Question

我的mysql计数有一个奇怪的问题。当我执行

   SELECT a.inc AS inc,
       a.cls AS cls,
       a.ord AS ord,
       a.fam AS fam,
       a.subfam AS subfam,
       a.gen AS gen,
       aspec AS spec,
       asubspec AS subspec
    FROM a
    WHERE (ainc = 11)

我获得：

没关系，因为我有2条记录。

执行时

 SELECT COUNT(DISTINCT a.inc) AS inc,
       COUNT(DISTINCT a.cls) AS cls,
       COUNT(DISTINCT a.ord) AS ord,
       COUNT(DISTINCT a.fam) AS fam,
       COUNT(DISTINCT asubfam) AS subfam,
       COUNT(DISTINCT a.gen) AS gen,
       COUNT(DISTINCT a.spec) AS spec,
       COUNT(DISTINCT a.subspec) AS subspec
    FROM a
    WHERE (a.inc = 11)
    GROUP BY a.inc

我获得了

这很奇怪，因为你看到gen, spec and subspec在一行上有0值。我知道count distinct不计算零值。我想要计算所有的值！= 0并且在计数不同之后我想得到

`1 | 2 | 2 | 2 | 2 | 1 | 1 | 1 |`

我也试试：

SELECT COUNT(DISTINCT a.inc) AS inc,
       SUM(if(a.cls <> 0, 1, 0)) AS cls,
       SUM(if(a.ord <> 0, 1, 0)) AS ord,
       SUM(if(a.fam <> 0, 1, 0)) AS fam,
       SUM(if(a.subfam <> 0, 1, 0)) AS subfam,
       SUM(if(a.gen <> 0, 1, 0)) AS gen,
       SUM(if(a.spec <> 0, 1, 0)) AS spec,
       SUM(if(a.subspec <> 0, 1, 0)) AS subspec
  FROM a
GROUP BY a.inc

和

SELECT COUNT(DISTINCT a.inc) AS inc,
           SUM(DISTINCT if(a.cls <> 0, 1, 0)) AS cls,
           SUM(DISTINCT if(a.ord <> 0, 1, 0)) AS ord,
           SUM(DISTINCT if(a.fam <> 0, 1, 0)) AS fam,
           SUM(DISTINCT if(a.subfam <> 0, 1, 0)) AS subfam,
           SUM(DISTINCT if(a.gen <> 0, 1, 0)) AS gen,
           SUM(DISTINCT if(a.spec <> 0, 1, 0)) AS spec,
           SUM(DISTINCT if(a.subspec <> 0, 1, 0)) AS subspec
FROM a
GROUP BY a.inc

但它不起作用，因为在第一种方法中没有区分并且将所有重复值加总大于0;在第二种情况下，它只给出1和0 那么，有人可以帮助我吗？先感谢您。利奥

Answer 1

我知道count distinct不计算零值。

我不知道你有什么想法，但这不正确。也许您正在考虑NULL值？获得所需结果的一种方法是在不同的计数中将0视为NULL。

尝试这样的事情（我也删除了小组，但没有帮助）：

SELECT COUNT(DISTINCT case when a.inc = 0 then null else a.inc end) AS inc,
       COUNT(DISTINCT case when a.cls = 0 then null else a.cls end) AS cls,
       COUNT(DISTINCT case when a.ord = 0 then null else a.ord end) AS ord,
       COUNT(DISTINCT case when a.fam = 0 then null else a.fam end) AS fam,
       COUNT(DISTINCT case when a.subfam = 0 then null else a.subfam end) AS subfam,
       COUNT(DISTINCT case when a.gen = 0 then null else a.gen end) AS gen,
       COUNT(DISTINCT case when a.spec = 0 then null else a.spec end) AS spec,
       COUNT(DISTINCT case when a.subspec = 0 then null else a.subspec end) AS subspec
    FROM a
    WHERE (a.inc = 11)

Answer 2

SELECT COUNT(DISTINCT a.inc) AS inc,
       COUNT(DISTINCT NULLIF(a.cls, 0)) AS cls,
       COUNT(DISTINCT NULLIF(a.ord, 0)) AS ord,
       COUNT(DISTINCT NULLIF(a.fam, 0)) AS fam,
       COUNT(DISTINCT NULLIF(a.subfam, 0)) AS subfam,
       COUNT(DISTINCT NULLIF(a.gen, 0)) AS gen,
       COUNT(DISTINCT NULLIF(a.spec, 0)) AS spec,
       COUNT(DISTINCT NULLIF(a.subspec, 0)) AS subspec
FROM a

再一次mysql“count distinct”

2 个答案: