創建SQL語句時出錯

Question

我編寫了以下代碼來嘗試獲取數據庫中3個表中的條目數。 當我在phpMyAdmin使用查詢時，查詢有效，但是運行此代碼時，我得到：

Error creating SQL statement

這是由if(!$stmt)語句生成的，我不知道為什么它不起作用。 提前致謝 ：）

<?php
include 'connection.php';   

$countArtists = $countAlbums = $countTracks = 0;

/* Create queries to get counts from each table */
$sql = "SELECT COUNT(*) FROM artist;";
$sql .= "SELECT COUNT(*) FROM cd;";
$sql .= "SELECT COUNT(*) FROM tracks;";

$stmt = $conn->prepare($sql);
if(!$stmt)
{
    echo "Error creating SQL statement";
    return 1;
}

$stmt->execute();

$stmt->bind_result($countArtists, $countAlbums, $countTracks);

echo "<li>Number of Artists: $countArtists</li><br>\n" .
    "<li>Number of Albums: $countAlbums</li><br>\n" .
    "<li>Number of Tracks: $countTracks</li><br>\n";
?>

Answer 1

我想您想要一個這樣的查詢：

$sql = "SELECT (SELECT COUNT(*) FROM artist) as countArtists, ".
              "(SELECT COUNT(*) FROM cd) as countAlbums, ".
              "(SELECT COUNT(*) FROM tracks) as countTracks";

這是一個具有三列的查詢，而不是三個單獨的查詢。