我正在建立一个网站，用户应该能够在其中将音乐艺术家和专辑输入数据库。我在musique数据库中有表artists和albums。
我试图从artists中选择artistId，并将其与用户试图输入的任何相册相关联。然而，artistId一直以0的形式出现。我认为我的SELECT声明有问题，但我不能完全确定。
有人知道发生这种事的原因吗？
inputalbum.php：
<?php
include "session.php";
include "db.php";
SessionClient::checkIfLoggedIn();
// Get list of artists to suggest
$conn = DB::connect();
$results = $conn->query("SELECT artistName FROM artists");
$artists = [];
while ($row = $results->fetch_assoc()) {
$artists[] = $row;
}
?>
<?php include "header.php"; ?>
<div class="container">
<h1>INSERT ALBUM</h1>
<form class="form" enctype="multipart/form-data" action="albumredir.php" method="POST">
<fieldset>
<label for ="artistName">Artist</label>
<input type="text" name="artistName">
<br>
<!-- <div>
Artists already in the database: <span>?</span>
</div> -->
<script>
// Transfer php array to js to use on the browser
var artists = <?php echo json_encode($artists) ?>;
// Grab the artist input field
var artistInput = document.querySelector('input[name="artists"]');
// Set an event for when they change to suggest artists
artistInput.oninput = function () {
var currentValue = artistInput.value;
var suggestedArtists = [];
artists.forEach(function (artist) {
var enteredArtists = currentValue.split(',');
if (artist.label.match(enteredTags[enteredArtists.length - 1].trim())) {
suggestedArtists.push(artist);
}
});
var suggestionString = suggestedArtists.map(t => t.label).join(',');
document.querySelector('div span').innerHTML = suggestionString;
}
</script>
<label for="albumName">Album Name:</label>
<input type="text" name="albumName" placeholder="Album One">
<br>
<label for="relDate">Release Date:</label>
<input type="date" name="relDate">
<br>
</fieldset>
<fieldset>
<input type="submit" name="submit" value="Submit">
</fieldset>
</form>
</div>
albumredir.php：
<?php
session_start();
$artistName = $_POST['artistName'];
$albumName = $_POST['albumName'];
$relDate = $_POST['relDate'];
$submit = $_POST['submit'];
include "db.php";
$conn = DB::connect();
$artistId = $conn->query("SELECT artistId FROM artists WHERE artistName = $artistName");
$stmt = $conn->prepare("INSERT INTO albums (artistId, userId, albumName, relDate) VALUES (?, ?, ?, ?)");
$stmt->bind_param(
"iiss",
$artistId,
$_SESSION['currentUser']['userId'],
$_POST['albumName'],
$_POST['relDate']
);
if(isset($_SESSION['currentUser']['userId']))
{
$currentUser = $_SESSION['currentUser']['userId'];
}
else
{
$currentUser = NULL;
}
if(isset($_POST['albumName']))
{
$albumName = $_POST['albumName'];
}
else
{
$albumName = NULL;
}
if(isset($_POST['relDate']))
{
$relDate = $_POST['relDate'];
}
else {
$relDate = NULL;
}
$stmt->execute();
// Close the connection
$conn->close();
// header('Location: index.php');
?>
$artistId = $conn->query正在返回一个结果集，因此稍后尝试时无法直接绑定到它：
$stmt->bind_param(
"iiss",
$artistId,
您需要首先从结果集中获取artistId。
在本例中，为了清楚起见，我将结果集的名称从$artistId更改为$result。
$result = $conn->query("SELECT artistId FROM artists WHERE artistName = $artistName");
// get row from result
$row = $result->fetch_assoc();
// get artistID from row
$artistId = $row["artistId"];
$stmt = $conn->prepare("INSERT INTO albums (artistId, userId, albumName, relDate) VALUES (?, ?, ?, ?)");
$stmt->bind_param(
"iiss",
$artistId,
$_SESSION['currentUser']['userId'],
$_POST['albumName'],
$_POST['relDate']
);