Question

想到打高尔夫球比赛我想计算一名球员在那里进行差点切换的次数（handicap_adjustment = -1），但我也希望包括那些没有被切断的球员。

PLAYER TABLE ...
PID Name
--- ----
1   Alan
2   Bob
3   Frank
4   Harry
5   Tom

CARD TABLE ...
PID HANDICAP_ADJUSTMENT
--- -------------------
1   -1
1   -1
2   0.2
2   0.2
2   0.2
2   0.2
3   0.2
3   0.2
4   -1
4   -1
4   -1
5   -1

RESULT I WANT IS ....
PID COUNT
--- -----
Alan   2
Bob    0
Frank  0
Harry  3
Tom    1

查询的任何想法？

我可以在牌表中获得带有handicap_adjustment = -1的pid计数，但不知道如何将其返回给玩家查询，该查询返回所有具有该计数的玩家：（

Answer 1

SELECT p.name,a.count FROM players p
LEFT JOIN (SELECT pid,COUNT(*) AS count FROM handicap_adjustment GROUP BY pid WHERE handicap_adjustment = -1 ) a ON a.pid = p.pid

Answer 2

您可以COUNT使用CASE表达式来获取总数。 CASE仅包含HANDICAP_ADJUSTMENT = -1：

的行

select p.name,
  count(case when HANDICAP_ADJUSTMENT = -1 then c.pid end) Total
from player p
left join card c
  on p.pid = c.pid
group by p.name

请参阅SQL Fiddle with Demo

或者您可以SUM()使用CASE来获取总数：

select p.name,
  sum(case when HANDICAP_ADJUSTMENT = -1 then 1 else 0 end) Total
from player p
left join card c
  on p.pid = c.pid
group by p.name

请参阅SQL Fiddle with Demo

两者都给出结果：

|  NAME | TOTAL |
-----------------
|  Alan |     2 |
|   Bob |     0 |
| Frank |     0 |
| Harry |     3 |
|   Tom |     1 |

MySQL连接表问题

2 个答案: