Question

我已经检查了其他堆栈溢出问题，但没有解释如何执行以下操作：

我有一个控制预订队列的应用程序，我需要列出每一行：

positionTotal - 表示基于ID较低的所有活动（状态）行的位置。 positionSamePeople - 表示基于所有具有较低ID和相同人数的活动（状态）行的排名。

预订表：

+---------------------------------------+
+ id INT AI                             +
+ status TINYINT (0-Inactive/1-Active)  +
+ people INT                            +
+---------------------------------------+

查询：

SELECT * FROM reservation
+----------------------+
+ id | status | people +
+----------------------+
+ 1  |    1   |    3   +
+ 2  |    0   |    1   +
+ 4  |    1   |    3   +
+ 6  |    1   |    1   +
+ 8  |    1   |    2   +
+ 9  |    0   |    3   +
+ 10 |    1   |    3   +
+ 11 |    1   |    2   +
+----------------------+

我想要的是什么：

+------------------------------------------------------------+
+ id | status | people | positionTotal | positionSamePeople  +
+------------------------------------------------------------+
+ 1  |    1   |    3   |       1       |          1          +
+ 4  |    1   |    3   |       2       |          2          +
+ 6  |    1   |    1   |       3       |          1          +
+ 8  |    1   |    2   |       4       |          1          +
+ 10 |    1   |    3   |       5       |          3          +
+ 11 |    1   |    2   |       6       |          2          +
+------------------------------------------------------------+

我可以查询SELECT * FROM reservation WHERE status=1，然后在PHP中计算positionTotal和positionSamePeople，但这似乎不是一个好选择。

Answer 1

子查询很容易做到：

SELECT r1.*,
  (SELECT COUNT(*) +1
   FROM reservation r2
   WHERE r2.status = 1
     AND r2.id < r1.id) AS positionTotal,
  (SELECT COUNT(*) +1
   FROM reservation r3
   WHERE r3.status = 1
     AND r3.people = r1.people
     AND r3.id < r1.id) AS positionSamePeople
FROM reservation r1

Mysql获取行位置

1 个答案: