简单的表格。很难要求

Question

我有一个MySQL表：

level  box 
1   1 
2   2 
3   2 
1   3 
2   1 
2   3 
2   3 
3   1 

我怎样才能得到这个表与MySQL查询？

  box1  box2   box3 
level1 1  0   1 
level2 1  1   2 
level3 1  1   0 

我有一张大桌子，所以请求应该很快。 谢谢。

Answer 1

我认为这应该做到这一点，虽然可能有更有效的方法或不需要聚合。

编辑：最初我误解了要求。改变了MAX()骨料COUNT()：

SELECT 
    CONCAT('level', level) AS levelname, 
    COUNT(CASE WHEN box = 1 THEN box ELSE NULL END) AS box1, 
    COUNT(CASE WHEN box = 2 THEN box ELSE NULL END) AS box2, 
    COUNT(CASE WHEN box = 3 THEN box ELSE NULL END) AS box3 
FROM tbl 
GROUP BY levelname 

使用JOIN是另一个方法：

SELECT 
    CONCAT('level', a.level) AS levelname, 
    b1.num AS box1, 
    b2.num AS box2, 
    b3.num AS box3 
FROM 
    table a 
    LEFT JOIN (SELECT level, COUNT(*) AS num FROM table WHERE box = 1) b1 ON a.level = b1.level 
    LEFT JOIN (SELECT level, COUNT(*) AS num FROM table WHERE box = 2) b2 ON a.level = b2.level 
    LEFT JOIN (SELECT level, COUNT(*) AS num FROM table WHERE box = 3) b3 ON a.level = b3.level