Question

我有这个问题，我的查询不会将数据从数据库拉到网站，我已经尝试了很多东西，但我得到一个偏移错误或根本没有出现。有人可以看一下吗？

带下拉菜单的网站 -

$dbc = @mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() );
//echo "Connected to MySQL<br>";
$q = "SELECT artistname FROM tbl_artists ORDER BY artistname ASC";      
$r = @mysqli_query ($dbc, $q);

    // Form start
echo '<form action="php/process_dropdown.php" method="post">

<select name="selection">
<option value=""> -- Please select an Artist -- </option>';


// Fetch and print all the records:
    while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
        echo '<option>' . $row['artistname'] . '</option>
        ';
    }

    echo '<p><input type="submit" value="Send"></p>

    </select>

</form>'; // End of form.

?>

下拉流程

$dbc = @mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() );

$name = $_POST['selection'];


$q = "SELECT tbl_songs.songtitle, tbl_songs.yearrelease, tbl_artists.genre, tbl_songs.price
FROM tbl_artists INNER JOIN tbl_songs ON tbl_artists.artistID = tbl_songs.artistID
WHERE (((tbl_artists.artistID)='$name'))";

$r = @mysqli_query ($dbc, $q) or die("Error: ".mysqli_error($dbc));

    // Table header.
    echo '<table>
    <tr>
    <th scope="col">Song</th>
    <th scope="col">Year Released</th>
    <th scope="col">Genre</th>
    <th scope="col">Price</th>
    </tr>
';

    // Fetch and print all the records:
    while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
        echo '<tr><td>' . $row['songtitle'] . '</td><td>' . $row['yearrelease'] . '</td><td>' . $row['genre'] . '</td><td>' . $row['price'] . '</td></tr>   ';
    }

echo '</table>'; // Close the table.

?>

谢谢。

Answer 1

修改：

$link = mysqli_connect('host', 'user', 'password', 'database');

转变为：

$dbc = mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() ); //echo "Connected to MySQL<br>"; $q = "SELECT artistname FROM tbl_artists ORDER BY artistname ASC"; $r = mysqli_query ($dbc, $q); ..................

看到这个：

WHERE (((tbl_artists.artistID)='$name'))";

你错过了：tbl_songs。在$name之前。

此外，使用@并不好，因为它隐藏了错误消息。（这是一个穷人脚本）

获取数组时没有出现结果 - PHP / MYSQL

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