选择查询计数根据病情(select query and count based on conditi

我想选择所有类别，子类别和计数业务属于子类别的数量。 这是SQL查询我使用。

SELECT
    c.id, 
    c.name,
    c.slug,
    sc.id,
    sc.name,
    sc.slug,
    COUNT(bsc.id) AS business_count
FROM 
    fi_category c
LEFT JOIN 
    fi_subcategory sc ON c.id = sc.category_id AND (sc.deleted_at IS NULL) 
LEFT JOIN 
    fi_business_subcategory bsc ON sc.id = bsc.subcategory_id AND (bsc.deleted_at IS NULL) 
WHERE 
    (c.deleted_at IS NULL) 
GROUP BY 
    c.id, sc.id

但有更我想做的事情，BUSINESS_COUNT应根据我想选择所有类别，子类别，但BUSINESS_COUNT应该有一个像子句他们属于中端，即全市过滤WHERE city.id = 1 ，为了这个，我想我必须使用数作为子查询对此我没有能够搞清楚。

下面是从关系结构fi_business_subcategory到fi_city 。

1） fi_business_subcategory

+----+----------------+-------------+
| id | subcategory_id | business_id |
+----+----------------+-------------+

2） fi_business

+----+---------+-----------+
| id | name    | suburb_id |
+----+---------+-----------+

3） fi_suburb

+-----+--------+---------+
| id  | name   | city_id |
+-----+--------+---------+

4） fi_city

+----+--------+
| id | name   |
+----+--------+

我想这样的事情，但是这似乎并没有工作

SELECT
    c.id, 
    c.name,
    c.slug,
    sc.id,
    sc.name,
    sc.slug,
    bsc.business_count
FROM 
    fi_category c
LEFT JOIN 
    fi_subcategory sc ON c.id = sc.category_id AND (sc.deleted_at IS NULL) 
LEFT JOIN (
    SELECT 
        COUNT(business_id) t1.business_count, t1.subcategory_id 
    FROM
        fi_business_subcategory t1
    LEFT JOIN
        fi_business t2 ON t2.id = t1.business_id
    LEFT JOIN
        fi_suburb t3 ON t3.id = t2.suburb_id
    LEFT JOIN
        fi_city t4 ON t4.id = t3.city_id
    WHERE
        t4.id = 1
    GROUP BY
        t1.subcategory_id
) bsc ON sc.id = bsc.subcategory_id AND (bsc.deleted_at IS NULL)
WHERE 
    (c.deleted_at IS NULL) 
GROUP BY 
    c.id, sc.id

我应该如何建立查询，以实现我想要什么？

Answer 1:

我看不出有什么理由你应该使用子查询。 我相信，你可以简单地结合起来fi_business和fi_business_subcategory到一个括号中的表的因素。

SELECT
    c.id, 
    c.name,
    c.slug,
    sc.id,
    sc.name,
    sc.slug,
    COUNT(bsc.id) AS business_count
FROM
    fi_category c
LEFT JOIN
    fi_subcategory sc ON c.id = sc.category_id AND (sc.deleted_at IS NULL)
LEFT JOIN (
        fi_business b
    INNER JOIN
        fi_business_subcategory bsc ON b.id = bsc.business_id AND (bsc.deleted_at IS NULL)
    INNER JOIN
        fi_suburb su ON su.id = b.suburb_id AND su.city_id = 1
    ) ON sc.id = bsc.subcategory_id
WHERE 
    (c.deleted_at IS NULL) 
GROUP BY 
    c.id, sc.id

我检查 ，这是对你的表结构有效的SQL。 我想有很好的机会，这将产生预期的结果，即使你的小提琴不包含任何数据。 见在JOIN语法手册上，您可以在一个连接使用括号的详细信息。

你也可以问问自己，如果你真的需要所有的连接要留给连接。 使用内部书写的东西加入就好办多了。

随着加入被执行左到右，你可能会做的内连接第一，其次是序列的正确连接。 这避免了括号：

SELECT
    c.id cat_id,
    c.name cat_name,
    c.slug cat_slug,
    sc.id sub_id,
    sc.name sub_name,
    sc.slug sub_slug,
    COUNT(bsc.id) AS business_count
FROM
    fi_business b
INNER JOIN
    fi_business_subcategory bsc ON b.id = bsc.business_id
    AND (b.deleted_at IS NULL) AND (bsc.deleted_at IS NULL)
INNER JOIN
    fi_suburb su ON su.id = b.suburb_id AND su.city_id = 1
RIGHT JOIN
    fi_subcategory sc ON sc.id = bsc.subcategory_id
RIGHT JOIN
    fi_category c ON c.id = sc.category_id AND (sc.deleted_at IS NULL)
WHERE
    (c.deleted_at IS NULL)
GROUP BY
    c.id, sc.id

Answer 2:

如果你想使用子查询，正确地表达出你与作为痘痘变化可能第二次查询是这样的：

SELECT
    c.id, 
    c.name,
    c.slug,
    sc.id,
    sc.name,
    sc.slug,
    IFNULL(bsc.business_count, 0)
          -- turn NULL from left join into 0
FROM 
    fi_category c
LEFT JOIN 
    fi_subcategory sc ON c.id = sc.category_id AND (sc.deleted_at IS NULL) 
LEFT JOIN (
    SELECT 
        COUNT(*) business_count, t1.subcategory_id
          -- removed table name from alias name,
          -- and improved performance by simply counting rows
    FROM
        fi_business_subcategory t1
    LEFT JOIN
        fi_business t2 ON t2.id = t1.business_id
    LEFT JOIN
        fi_suburb t3 ON t3.id = t2.suburb_id
    LEFT JOIN
        fi_city t4 ON t4.id = t3.city_id
    WHERE
        t4.id = 1 AND (t1.deleted_at IS NULL)
          -- check deletion in subquery for performance
    GROUP BY
        t1.subcategory_id
) bsc ON sc.id = bsc.subcategory_id
          -- no longer need to check deletion here
WHERE 
    (c.deleted_at IS NULL) 
GROUP BY 
    c.id, sc.id

小提琴这里 。

Answer 3:

尝试这个

select 
    c.id,
    c.name,
    count(sc.name) as Count
from fi_category as c
left join fi_subcategory as sc on sc.category_id = c.id
left join fi_business_subcategory as  fbs on fbs.subcategory_id = sc.id
inner join (
select 
    fb.name,
    fs.id,
    fs.city_id

from fi_business as fb 
inner join fi_suburb as fs on fs.id = fb.suburb_id
where fs.city_id = 1



) as  fb on fb.id = fbs.business_id
group by c.id

文章来源: select query and count based on condition