內部聯接和子查詢的mysql限制

Question

我有以下查詢：

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 

查詢的目的是提供表saturday_combinations中列ONE，TWO，THREE，FOUR，FIVE，SIX和SEVEN中包含的不同值的計數。 但是我想設置一個desc限制4，以便它只根據最后4行（最后4個最大索引）執行計數。 但我沒有讓它與工會合作。 最后添加順序和限制僅限於最終選擇，而不是獲取最后4行並計算它們的分布。 有小費嗎？

表模式如下：

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

索引是自動遞增，ONE-SEVEN具有整數值。

表中大約有3000行，我想根據最后n行計算每個值的出現次數。

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，則其計數應該增加。 如果我可以持續10行，則計數應該增加，其他數字應該隨着計數而出現。

這是一個真實表樣本的鏈接。 http://sqlfiddle.com/#!2/d035b

Answer 1

如果回答我的評論是yes那么您可以嘗試以下方法。 當你需要添加limit, order by union selects你需要用括號()包裝union queries 。

• SQLFIDDLE DEMO

碼：

(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)

如果對我的評論的回答是no ，那么請澄清。

以下是基於您的樣本日期的代碼：

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

根據OP編輯澄清並更新了問題。

引用： “但是我想限制它，以便它首先占用最后n行，然后對這n行中的值進行計數。這意味着，如果我有3列，其中3000行和35個整數隨機出現在這3000行中它應該計算每個整數出現的次數。“

• SQLFIDDLE DEMO2

查詢：

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;

輸出：

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

Answer 2

也許我在你的請求中遺漏了一些東西，但根據你想要的結果，為什么不只是取消數據並執行計數。

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

請參閱SQL Fiddle with Demo

您的樣本的結果是：

| 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 |

編輯＃1：根據您的更新，這可能需要您：

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

請參閱SQL Fiddle with Demo

結果：

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

Answer 3

如果您想查看最終解決方案： http ：//sqlfiddle.com/#！2/867a6/13感謝@bonCodigo的所有幫助