php - 内部联接上的未定义索引

尝试访问CDID以在查询中进行设置，然后稍后使用GET在下一页上检索它。

我总是觉得INNER JOIN cdreview ON cdreview.CDID=cd.CDID会合并CDID，因为它们是相同的值，然后我可以通过在查询中将其设置为$cdid = $row['CDID'];来访问该值，但我总是收到Undefined index: CDID错误消息。

我是菜鸟，因此可以提供任何帮助。

<?php

    require_once 'database_conn.php';

    $userid = $_SESSION['userSession'];

$sql = "SELECT cdreview.reviewDate, cdreview.reviewText, cd.CDTitle FROM cd
INNER JOIN cdreview ON cdreview.CDID=cd.CDID AND cdreview.userID='$userid' ORDER BY cdreview.reviewDate;" or die;


$result = mysqli_query($conn, $sql) or die(mysqli_error($conn));

   if (mysqli_num_rows($result) > 0) {

    while ($row = mysqli_fetch_assoc($result)){

    $date = $row['reviewDate'];
    $album = $row['CDTitle'];
    $review = $row['reviewText'];
    $cdid = $row['CDID'];

            echo"<table align='center'>
            <tr align='center'>
                <td colspan='5'>
                    <h2>View All Reviews</h2>
                </td>
            </tr>

            <tr align='center'>
                <th>Date</th>
                <th>Album</th>
                <th>Reviews</th>
                <th>Edit Review</th>
                <th>Delete Review</th>
            </tr>

            <tr align='center'>
                    <td>$date</td>
                    <td>$album</td>
                    <td>$review</td>
                    <td><a href=\"album_reviews.php?id=$cdid\"></a></td>
                    <td><a href=\"review.php?id=$cdid\"></a></td>
                </tr>
            </table>";
    }
} else {
    echo "<script>alert('You have not left any reviews!')</script>";
            echo "<script>window.open('home.php', '_self')</script>";
}
mysqli_close($conn);
?>

最佳答案

您正在尝试获取未选择的值

  $date = $row['reviewDate'];
  $album = $row['CDTitle'];
  $review = $row['reviewText'];
  $cdid = $row['CDID'];

但您只选择

 SELECT cdreview.reviewDate, cdreview.reviewText, cd.CDTitle  from

您的选择中没有CDID列