php - SQL查询MySQL到JSON

Question

我在创建 SQL 语句时遇到问题

我有 2 张 table :球队和杯赛

table Teams :team id, team
table Cups:team_id, EuroCup, Worldcup

目前我有:

select t.team as Stats,sum(Eurocups) as Value
  from cups c 
    join team t
       on c.team_id=t.team_id
  where c.team_id = 3 or c.team_id=1
 group by t.team

这给了我

Stats    Value
Spain     2
France    3

我希望创建 sql 查询，即比较 2 个不同的团队，例如:

Stats     Value1 Value2
Eurocups     2     3

所有这些都是针对 Google 图表堆叠条形图，它将从 MySQL 获取数据并转换为 JSON，例如

var data = google.visualization.arrayToDataTable([
['Stats',   'Value1', 'Value2'],
['Eurocups',    2,       3     ],
['Worldcups',   1,       2     ]

]);

Answer 1

类似这样的事情:

select t.team as Stats,
(select sum(t1.Eurocups) from cups t1 where t1.team_id = 3) as Value1,
(select sum(t2.Eurocups) from cups t2 where t2.team_id = 1) as Value2
 from cups c 
join team t
   on c.team_id=t.team_id
where c.team_id = 3 or c.team_id=1
group by t.team