MySQL查询-复杂计数条件_Mysql_Optimization

MySQL查询-复杂计数条件

mysql optimization

MySQL查询-复杂计数条件,mysql,optimization,Mysql,Optimization,我有一个带有以下结构的表“位置”： id | property_id | location_type 1 | 1 | 1 2 | 1 | 2 3 | 2 | 1 4 | 3 | 2 5 | 4 | 1 6 | 4 | 2 id | property_id | amenity_type 1 | 1 | 1 2 | 1

我有一个带有以下结构的表“位置”：

id  | property_id | location_type
1   | 1           | 1
2   | 1           | 2
3   | 2           | 1
4   | 3           | 2
5   | 4           | 1
6   | 4           | 2

id  | property_id | amenity_type
1   | 1           | 1
2   | 1           | 3
3   | 2           | 2
4   | 3           | 4
5   | 4           | 1
6   | 4           | 3

id  | property_id | property_type
1   | 1           | 2
2   | 1           | 3
3   | 2           | 2
4   | 3           | 4
5   | 4           | 2
6   | 4           | 3

我还有另一张桌子“便利设施”，其结构如下：

id  | property_id | location_type
1   | 1           | 1
2   | 1           | 2
3   | 2           | 1
4   | 3           | 2
5   | 4           | 1
6   | 4           | 2

id  | property_id | amenity_type
1   | 1           | 1
2   | 1           | 3
3   | 2           | 2
4   | 3           | 4
5   | 4           | 1
6   | 4           | 3

id  | property_id | property_type
1   | 1           | 2
2   | 1           | 3
3   | 2           | 2
4   | 3           | 4
5   | 4           | 2
6   | 4           | 3

我有另一个表“property”，其结构如下：

id  | property_id | location_type
1   | 1           | 1
2   | 1           | 2
3   | 2           | 1
4   | 3           | 2
5   | 4           | 1
6   | 4           | 2

id  | property_id | amenity_type
1   | 1           | 1
2   | 1           | 3
3   | 2           | 2
4   | 3           | 4
5   | 4           | 1
6   | 4           | 3

id  | property_id | property_type
1   | 1           | 2
2   | 1           | 3
3   | 2           | 2
4   | 3           | 4
5   | 4           | 2
6   | 4           | 3

id-是相应表的主键。property_id是我的数据库的属性id（外键）。位置类型为

beach (value - 1), mountain (value - 2).

amenity_type is car (value - 1), bike (value - 2), football (value - 3).

property_type is villa (value - 2), house (value - 3)

我正在使用以下SQL查询来选择位置为1、位置为2、舒适度为1、属性为3、属性为1的属性id，即属性具有海滩、山脉、汽车、别墅和房屋：

SELECT p.id  
FROM 
    property AS p           
  JOIN
    location AS l1
        ON  l1.property_id = p.id  
        AND l1.location_type = 1 
  JOIN
    location AS l2
        ON  l2.property_id = p.id  
        AND l2.location_type = 2 
  JOIN                      
    amentities AS a1
        ON  a1.property_id = p.id
        AND a1.amenity_type = 2                 
  JOIN
    properties AS p1
        ON  p1.property_id = p.id  
        AND p1.property_type = 3 
  JOIN
    properties AS p2 
        ON  p2.property_id = p.id  
        AND p2.property_type = 1

假设我将（位置为1、位置为2、设施为1、属性为3、属性为1的属性为id）的计数设为1500。我需要获得具有相同条件和其他属性类型、位置类型、设施类型的计数

但我无法使用上述查询获取以下条件的计数：

（位置类型为1、位置类型为2、设施类型为1、属性类型为3、属性类型为1的属性id）和位置类型为3的计数

（位置类型为1、位置类型为2、设施类型为1、属性类型为3、属性类型为1的属性id）和位置类型为4的计数

（位置类型为1，位置类型为2，设施类型为1，属性类型为3，属性类型为1的属性id）和设施类型为2的计数

（位置类型为1，位置类型为2，设施类型为1，属性类型为3，属性类型为1的属性id）和设施类型为3的计数

是否有任何有效的方法来获取不同位置类型、舒适度类型等的计数

请参阅我前面的问题-

由于以下原因，您的查询失败：

property_type = 3 AND property_type = 1

单个字段不能同时具有两个值。应该是

... AND ... (property_type = 3 OR property_type = 1) AND ...

或者更具可读性：

... AND ... property_type IN (1, 3) AND ...

对于

位置类型

和

便利设施类型

字段也是如此。

由于以下原因，您的查询失败：

property_type = 3 AND property_type = 1

单个字段不能同时具有两个值。应该是

... AND ... (property_type = 3 OR property_type = 1) AND ...

或者更具可读性：

... AND ... property_type IN (1, 3) AND ...

location\u type

和

amerity\u type

字段也是如此。

我理解这个问题有点困难，但我想您是在询问有关与基本过滤器匹配的属性的更多信息

您可以尝试以下方法：

SELECT p.id, l.location_type, a.amenity_type, count(*) property_count
FROM (
    SELECT p.id
    FROM property AS p           
        INNER JOIN location AS l1 ON l1.property_id = p.id AND l1.location_type = 1 
        INNER JOIN location AS l2 ON l2.property_id = p.id AND l2.location_type = 2 
        INNER JOIN amentities AS a1 ON  a1.property_id = p.id AND a1.amenity_type = 2                 
        INNER JOIN properties AS p1 ON  p1.property_id = p.id AND p1.property_type = 3 
        INNER JOIN properties AS p2 ON  p2.property_id = p.id AND p2.property_type = 1
) p
INNER JOIN location AS l ON l.property_id = p.id
INNER JOIN amentities AS a ON  a.property_id = p.id
GROUP BY p.id, l.location_type, a.amenity_type
WITH ROLLUP

SELECT COUNT(*) AS BaseCount
     , ( SELECT COUNT(*)
         FROM location AS l3
         WHERE l3.property_id = p.id  
           AND l3.location_type = 3
       ) AS CountLocation3
     , ( SELECT COUNT(*)
         FROM location AS l4
         WHERE l4.property_id = p.id  
           AND l4.location_type = 4
       ) AS CountLocation4
     , ( SELECT COUNT(*)
         FROM amenities AS a2
         WHERE a2.property_id = p.id  
           AND a2.amenity_type = 2
       ) AS CountAmenity4
     , ...
FROM 
    property AS p           
  JOIN
    location AS l1
        ON  l1.property_id = p.id  
        AND l1.location_type = 1 
  JOIN
    location AS l2
        ON  l2.property_id = p.id  
        AND l2.location_type = 2 
  JOIN                      
    amentities AS a1
        ON  a1.property_id = p.id
        AND a1.amenity_type = 1                 
  JOIN
    properties AS p3
        ON  p3.property_id = p.id  
        AND p3.property_type = 3 
  JOIN
    properties AS p1 
        ON  p1.property_id = p.id  
        AND p1.property_type = 1

您可以从此处展开联接列表和条件

编辑：根据您的评论，我认为您需要添加逻辑分组…

我理解这个问题有点困难，但我认为您是在询问有关与基本筛选器匹配的属性的附加信息

您可以尝试以下方法：

SELECT p.id, l.location_type, a.amenity_type, count(*) property_count
FROM (
    SELECT p.id
    FROM property AS p           
        INNER JOIN location AS l1 ON l1.property_id = p.id AND l1.location_type = 1 
        INNER JOIN location AS l2 ON l2.property_id = p.id AND l2.location_type = 2 
        INNER JOIN amentities AS a1 ON  a1.property_id = p.id AND a1.amenity_type = 2                 
        INNER JOIN properties AS p1 ON  p1.property_id = p.id AND p1.property_type = 3 
        INNER JOIN properties AS p2 ON  p2.property_id = p.id AND p2.property_type = 1
) p
INNER JOIN location AS l ON l.property_id = p.id
INNER JOIN amentities AS a ON  a.property_id = p.id
GROUP BY p.id, l.location_type, a.amenity_type
WITH ROLLUP

SELECT COUNT(*) AS BaseCount
     , ( SELECT COUNT(*)
         FROM location AS l3
         WHERE l3.property_id = p.id  
           AND l3.location_type = 3
       ) AS CountLocation3
     , ( SELECT COUNT(*)
         FROM location AS l4
         WHERE l4.property_id = p.id  
           AND l4.location_type = 4
       ) AS CountLocation4
     , ( SELECT COUNT(*)
         FROM amenities AS a2
         WHERE a2.property_id = p.id  
           AND a2.amenity_type = 2
       ) AS CountAmenity4
     , ...
FROM 
    property AS p           
  JOIN
    location AS l1
        ON  l1.property_id = p.id  
        AND l1.location_type = 1 
  JOIN
    location AS l2
        ON  l2.property_id = p.id  
        AND l2.location_type = 2 
  JOIN                      
    amentities AS a1
        ON  a1.property_id = p.id
        AND a1.amenity_type = 1                 
  JOIN
    properties AS p3
        ON  p3.property_id = p.id  
        AND p3.property_type = 3 
  JOIN
    properties AS p1 
        ON  p1.property_id = p.id  
        AND p1.property_type = 1

您可以从此处展开联接列表和条件

编辑：根据您的评论，我认为您需要添加逻辑分组…

您可以使用基本查询获取所有

属性.id

（并计算它们）和

选择列表中的子查询来计算附加条件，如下所示：
SELECT p.id, l.location_type, a.amenity_type, count(*) property_count
FROM (
    SELECT p.id
    FROM property AS p           
        INNER JOIN location AS l1 ON l1.property_id = p.id AND l1.location_type = 1 
        INNER JOIN location AS l2 ON l2.property_id = p.id AND l2.location_type = 2 
        INNER JOIN amentities AS a1 ON  a1.property_id = p.id AND a1.amenity_type = 2                 
        INNER JOIN properties AS p1 ON  p1.property_id = p.id AND p1.property_type = 3 
        INNER JOIN properties AS p2 ON  p2.property_id = p.id AND p2.property_type = 1
) p
INNER JOIN location AS l ON l.property_id = p.id
INNER JOIN amentities AS a ON  a.property_id = p.id
GROUP BY p.id, l.location_type, a.amenity_type
WITH ROLLUP

SELECT COUNT(*) AS BaseCount
     , ( SELECT COUNT(*)
         FROM location AS l3
         WHERE l3.property_id = p.id  
           AND l3.location_type = 3
       ) AS CountLocation3
     , ( SELECT COUNT(*)
         FROM location AS l4
         WHERE l4.property_id = p.id  
           AND l4.location_type = 4
       ) AS CountLocation4
     , ( SELECT COUNT(*)
         FROM amenities AS a2
         WHERE a2.property_id = p.id  
           AND a2.amenity_type = 2
       ) AS CountAmenity4
     , ...
FROM 
    property AS p           
  JOIN
    location AS l1
        ON  l1.property_id = p.id  
        AND l1.location_type = 1 
  JOIN
    location AS l2
        ON  l2.property_id = p.id  
        AND l2.location_type = 2 
  JOIN                      
    amentities AS a1
        ON  a1.property_id = p.id
        AND a1.amenity_type = 1                 
  JOIN
    properties AS p3
        ON  p3.property_id = p.id  
        AND p3.property_type = 3 
  JOIN
    properties AS p1 
        ON  p1.property_id = p.id  
        AND p1.property_type = 1 

您可以使用基本查询获取所有property.id
（并计算它们）和SELECT
列表中的子查询，以计算附加条件，如下所示：
SELECT p.id, l.location_type, a.amenity_type, count(*) property_count
FROM (
    SELECT p.id
    FROM property AS p           
        INNER JOIN location AS l1 ON l1.property_id = p.id AND l1.location_type = 1 
        INNER JOIN location AS l2 ON l2.property_id = p.id AND l2.location_type = 2 
        INNER JOIN amentities AS a1 ON  a1.property_id = p.id AND a1.amenity_type = 2                 
        INNER JOIN properties AS p1 ON  p1.property_id = p.id AND p1.property_type = 3 
        INNER JOIN properties AS p2 ON  p2.property_id = p.id AND p2.property_type = 1
) p
INNER JOIN location AS l ON l.property_id = p.id
INNER JOIN amentities AS a ON  a.property_id = p.id
GROUP BY p.id, l.location_type, a.amenity_type
WITH ROLLUP

SELECT COUNT(*) AS BaseCount
     , ( SELECT COUNT(*)
         FROM location AS l3
         WHERE l3.property_id = p.id  
           AND l3.location_type = 3
       ) AS CountLocation3
     , ( SELECT COUNT(*)
         FROM location AS l4
         WHERE l4.property_id = p.id  
           AND l4.location_type = 4
       ) AS CountLocation4
     , ( SELECT COUNT(*)
         FROM amenities AS a2
         WHERE a2.property_id = p.id  
           AND a2.amenity_type = 2
       ) AS CountAmenity4
     , ...
FROM 
    property AS p           
  JOIN
    location AS l1
        ON  l1.property_id = p.id  
        AND l1.location_type = 1 
  JOIN
    location AS l2
        ON  l2.property_id = p.id  
        AND l2.location_type = 2 
  JOIN                      
    amentities AS a1
        ON  a1.property_id = p.id
        AND a1.amenity_type = 1                 
  JOIN
    properties AS p3
        ON  p3.property_id = p.id  
        AND p3.property_type = 3 
  JOIN
    properties AS p1 
        ON  p1.property_id = p.id  
        AND p1.property_type = 1 

抱歉，我之前没有澄清，但计数条件不是查询，它们只是定义我想要的属性\u id的计数，该属性的位置\u type=1、位置\u type=2、舒适性\u type=1、属性\u type=3、属性\u type=1（所有强制条件）-这是一个搜索应用程序。对不起，我之前没有澄清，但是计数条件不是查询，它们只是定义我想要的属性id的计数，它的位置类型=1，位置类型=2，舒适类型=1，属性类型=3，属性类型=1（所有强制条件）-这是一个搜索应用程序。谢谢Doug，但我不想选择属性，但我想用我的基本过滤器和每个位置类型、设施类型选择计数。听起来你需要使用分组方式。。。使用ROLLUP。重新考虑上述内容，您可能希望从SELECT列表和GROUP BY子句中删除p.id，以统计基本筛选器中的所有属性（而不是每个属性的位置/便利设施计数，这似乎没有帮助）。谢谢Doug，但我不想选择属性，但我想用我的基本过滤器和每个位置类型、设施类型选择计数。听起来你需要使用分组方式。。。使用ROLLUP.REFINATING，您可能希望从SELECT列表和GROUP BY子句中删除p.id，以统计基本筛选器中的所有属性（而不是每个属性的位置/便利设施计数，这似乎没有帮助）.因此，您需要一些基本情况下的计数，以及在每个不同情况下您将获得的计数，在这些情况下，您保持所有内容不变，但改变一个参数？所以这会产生很多不同的计数？你真的需要所有这些不同的组合吗？或者它只是一些子集？是的，我需要基本计数和基于基本条件的计数，并添加另一个参数，以便我可以向用户显示如果您也选择此位置类型，有多少属性可用。我们真的需要展示所有不同的组合。与之非常相似的是，在这里你可以看到左边过滤器上的计数。那么你想要一些基本情况下的计数，以及在每个不同情况下得到的计数，在这些情况下，你保持所有内容不变，但改变一个参数？所以这会产生很多不同的计数？你真的需要所有这些不同的组合吗？或者它只是一些子集？是的，我需要基本计数和基于基本条件的计数，并添加另一个参数，以便我可以向用户显示如果您也选择此位置类型，有多少属性可用。我们真的需要展示所有不同的组合