带有内部连接和子查询的mysql限制

Question

我有以下疑问：

SELECT saturday_combinations.index, v.val AS  `row` , COUNT( * )  AS  `count` 
FROM saturday_combinations  
INNER JOIN (

SELECT ONE AS val
FROM saturday_combinations 
WHERE ONE IS NOT NULL 

UNION 
SELECT TWO AS val
FROM saturday_combinations
WHERE TWO IS NOT NULL 

UNION 
SELECT THREE AS val
FROM saturday_combinations
WHERE THREE IS NOT NULL 

UNION 
SELECT FOUR AS val
FROM saturday_combinations
WHERE FOUR IS NOT NULL 

UNION 
SELECT FIVE AS val
FROM saturday_combinations
WHERE FIVE IS NOT NULL


UNION 
SELECT SIX AS val
FROM saturday_combinations
WHERE SIX IS NOT NULL 

UNION 
SELECT SEVEN AS val
FROM saturday_combinations
WHERE SEVEN IS NOT NULL 

) v  ON v.val = saturday_combinations.ONE
  OR v.val = saturday_combinations.TWO
  OR v.val = saturday_combinations.THREE
  OR v.val = saturday_combinations.FOUR
  OR v.val = saturday_combinations.FIVE
  OR v.val = saturday_combinations.SIX
  OR v.val = saturday_combinations.SEVEN 
  GROUP BY v.val 

查询的目的是提供表 saturday_combinations 中 ONE、TWO、THREE、FOUR、FIVE、SIX 和 SEVEN 列中包含的不同值的计数。但是我想设置一个 desc 限制 4 以便它只根据最后 4 行（最后四个最大索引）执行计数。但我没有让它与工会合作。在最后添加 order 和 limit 仅限制最终选择，而不是获取最后 4 行并计算它们的分布。有什么建议吗？

表架构如下：

index | ONE|TWO|THREE|FOUR|FIVE|SIX|SEVEN
 1       1   3   7     10    11  12  13
 2       3   4   5     30    31  22  23
 3       1   2   3      4     5   6   7
 4       1   2   3      4     5   6   7
 5       1   2   3      4     5   6   7
 6       1   2   3      4     5   6   7
 7       1   2   3      4     5   6   7
 8       1   2   3      4     5   6   7
 9       1   2   3      4     5   6   7
 10      1   2   3      4     5   6   7

索引是自动递增的，ONE-SEVEN 具有整数值。

表中有大约 3000 行，我想根据最后 n 行计算每个值的出现次数。

Ideal result for the last n rows where n = last 3 rows should be
Numbers|Count
   1       3
   2       3
   3       3
   4       3
   5       3
   6       3
   7       3

如果我增加 n 以包括最后 6 行，它们的计数应该会增加。如果我可以持续 10 行，那么计数应该会增加，并且其他数字应该与它们的计数一起出现。

这里是一个真实表格示例的链接。 http://sqlfiddle.com/#!2/d035b

Answer 1

如果对我的评论的回答是yes，那么您可以尝试以下方法。当您需要将limit, order by 添加到union selects 时，您需要用括号() 包裹union queries。

• SQLFIDDLE DEMO

代码：

(SELECT ONE AS val
FROM saturday_combinations 
WHERE ONE IS NOT NULL 
order by ONE desc limit 4)

UNION 
(SELECT TWO AS val
FROM saturday_combinations
WHERE TWO IS NOT NULL 
order by TWO desc limit 4)

UNION 
(SELECT THREE AS val
FROM saturday_combinations
WHERE THREE IS NOT NULL 
order by THREE desc limit 4)

如果对我的评论的回答是no，请澄清。

这是基于您的示例日期的代码：

select distinct x.one as uniqunumbers,
count(x.one) as counts
from(
sELECT DISTINCT 'one' 
AS col1, one FROM sat_comb
UNION ALL
SELECT DISTINCT 'two' 
AS col1, two FROM sat_comb
UNION ALL
SELECT DISTINCT 'three' 
AS col1, three FROM sat_comb
) as x
group by x.one;

UNIQUNUMBERS    COUNTS
1               1
3               2
4               1
5               1
7               1

根据 OP 编辑​​已澄清并更新了问题。

引用：“但是我想限制它，以便它首先获取最后 n 行，然后对这 n 行中的值进行计数。这意味着，如果我有 3 列，3000 行和 35整数随机出现在这 3000 行中，它应该计算每个整数出现的次数。”

• SQLFIDDLE DEMO2

查询：

select x.one as uniqunumbers,
    count(x.one) as counts
    from(
    (sELECT DISTINCT 'one' 
    AS col1, one FROM sat_comb
     order by id desc limit 4)
    UNION ALL
    (SELECT DISTINCT 'two' 
    AS col1, two FROM sat_comb
     order by id desc limit 4)
    UNION ALL
    (SELECT DISTINCT 'three' 
    AS col1, three FROM sat_comb
     order by id desc limit 4)
      UNION ALL
    (SELECT DISTINCT 'four' 
    AS col1, four FROM sat_comb
     order by id desc limit 4)
      UNION ALL
    (SELECT DISTINCT 'five' 
    AS col1, five FROM sat_comb
     order by id desc limit 4)
      UNION ALL
    (SELECT DISTINCT 'six' 
    AS col1, six FROM sat_comb
     order by id desc limit 4)
      UNION ALL
    (SELECT DISTINCT 'seven' 
    AS col1, seven FROM sat_comb
     order by id desc limit 4)
    ) as x
    group by x.one;

输出：

UNIQUNUMBERS    COUNTS
2               4
3               3
4               3
5               4
6               4
8               3
9               4
20              3

Answer 2

也许我在您的请求中遗漏了一些东西，但根据您想要的结果，为什么不直接unpivot 数据并执行计数。

select value, count(*) Total
from
(
  select 'one' col, one value
  from saturday_combinations
  union all
  select 'two' col, two value
  from saturday_combinations
  union all
  select 'three' col, three value
  from saturday_combinations
  union all
  select 'four' col, four value
  from saturday_combinations
  union all
  select 'five' col, five value
  from saturday_combinations
  union all
  select 'six' col, six value
  from saturday_combinations
  union all
  select 'seven' col, seven value
  from saturday_combinations
) src
group by value

见SQL Fiddle with Demo

您的样本结果是：

| VALUE | TOTAL |
-----------------
|     1 |     1 |
|     3 |     2 |
|     4 |     1 |
|     5 |     1 |
|     7 |     1 |
|    10 |     1 |
|    11 |     1 |
|    12 |     1 |
|    13 |     1 |
|    22 |     1 |
|    23 |     1 |
|    30 |     1 |
|    31 |     1 |

编辑 #1：根据您的更新，这可能需要您什么：

select value, count(*)
from
(
  select col, value
  from
  (
    select 'one' col, one value
    from saturday_combinations
    order by one 
    limit 3
  ) one
  union all
  select col, value
  from
  (
    select 'two' col, two value
    from saturday_combinations
    order by two desc
    limit 3
  ) two
  union all
  select col, value
  from
  (
    select 'three' col, three value
    from saturday_combinations
    order by three 
    limit 3
  ) three
  union all
  select col, value
  from
  (
    select 'four' col, four value
    from saturday_combinations
    order by four 
    limit 3
  ) four
  union all
  select col, value
  from
  (
    select 'five' col, five value
    from saturday_combinations
    order by five 
    limit 3
  ) five
  union all
  select col, value
  from
  (
    select 'six' col, six value
    from saturday_combinations
    order by six 
    limit 3
  ) six
  union all
  select col, value
  from
  (
    select 'seven' col, seven value
    from saturday_combinations
    order by seven 
    limit 3
  ) seven
) src
group by value

见SQL Fiddle with Demo

结果：

| VALUE | COUNT(*) |
--------------------
|     1 |        3 |
|     2 |        1 |
|     3 |        4 |
|     4 |        4 |
|     5 |        3 |
|     6 |        3 |
|     7 |        3 |

Answer 3

如果您想查看最终解决方案： http://sqlfiddle.com/#!2/867a6/13 感谢@bonCodigo 的所有帮助