Question

我有table1（id，name），table2（id [FK to table1.id]，user_id，visit）。对于每个table1.id，我需要从table1显示其名称，计算用户并显示属于此id的user_id，当table2.visit = 1时，计算用户并显示属于此id的user_id以及table2.visit = 0。此外，table2中的某些id可能在table2中不存在，对于这些id我想在计数器中有0。

结果表应如下所示：

id|name|count_visitors|list_of_visitors_id|count_non_visitors|list_of_non_visitors_id|
1 |a   |2             |5,10               |1                 |4                      |

你能帮忙查询吗？

Answer 1

SELECT t1.name, 
       COALESCE( visitors.visitors_count, 0) AS `count_visitors`,
       COALESCE( visitors.visitors_list, '') AS `list_of_visitors_id`, 
       COALESCE( nonvisitors.nonvisitors_count, 0 ) AS `count_non_visitors`,
       COALESCE( nonvisitors.nonvisitors_list, '') AS `list_of_non_visitors_id`
FROM table1 t1
LEFT JOIN
( SELECT id, GROUP_CONCAT( user_id ) AS visitors_list, COUNT(*) AS visitors_count
    FROM table2
    WHERE visit = 1
    GROUP BY id ) visitors
ON t1.id = visitors.id
LEFT JOIN
  ( SELECT id, GROUP_CONCAT( user_id ) AS nonvisitors_list, COUNT(*) AS nonvisitors_count
    FROM table2
    WHERE visit = 0
    GROUP BY id ) nonvisitors
ON t1.id = nonvisitors.id

亲自尝试：Sqlfiddle

如果你需要一个用户，只需在最后添加一行：

WHERE t1.id = 1

根据mysql列和组结果递增一个值或另一个值

1 个答案: