MySql 查询帮助 : joints with sums and counts

Question

我有这样的数据库结构:

TBL_A  |  TBL_B  |  TBL_C  |  TBL_D  | TBL_E
-------+---------+---------+---------+----------
id     | id_tbla | id_tbla | id_tbla | id
name   | id_user | id_user | id_user | name_tbla
...    | is_bool |         | weight  | id_user

这是我要实现的目标:

SELECT 
    a.id, 
    a.name, 
    b.is_bool,
    count(c.id_user) AS nb_views, 
    sum(d.weight) AS total_weight,
    count(distinct e.id_user) AS distinct_users,
FROM TBL_A AS a 
LEFT JOIN (TBL_B AS b) on (b.id_tbla = a.id)
LEFT JOIN (TBL_C AS c) on (c.id_tbla = a.id)
LEFT JOIN (TBL_D AS d) on (d.id_tbla = a.id)
LEFT JOIN (TBL_E AS e) on (e.name_tbla = a.name)
where a.id = 1 and e.id_user = 1

执行了查询，但结果(nb_views、total_weight、distinct_users)是错误的。知道为什么吗？

Answer 1

您试图在一个查询中计算太多聚合。

Enita non sunt multiplicanda praeter necessitatem

(Latin, "entities are not to be multiplied beyond necessity")

您的表 B、C、D 和 E 已生成 Cartesian Products互相反对。假设A匹配中的给定行:

• B 中 3 行
• C 中的 6 行
• D 中 4 行
• E 中的 1 行

结果中的总行数为 3 * 6 * 4 * 1 = 72 行。所以你的 count(c.id_user) 是它应该的 12 倍，你的 sum(d.weight) 是它应该的 18 倍，等等。

最简单的补救措施是在单独的查询中计算这些聚合中的每一个:

SELECT a.id, a.name, COALESCE(b.is_bool, FALSE) AS is_bool
FROM TBL_A AS a LEFT JOIN TBL_B AS b ON (b.id_tbla = a.id)
WHERE a.id = 1;

SELECT a.id, COUNT(c.id_user) AS nb_views
FROM TBL_A AS a LEFT JOIN TBL_C AS c ON (c.id_tbla = a.id)
WHERE a.id = 1;

SELECT a.id, SUM(d.weight) AS total_weight,
FROM TBL_A AS a LEFT JOIN TBL_D AS d ON (d.id_tbla = a.id)
WHERE a.id = 1;

SELECT a.id, COUNT(DISTINCT e.id_user) AS distinct_users,
FROM TBL_A AS a LEFT JOIN TBL_E AS e 
  ON (e.name_tbla = a.name AND e.id_user = 1)
WHERE a.id = 1;