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我试图让这个内部连接显示输出标题和日期。仅返回提供对象详细信息的 sql 对象。这是结果。我可以访问 num_rows 但只给我 int(11) 的记录数,不知道如何访问字段中的信息。这就是结果。object(mysqli_result)#6 (5) { ["current_field"]=> int(0) ["field_count"]=> int(2) ["lengths"]=> NULL ["num_rows"]=> int(11 ) [“类型”]=> int(0) }

感谢帮助。

$sql  = "SELECT  title, artist_name"; 
$sql .= " FROM follows";
$sql .= " INNER JOIN artworks";
$sql .= " ON follows.user_id_followed = artworks.artist_id";
$sql .= " AND follows.user_id_follower='12'";

$result_set = $database->query($sql);

echo var_dump($result_set);
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1 回答 1

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你需要一个 WHERE

$sql  = "SELECT  title, artist_name"; 
$sql .= " FROM follows";
$sql .= " INNER JOIN artworks";
$sql .= " ON follows.user_id_followed = artworks.artist_id";
$sql .= " WHERE follows.user_id_follower='12'";

你也可能不想LEFT JOININNER JOIN`

于 2015-12-22T22:10:59.180 回答