser_id(); 这就是获取最新 ID 的代码，如果你使用准备好的 stament 然后我们使用变量 $latestID 在您的 track_rating 表中插入 track_ID..
关于php - 需要帮助将 ID 值从一个表插入到 PHP 中的下一个表中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40850162/