SQL优化案例-从执行计划定位SQL问题（三）
原创
当SQL出现问题，能从执行计划中快速的定位哪部分出现问题很重要，SQL文本如下（为保证客户隐私，已经将注释和文字部分去掉）：
SELECT /*+ index(i IDX_INVM_BEC)*/ RQ,JGM,BZ,CUSTOMER_TYPE,
B.CUSTOMER_NO,
B.CUSTOMER_NAME AS DKHM,
B.ACCT_NO DKZH,
B.STATUS,
B.LOAN_BAL,
P,
LX,
NVL((SELECT A.ACCT_NO_DESC || A.ACCT_NO_ALL2
FROM INVM_ZMQ A
WHERE A.ACCT_NO=I.ACCT_NO
AND A.ZHLB='3'),I.ACCT_NO) AS CKZH,
I.CURR_VAL,
(CASE WHEN B.TRANSFER_ACCT=I.ACCT_NO THEN '嘻嘻嘻' ELSE '' END) AS SM
FROM(
SELECT B.EXTDATE AS RQ,
B.BRANCH_NO AS JGM,
B.CURRENCY AS BZ,
C.CUSTOMER_TYPE,
B.CUSTOMER_NO,
C.CUSTOMER_NAME,
B.ACCT_NO,
'啊啊' AS STATUS,
B.LOAN_BAL,
B.UNPD_PRIN_BAL AS P,
ROUND(B.CAP_UNPD_INT,2)+
(CASE WHEN B.REPAY_SCHED IN ('M','G') OR L.REPAY_SCHED IN ('M','G') THEN 0
ELSE (CASE WHEN L.PIA_CAPN_FREQ='S' AND L.PAST_DUE_CAPN_FREQ='S'
THEN ROUND(B.THEO_UNPD_ARR_PRN,2)
+(CASE WHEN L.FINE1_COMPD_OPT='Y' THEN ROUND(B.THEO_UNPD_ARRS_INT,2) ELSE 0 END)
+(CASE WHEN L.FINE2_COMPD_OPT='Y' THEN ROUND(B.THEO_UNPD_INT_ARR,2) ELSE 0 END)
ELSE 0
END)
END) AS LX,
B.TRF_ACCT_NO AS TRANSFER_ACCT
FROM BORM PARTITION("BORM_2018-06-13") B
INNER JOIN LONP L ON L.ACCT_TYPE=B.ACCT_TYPE AND L.INT_CAT=B.INT_CAT
INNER JOIN JGDY J ON B.BRANCH_NO = J.JGM AND (J.JGM='1700' OR J.SJJGM='1700')
LEFT JOIN CB_ACCT C ON B.ACCT_NO = C.ACCT_NO AND C.SYS_ID = 'BOR'
WHERE B.BAD_DEBT_IND in ('02','52')
AND B.STAT<>'40'
) B
LEFT JOIN INVM PARTITION("INVM_2018-06-13") I
ON B.CUSTOMER_NO=I.CUSTOMER_NO
AND I.ACCT_DESC='S'
AND I.CURR_VAL<>0
AND I.BRANCH_NO IN (SELECT JGM FROM JGDY WHERE JGM='1700' OR SJJGM='1700')
AND I.EXTDATE = DATE'2018-06-13'
AND I.CURR_STATUS='00'
WHERE (B.P>0 OR B.LX>0.01);执行计划如下：
可以一眼定位到view部分导致整个执行缓慢，那么我们仔细分析下view部分是怎么执行的。INVM TABLE ACCESS BY LOCAL INDEX ROWID执行11分钟，总计13分钟执行完。
寻找view部分执行计划的入口，ID18和ID19做nested loop，返回结果17与21做NESTED LOOP，可以得知最先执行的是ID18，ID18走的iffs，且A-rows返回记录6256行数据，查看ID18谓词信息
18 - filter(("SJJGM"='1700' OR "JGM"='1700')) 从这部分再回到SQL文本寻找SQL代码是AND I.BRANCH_NO IN (SELECT JGM FROM JGDY WHERE JGM='1700' OR SJJGM='1700')也就是【SELECT JGM FROM JGDY WHERE JGM='1700' OR SJJGM='1700'】，难道这部分返回结果真的是6256行数据吗？带着疑问我查询了一下。
居然只返回了34行数据，为什么会这样子？
肯定是此处的JGDY_IDX3有什么问题，那么会有什么问题呢？往上看ID为8的JGDY_IDX3正确的返回了34行数据，又仔细看了下ID为18的JGDY_IDX3，starts184次，正好6256/184=34，那么原因找到了，正是因为ID4和ID15做NESTED LOOP，导致视图里面所有的部分都要多执行184次。按照上面的分析思路看ID4里面的执行计划都很正确，但是返回结果184行记录且ID4和ID15做nested loop，导致整个view部分缓慢。
那么就很好办了，ID4和ID15应该走hash join，查看outline data信息，还没办法使用db_name信息引导执行计划走hash join，那么只能改写SQL。
改写SQL如下：
SELECT RQ,JGM,BZ,CUSTOMER_TYPE,
B.CUSTOMER_NO,
B.CUSTOMER_NAME AS DKHM,
B.ACCT_NO DKZH,
B.STATUS,
B.LOAN_BAL,
P,
LX,
NVL((SELECT A.ACCT_NO_DESC || A.ACCT_NO_ALL2
FROM INVM_ZMQ A
WHERE A.ACCT_NO=C.ACCT_NO
AND A.ZHLB='3'),C.ACCT_NO) AS CKZH,
C.CURR_VAL,
(CASE WHEN B.TRANSFER_ACCT=C.ACCT_NO THEN '嘻嘻嘻' ELSE '' END) AS SM
FROM(
SELECT B.EXTDATE AS RQ,
B.BRANCH_NO AS JGM,
B.CURRENCY AS BZ,
C.CUSTOMER_TYPE,
B.CUSTOMER_NO,
C.CUSTOMER_NAME,
B.ACCT_NO,
'啊啊' AS STATUS,
B.LOAN_BAL,
B.UNPD_PRIN_BAL AS P,
ROUND(B.CAP_UNPD_INT,2)+
(CASE WHEN B.REPAY_SCHED IN ('M','G') OR L.REPAY_SCHED IN ('M','G') THEN 0
ELSE (CASE WHEN L.PIA_CAPN_FREQ='S' AND L.PAST_DUE_CAPN_FREQ='S'
THEN ROUND(B.THEO_UNPD_ARR_PRN,2)
+(CASE WHEN L.FINE1_COMPD_OPT='Y' THEN ROUND(B.THEO_UNPD_ARRS_INT,2) ELSE 0 END)
+(CASE WHEN L.FINE2_COMPD_OPT='Y' THEN ROUND(B.THEO_UNPD_INT_ARR,2) ELSE 0 END)
ELSE 0
END)
END) AS LX,
B.TRF_ACCT_NO AS TRANSFER_ACCT
FROM BORM PARTITION("BORM_2018-06-13") B
INNER JOIN LONP L ON L.ACCT_TYPE=B.ACCT_TYPE AND L.INT_CAT=B.INT_CAT
INNER JOIN JGDY J ON B.BRANCH_NO = J.JGM AND (J.JGM='1700' OR J.SJJGM='1700')
LEFT JOIN CB_ACCT C ON B.ACCT_NO = C.ACCT_NO AND C.SYS_ID = 'BOR'
WHERE B.BAD_DEBT_IND in ('02','52')
AND B.STAT<>'40'
) B
LEFT JOIN (SELECT /*+ index(I IDX_INVM_BEC) */ CUSTOMER_NO,ACCT_NO,CURR_VAL FROM
INVM PARTITION("INVM_2018-06-13") I
INNER JOIN JGDY ON JGDY.JGM=I.BRANCH_NO AND (JGM='1700' OR SJJGM='1700')
AND I.ACCT_DESC='S'
AND I.CURR_VAL<>0
AND I.EXTDATE = DATE'2018-06-13'
AND I.CURR_STATUS='00') C ON B.CUSTOMER_NO=C.CUSTOMER_NO
WHERE (B.P>0 OR B.LX>0.01); 
改写完之后SQL由13分钟变为5秒钟执行完，看ID19还要执行999k次，查询ID18谓词信息对应SQL如下，确实是要返回999k行数据。
10:59:23 report.QData>SELECT COUNT(*) FROM INVM PARTITION("INVM_2018-06-13") I WHERE I.BRANCH_NO IN (SELECT JGM FROM JGDY WHERE JGM='1700' or SJJGM='1700') AND I.EXTDATE=TO_DATE(' 2018-06-13 00:00:00', 'syyyy-mm-dd hh24:mi:ss') AND I.CURR_STATUS='00';
COUNT(*)
----------
999424
| 作者简介
姚崇·沃趣科技高级数据库技术专家
熟悉Oracle数据库内部机制，丰富的数据库及RAC集群层故障诊断、性能调优、OWI、数据库备份恢复及迁移经验。
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原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。
如有侵权，请联系 cloudcommunity@tencent.com 删除。