如何改进以下 mysql 选择？

Question

我目前正在重构一个遗留应用程序并逐步转换为 zend 框架 1.12。

我正在为如何将其转换为 zend db 而摸不着头脑，有没有一种方法可以在一个查询中完成？

现在我看到它首先获取文件夹列表，然后对每个文件夹运行一个附加查询... 将其作为一个查询运行会提高性能，对吗？

$folders_query = DB::Query("select * from contacts_folders order by sort_order, name");
while($folders = DB::FetchArray($folders_query)){
    $counts_total = DB::QueryOne("SELECT count(cm.messages_id) AS total
    FROM contacts_basics cb, contacts_messages cm
    WHERE cb.contacts_id = cm.contacts_id
    AND cm.folders_id =  '" . $folders['folders_id'] . "'
    AND cm.status = '1'
    AND cm.mark =  '0'");


    if ($counts_total >0){ 
        $folders_name = '<strong>' . $folders['name'] . ' (' . $counts_total . ')</strong>'; 
    } else { 
        $folders_name = $folders['name']; 
    }

    echo '<li><a href="messages.php?fID=' . $folders['folders_id'] . '">';

    echo $folders_name;

    echo '</a></li>';
}

Answer 1

是的，您可以在同一个查询中同时执行这两项操作

SELECT cf.*, count(cm.messages_id) AS total
FROM contacts_folders cf left outer join
     contacts_messages cm
     on cf.id = cm.folders_id and
        cm.status = '1' AND cm.mark =  '0' left outer join
     contacts_basics cb
     on cb.contacts_id = cm.contacts_id
group by cf.folders_id
order by cf.sort_order, cf.name;

这使用left outer join 来确保您获得所有文件夹，即使没有消息也是如此（原始代码的工作方式）。由于left outer join，需要将条件移入on 子句中。

它还从文件夹中获取所有信息以及总数。如果没有消息，则该文件夹应返回 0。

Answer 2

戈登的回答中有一个小错误，但多亏了他，我才弄明白。

我变了

contacts_basics cb left outer join

收件人：

left outer join contacts_basics cb

以下代码按预期工作：

public function getMenuCounts(){
    $raw = "SELECT cf.*, count(cm.messages_id) AS total
    FROM contacts_folders cf left outer join
    contacts_messages cm
    on cf.folders_id = cm.folders_id and
    cm.status = '1' AND cm.mark =  '0'
    left outer join contacts_basics cb
    on cb.contacts_id = cm.contacts_id
    group by cf.folders_id
    order by cf.sort_order, cf.name;";
    $db = Zend_Db_Table::getDefaultAdapter();
    $stmt = $db->query($raw);
    return $stmt->fetchAll();
}