我正在尝试在一个查询中COUNT()多个表,但我无法让它工作。这是我到目前为止所拥有的。
TABLES
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
我从以上查询得到的结果:
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
任何帮助将不胜感激
答案 0 :(得分:1)
您需要使用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fiddle。
说明:DISTINCT
关键字会消除所有重复值,从而产生唯一值列表。
如果您在没有COUNT()
和SUM()
的情况下运行查询,则会获得:
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()
和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的值,这将导致:
name table1_id table2_id table3_id size test 1 1 1 1224
现在,如果您在没有COUNT()
和SUM()
的情况下运行我的查询,则会得到:
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)
如果您添加COUNT()
和SUM()
,则会得到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,则得到:
name table1_id table2_id table3_id size test 3 1 2 1224
答案 1 :(得分:0)
你能这样试试吗
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