Question

我们如何在此处删除具有良好性能的子查询。如果我附加此子查询，它会提供正确的输出，但由于可用记录的数量而减慢了搜索速度。但在删除它之后，它会增加多倍。任何人都可以帮助省略这个子查询或任何改进以下查询的内容。

SELECT insurance_folder_cnt, MAX(lapsed_date) lapseddate 
FROM insurance_file ifl 
WHERE insurance_file_status_id=1 AND insurance_file_type_id in (2, 5, 6, 8, 9) AND (ISNULL(Base_Insurance_File_Cnt,0) = 0 OR ISNULL(Base_Insurance_File_Cnt, 0) = insurance_file_cnt) AND ISNULL(out_of_sequence_replaced, 0) = 0 
AND policy_version = (SELECT MAX(policy_version) FROM Insurance_File ifld WHERE ifl.insurance_folder_cnt=ifld.insurance_folder_cnt) GROUP BY insurance_folder_cnt having policy_version = max(policy_version)

Answer 1

Please use this for better performance:

SELECT  insurance_folder_cnt ,
        MAX(lapsed_date) lapseddate
FROM    insurance_file ifl
        CROSS APPLY ( SELECT    MAX(policy_version) AS maxVersion
                      FROM      Insurance_File ifld
                    ) TmpMaxVersion
WHERE   insurance_file_status_id = 1
        AND insurance_file_type_id IN ( 2, 5, 6, 8, 9 )
        AND ( ISNULL(Base_Insurance_File_Cnt, 0) = 0
              OR ISNULL(Base_Insurance_File_Cnt, 0) = insurance_file_cnt
            )
        AND ISNULL(out_of_sequence_replaced, 0) = 0
        AND policy_version = TmpMaxVersion.maxVersion
GROUP BY insurance_folder_cnt
HAVING  policy_version = MAX(policy_version)

Answer 2

也许：

SELECT insurance_folder_cnt, MAX(lapsed_date) lapseddate 
FROM insurance_file ifl 
join (select insurance_folder_cnt inffld, max(policy_version) pol_ver from insurace_file group by insurance_folder_cnt) ifld 
on (ifld.inffld = ifl.insurance_folder_cnt and pol_ver = ifl.policy_version)
WHERE insurance_file_status_id=1 AND insurance_file_type_id in (2, 5, 6, 8, 9)
 AND (ISNULL(Base_Insurance_File_Cnt,0) = 0 OR ISNULL(Base_Insurance_File_Cnt, 0) = insurance_file_cnt) 
 AND ISNULL(out_of_sequence_replaced, 0) = 0 
 GROUP BY insurance_folder_cnt

你也可以替换ISNULL函数：

(Base_Insurance_File_Cnt is null OR Base_Insurance_File_Cnt = insurance_file_cnt)

我们如何以良好的性能删除子查询

2 个答案: