对于一份报告,我不得不写一个递归的存储过程GET_RECIPE_STEPS_ID(recipe_id)
。它返回类型为recipe的步骤ID。 I.E.
SELECT GET_RECIPE_STEPS_ID.ID FROM GET_RECIPE_STEPS_ID(3189)
It Returns
3189
3190
3191
3192
当我自己运行它时它很快(比如0.031秒的执行时间)。但是如果要在查询中使用IN子句则需要很长时间。像下面的查询花了差不多12分钟。
SELECT rs.RECIPEID
FROM RECIPESTEPS rs
WHERE rs.RECIPEID IN (select GET_RECIPE_STEPS_ID.ID from GET_RECIPE_STEPS_ID(3189))
这相当于以下查询,几乎和存储过程本身一样快(0.038sec)
Select rs.RECIPEID
FROM RECIPESTEPS rs
WHERE rs.RECIPEID IN (3189, 3190, 3191, 3192)
存储过程
CREATE OR ALTER PROCEDURE GET_RECIPE_STEPS_ID
(recipe_id integer)
RETURNS
(id integer)
AS
declare variable coType integer;
BEGIN
/* Recursive Procedure
* For Passed Recipe 'Recipe_id', it Returns the step's which are of type Recipe again.
*
* If any step is of type Recipe(i.e COTYPE = 1)
* Then it calls itself again for that step(Recipe)
*/
id =: recipe_id;
SUSPEND;
FOR SELECT rs.COMMODITYID, c.COTYPE
FROM RECIPESTEPS rs
LEFT JOIN COMMODITIES c ON c.COMMODITYID = rs.COMMODITYID
WHERE rs.RECIPEID =: recipe_id INTO :id, :coType
Do
BEGIN
IF(coType = 1)
THEN
FOR SELECT r.RECIPEID FROM RECIPES r WHERE r.LATEST = 1 AND r.COMMODITYID =:id into :id
DO
BEGIN
FOR SELECT GET_RECIPE_STEPS_ID.ID
FROM GET_RECIPE_STEPS_ID(:id) INTO :id
DO
BEGIN
SUSPEND;
END
END
END
END^
问题是双重的:
IN
开始并没有很好的表现如果您将查询转换为使用INNER JOIN
而不是IN
,它可能会表现得更好:
select rs.RECIPEID
from GET_RECIPE_STEPS_ID(3189) grs
inner join RECIPESTEPS rs
on rs.RECIPEID = grs.ID
我假设您的真实查询可能更复杂,因为否则只需select ID from GET_RECIPE_STEPS_ID(3189)
即可。
上面的查询将与IN
略有不同,例如,如果ID
在存储过程输出中多次出现,它现在也会产生多行。您可能需要相应调整。