MySQL 在多个表上加入和 COUNT()

MySQL join and COUNT() on multiple tables

我正在尝试 COUNT() 在一个查询中对多个表进行查询,但我无法让它工作。这是我目前所拥有的:

表格:

table1
---------------------
id | name
---------------------
 1 | test
 2 | test2


table2
---------------------
id | table1_id
---------------------
 1 | 1
 2 | 1
 3 | 1


table3
---------------------
id | table2_id
---------------------
 1 | 1


table4
---------------------
id | size | table3_id
---------------------
 1 | 1024 | 1
 1 | 200  | 1

SQL:

SELECT
    table1.name,
    COUNT(table2.table1_id) AS table2_count,
    COUNT(table3.table2_id) AS table3_count,
    COUNT(table4.table3_id) AS table4_count,
    SUM(table4.size) AS table4_size
FROM
    table1
LEFT JOIN table2
    ON table1.id = table2.table1_id
LEFT JOIN table3
    ON table2.id = table3.table2_id
LEFT JOIN table4
    ON table3.id = table4.table3_id
WHERE
    table1.id = 1

我从上述查询中得到的结果:

name | table2_count | table3_count | table4_count | table4_size
---------------------------------------------------------------
test |      4       |      2       |      2       |    1224

我应该得到的结果:

name | table2_count | table3_count | table4_count | table4_size
---------------------------------------------------------------
test |      3       |      1       |      2       |    1224

您将需要使用 DISTINCT,但您还需要计算 ID,而不是外键:

SELECT
    table1.name,
    COUNT(DISTINCT table2.id) AS table2_count,
    COUNT(DISTINCT table3.id) AS table3_count,
    COUNT(DISTINCT table4.id) AS table4_count,
    SUM(table4.size) AS table4_size
FROM table1
LEFT JOIN table2 ON table1.id = table2.table1_id
LEFT JOIN table3 ON table2.id = table3.table2_id
LEFT JOIN table4 ON table3.id = table4.table3_id
WHERE table1.id = 1

这是一个fiddle.

解释: DISTINCT 关键字消除了所有重复值,生成了一个唯一值列表。

如果您 运行 您的查询没有 COUNT()SUM(),您将得到:

name  table1_id  table2_id  table3_id  size
test  1          1          1          1024 
test  1          1          1          200 
test  1          (null)     (null)     (null) 
test  1          (null)     (null)     (null) 

因此,如果您添加 COUNT()SUM(),您显然会得到:

name  table1_id  table2_id  table3_id  size
test  4          2          2          1224 

但是,在您的查询中使用 DISTINCT 将无济于事,因为您可以清楚地看到重复值,这将导致:

name  table1_id  table2_id  table3_id  size
test  1          1          1          1224 

现在,如果您 运行 我的查询没有 COUNT()SUM(),您将得到:

name  table1_id  table2_id  table3_id  size
test  1          1          1          1024 
test  1          1          2          200 
test  2          (null)     (null)     (null) 
test  3          (null)     (null)     (null) 

如果添加 COUNT()SUM(),您将获得与查询完全相同的结果:

name  table1_id  table2_id  table3_id  size
test  4          2          2          1224 

但是,因为这次您有不同的值(即并非所有值都是 1),所以现在如果您使用 DISTINCT 计算唯一值,您将得到:

name  table1_id  table2_id  table3_id  size
test  3          1          2          1224 

你可以这样试试吗

SELECT  table1.name,
    (SELECT COUNT(table2.table1_id) WHERE  table1.id = table2.table1_id ) AS table2_count,
    (SELECT COUNT(table3.table2_id) WHERE  table2.id = table3.table1_id ) AS table3_count,
    (SELECT COUNT(table4.table3_id) WHERE  table3.id = table4.table1_id ) AS table4_count,
    (SELECT SUM(table4.size) WHERE  table3.id = table4.table1_id ) AS table4_size
    FROM
    table1
    WHERE
    table1.id = 1