Oracle如何简化（子查询）的选择计数（*）？

Question

我被要求尝试简化count()查询，但是我不知道从哪里开始，查询是这样的：

SELECT COUNT( 1 )
FROM (
        SELECT DISTINCT a.col,b.colx,c.coly
        FROM  a
            JOIN b on a.id = b.id
            JOIN c on b.id = c.id
        WHERE a.xyz = 'something'
        AND   b.hijk = 'something else'
        AND c.id IN (
                SELECT cid
                FROM cwa
                WHERE csid = 22921
            )
        ORDER BY
            e.create_timestamp DESC
    );

有人告诉我可以简化SELECT COUNT(1) FROM (subquery) ，这怎么做？

我已经尝试了几件事，但是结果与上面的查询不同。

Answer 1

除非您在rownum上过滤结果，否则子查询中的order-by不会有用（有时会出错，具体取决于上下文）。 您可以用联接替换内部子查询：

SELECT COUNT(*)
FROM (
    SELECT DISTINCT a.col,b.colx,c.coly
    FROM  a
    JOIN b on a.id = b.id
    JOIN c on b.id = c.id
    JOIN cwa on c.id cwa.cid
    WHERE a.xyz = 'something'
    AND   b.hijk = 'something else'
    AND   cwa.csid = 22921
);

如果您可以识别出您选择的三列中没有出现的字符，甚至可以不使用子查询就可以做到这一点，因此您可以将其用作分隔符。 例如，如果您永远不会出现波浪号，则可以执行以下操作：

SELECT COUNT(DISTINCT a.col ||'~'|| b.colx ||'~'|| c.coly)
FROM  a
JOIN b on a.id = b.id
JOIN c on b.id = c.id
JOIN cwa on c.id cwa.cid
WHERE a.xyz = 'something'
AND   b.hijk = 'something else'
AND   cwa.csid = 22921;

尽管这是更简单还是更清晰是一个意见问题。

---

由于count()仅接受单个参数，并且您希望计算这三列的（不同的）组合，因此该机制将所有三个列连接为一个字符串，然后计算该字符串的出现次数。 添加了定界符，以便您可以区分歧义的列值，例如在CTE中使用人为的示例：

with cte (col1, col2) as (
  select 'The', 'search' from dual
  union all select 'These', 'arch' from dual
)
select col1, col2,
  col1 || col2 as bad,
  col1 ||'~'|| col2 as good
from cte;

COL1  COL2   BAD         GOOD        
----- ------ ----------- ------------
The   search Thesearch   The~search  
These arch   Thesearch   These~arch  

通过简单的“坏”串联，两行看起来是相同的。 通过添加定界符以制作“好”版本，您仍然可以区分它们，因此计算不同的串联值可获得正确的答案：

with cte (col1, col2) as (
  select 'The', 'search' from dual
  union all select 'These', 'arch' from dual
)
select count(distinct col1 || col2) as bad_count,
  count (distinct col1 ||'~'|| col2) as good_count
from cte;

 BAD_COUNT GOOD_COUNT
---------- ----------
         1          2

如果col1以波浪号结尾，或者col2以波浪号开头，您将回到歧义：

with cte (col1, col2) as (
  select 'The~', 'search' from dual
  union all select 'The', '~search' from dual
)
select col1, col2,
  col1 || col2 as bad,
  col1 ||'~'|| col2 as still_bad
from cte;

COL1 COL2    BAD         STILL_BAD   
---- ------- ----------- ------------
The~ search  The~search  The~~search 
The  ~search The~search  The~~search 

因此分隔符必须是在任何值中都找不到的东西。

Answer 2

尝试使用

    SELECT count(DISTINCT a.col)
    FROM  a
        JOIN b on a.id = b.id
        JOIN c on b.id = c.id
    WHERE a.xyz = 'something'
    AND   b.hijk = 'something else'
    AND c.id IN (
            SELECT cid
            FROM cwa
            WHERE csid = 22921
        );

因为order by会增加您不必要的执行时间