簡單的表格。很難要求

Question

我有一個MySQL表：

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

我怎樣才能得到這個表與MySQL查詢？

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

我有一張大桌子，所以請求應該很快。 謝謝。

Answer 1

我認爲這應該做到這一點，雖然可能有更有效的方法或不需要聚合。

編輯：最初我誤解了要求。改變了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是另一個方法：

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