SQL优化案例-从执行计划定位SQL问题(三)

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。本文将结合实际案例,给出优化方案。

作者 姚崇·沃趣科技高级数据库技术专家
出品 沃趣科技

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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