Question

任何人都可以告诉我为什么这段代码：

SELECT t.prob FROM 
(SELECT e1.evaled/IFNULL(NULLIF(e2.total,0),1) AS prob FROM 
    (SELECT COUNT(*) AS evaled FROM els WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els) AS e2) AS t
 UNION ALL (SELECT e1.evaled/IFNULL(NULLIF(e2.total,0),1) AS prob FROM 
    (SELECT COUNT(*) AS evaled FROM els2 WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els2) AS e2)
 UNION ALL (SELECT e1.evaled/IFNULL(NULLIF(e2.total,0),1) AS prob FROM 
    (SELECT COUNT(*) AS evaled FROM els3 WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els3) AS e2);

产生与此相同的输出而不是'prob'的总和？

SELECT SUM(t.prob) FROM 
(SELECT e1.evaled/IFNULL(NULLIF(e2.total,0),1) AS prob FROM 
    (SELECT COUNT(*) AS evaled FROM els WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els) AS e2) AS t
 UNION ALL (SELECT e1.evaled/IFNULL(NULLIF(e2.total,0),1) AS prob FROM 
    (SELECT COUNT(*) AS evaled FROM els2 WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els2) AS e2)
 UNION ALL (SELECT e1.evaled/IFNULL(NULLIF(e2.total,0),1) AS prob FROM 
    (SELECT COUNT(*) AS evaled FROM els3 WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els3) AS e2);

（代码基本上创建了一个列prob，它只包含三个表els，els2和els3中的每一个的一个值，然后将它们中的三个组合成一个列，我想要它的3个元素的总和）

我想出了其他代码。它起作用，它是一个更清晰的陈述，所以没关系：

SELECT SUM(t.evaled/IFNULL(NULLIF(t.total,0),1)) as sumatory FROM
(SELECT evaled,total FROM
 (SELECT COUNT(*) AS evaled FROM els WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els) AS e2
UNION ALL SELECT * FROM
 (SELECT COUNT(*) AS evaled FROM els2 WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els2) AS e2
UNION ALL SELECT * FROM
 (SELECT COUNT(*) AS evaled FROM els3 WHERE evals=0) AS e1 
     INNER JOIN (SELECT COUNT(*) AS total FROM els3) AS e2) as t;

Answer 1

不确定您的解决方案出了什么问题，但我认为我得到了您想要实现的目标。这样的方法怎么样：

SELECT 
 (SELECT COUNT(*)/(SELECT COUNT(*) FROM els) AS evaled FROM els WHERE evals=0)
 +
 (SELECT COUNT(*)/(SELECT COUNT(*) FROM els2) AS evaled FROM els2 WHERE evals=0) 
 +
 (SELECT COUNT(*)/(SELECT COUNT(*) FROM els3) AS evaled FROM els3 WHERE evals=0)
 /3

编辑：如果你想要平均值，我猜你最后需要一个/3 ..

mysql SUM（列）不会对列union上的元素求和

1 个答案: