Question

我有两张桌子

TEAMS - team_id和team_name（以及其他内容）
SCHEDULE - game_id，team_a，team_b，team_a_id，team_a_id（以及其他内容）

我正在尝试创建一个结果，其中我发现team_name（或team_id）在team_a或team_b列中显示的次数（或{ {1}}或team_a_id）

TEAMS
team_id team_name
1001纽约
克利夫兰1011年 1021年底特律 1031休斯敦

SCHEDULE
game_id team_a team_b team_a_id team_b_id
1纽约克利夫兰1001 1011
2纽约底特律1001 1021
3 Cleveland Houston 1011 1031

答案：纽约2 克利夫兰2 底特律1 休斯顿1

Answer 1

单程

Select t.team_name,Sum(allteams.teamcount) as NumberOfTimes
From teams t
inner join(
select team_a_id as team_ID, 1 as teamcount From Schedule
Union 
Select team_b_id, 1 From Schedule
) allteams
on allteams.Team_ID = t.Team_ID
Group By t.team_name

Answer 2

也许

SELECT count(*)
FROM teams
WHERE $teamID IN (team_a, team_b)

Answer 3

看起来很简单，除非我遗漏了什么。只需选择一个计数，然后使用WHERE OR clause。

SELECT COUNT(*)
FROM schedule
WHERE team_a = 'Team Name'
OR team_b = 'Team Name'

可替换地：

WHERE team_a_id = 'Team ID'
OR team_b_id = 'Team ID'

Answer 4

尝试这样的事情：

select count(*)
from teams t
join schedule s on (s.team_a = t.team_name OR s.team_b = t.team_name)

MySQL两个表两列COUNT

4 个答案: