Question

我一直在思考这个问题，并且我认为最好先问一下并倾听其他人的想法。

我建立了一个在Mysql上存储位置的系统。每个位置都有一个类型，一些位置有多个地址。

表格看起来像这样

location
  - location_id (autoincrement)
  - location_name
  - location_type_id 

location_types
  - type_id
  - type_name (For example "Laundry")

location_information
  - location_id (Reference to the location table)
  - location_address
  - location_phone

因此，如果我想查询最近添加的10个数据库，我会选择以下内容：

SELECT l.location_id, l.location_name,
       t.type_id, t.type_name,
       i.location_address, i.location_phone
FROM location AS l
LEFT JOIN location_information AS i ON (l.location_id = i.location_id)
LEFT JOIN location_types AS t ON (l.location_type_id = t.type_id)
ORDER BY l.location_id DESC
LIMIT 10

右？但问题是，如果一个位置有超过1个地址，限制/分页将不会被激活，除非我“GROUP BY l.location_id”，但这将只显示每个地方的一个地址..会发生什么有多个地址的地方？

所以我认为解决这个问题的唯一方法是在循环内部进行查询..这样的事情（伪代码）：

$db->query('SELECT l.location_id, l.location_name,
            t.type_id, t.type_name
            FROM location AS l
            LEFT JOIN location_types AS t ON (l.location_type_id = t.type_id)
            ORDER BY l.location_id DESC
            LIMIT 10');

 $locations = array();
 while ($row = $db->fetchRow())
 {
     $db->query('SELECT i.location_address, i.location_phone
                 FROM location_information AS i
                 WHERE i.location_id = ?', $row['location_id']);

     $locationInfo = $db->fetchAll();
     $locations[$row['location_id']] = array('location_name' => $row['location_name'],
                                             'location_type' => $row['location_type'],
                                             'location_info' => $locationInfo);

 }

现在我获得了最后10个位置，但通过这样做我最多得到10个查询，我认为这不会有助于应用程序的性能。

有没有更好的方法来实现我正在寻找的东西？（准确的分页）。

Answer 1

这是您的原始查询

SELECT l.location_id, l.location_name, 
       t.type_id, t.type_name, 
       i.location_address, i.location_phone 
FROM location AS l 
LEFT JOIN location_information AS i ON (l.location_id = i.location_id) 
LEFT JOIN location_types AS t ON (l.location_type_id = t.type_id) 
ORDER BY l.location_id DESC 
LIMIT 10

你最后执行分页。如果您重构此查询，则可以更早地执行分页。

SELECT l.location_id, l.location_name, 
       t.type_id, t.type_name, 
       i.location_address, i.location_phone 
FROM
    (SELECT location_id,location_type_id FROM location
    ORDER BY location_id LIMIT 10) AS k
    LEFT JOIN location AS l ON (k.location_id = l.location_id)
    LEFT JOIN location_information AS i ON (k.location_id = i.location_id) 
    LEFT JOIN location_types AS t ON (l.location_type_id = t.type_id) 
;

注意我创建了一个名为k的子查询。 10个钥匙被拿起并命令FIRST !!!

然后JOIN可以从那里开始，希望只使用10个location_ids。

什么有助于子查询k是一个带有location_id和location_type_id的索引

ALTER TABLE location ADD INDEX id_type_ndx (location_id,location_type_id);

以下是您对此方法的其他看法

您如何查询下一个10个ID（ID 11 - 20）？像这样：

SELECT l.location_id, l.location_name, 
       t.type_id, t.type_name, 
       i.location_address, i.location_phone 
FROM
    (SELECT location_id,location_type_id FROM location
    ORDER BY location_id LIMIT 10,10) AS k
    LEFT JOIN location AS l ON (k.location_id = l.location_id)
    LEFT JOIN location_information AS i ON (k.location_id = i.location_id) 
    LEFT JOIN location_types AS t ON (l.location_type_id = t.type_id) 
;

您需要做的就是在每个新页面中更改子查询LIMIT中的k子句。

LIMIT 20,10
LIMIT 30,10
等......

我可以通过删除位置表来改进重构，并让子查询k携带所需的字段，如下所示：

SELECT k.location_id, k.location_name, 
       t.type_id, t.type_name, 
       i.location_address, i.location_phone 
FROM
    (SELECT location_id,location_type_id,location_name
    FROM location ORDER BY location_id LIMIT 10,10) AS k
    LEFT JOIN location_information AS i ON (k.location_id = i.location_id) 
    LEFT JOIN location_types AS t ON (k.location_type_id = t.type_id) 
;

此版本不需要额外索引。

试一试!!!

Answer 2

比循环和10个查询更好，您可以查询分页的location.location_id限制10，将其连接成逗号分隔的字符串，然后查询完整查询以获取WHERE location.location_id IN (1,2,3...{list of ids})

Answer 3

您可以按照您对location_id进行分组的原始想法，然后使用group_concat函数将该位置的所有地址显示为1个字段。

SELECT l.location_id, l.location_name,
   t.type_id, t.type_name,
   group_concat(concat("Address: ",i.location_address, " Phone: ", i.location_phone)) as addresses
FROM location AS l
LEFT JOIN location_information AS i ON (l.location_id = i.location_id)
LEFT JOIN location_types AS t ON (l.location_type_id = t.type_id)
GROUP BY l.location_id
ORDER BY l.location_id DESC
LIMIT 10

Answer 4

有几种方法可以解决这个问题：

您可以在IsPrimary表中添加location_information位列，并添加触发器以确保每个位置始终只有一个location_information记录，并将此值设置为1。
如果您没有location_information或location_id列，则可以使用DateCreated列选择最早或最新的DateModified记录（MIN / MAX）。

使用左连接准确分页

4 个答案: