本文主要是介绍记一次oracle 连续记录合并处理 （LAG，LEAD，row_number()），对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

需求

下图中的数据,要求 取出 seq 1,2 ;合并3~6，取最早生效时间，和 最晚失效时间 ;合并7~10，取最早生效时间，和 最晚失效时间。

以上需求理解为:  连续的cust_id 需要合并

 测试数据

create table TMP_MIG_ACCOUNT

(

  acct_id  NUMBER(18) not null,

  seq_nbr  NUMBER(5) not null,

  cust_id  NUMBER(16),

  eff_date DATE not null,

  exp_date DATE not null

);

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 1, 700752288, to_date('01-01-2000', 'dd-mm-yyyy'), to_date('07-03-2013 17:39:55', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 2, 701059792, to_date('07-03-2013 17:39:56', 'dd-mm-yyyy hh24:mi:ss'), to_date('07-03-2013 17:40:13', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 3, 701059793, to_date('07-03-2013 17:40:14', 'dd-mm-yyyy hh24:mi:ss'), to_date('09-03-2013 13:02:04', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 4, 701059793, to_date('09-03-2013 13:02:05', 'dd-mm-yyyy hh24:mi:ss'), to_date('09-03-2013 13:35:01', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 5, 701059793, to_date('09-03-2013 13:35:02', 'dd-mm-yyyy hh24:mi:ss'), to_date('02-08-2014 12:14:22', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 6, 701059793, to_date('02-08-2014 12:14:23', 'dd-mm-yyyy hh24:mi:ss'), to_date('15-12-2014 15:05:08', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 7, 700752288, to_date('15-12-2014 15:05:09', 'dd-mm-yyyy hh24:mi:ss'), to_date('15-12-2014 15:06:08', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 8, 700752288, to_date('15-12-2014 15:06:09', 'dd-mm-yyyy hh24:mi:ss'), to_date('01-01-2019 21:59:23', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 9, 700752288, to_date('01-01-2019 21:59:24', 'dd-mm-yyyy hh24:mi:ss'), to_date('01-02-2019 19:04:08', 'dd-mm-yyyy hh24:mi:ss'));

insert into TMP_MIG_ACCOUNT (ACCT_ID, SEQ_NBR, CUST_ID, EFF_DATE, EXP_DATE)

values (90174285, 10, 700752288, to_date('01-02-2019 19:04:09', 'dd-mm-yyyy hh24:mi:ss'), to_date('01-01-9999', 'dd-mm-yyyy'));

处理过程

构建level 1 临时表

DROP TABLEMIG_ACCOUNT_2;

CREATE TABLE MIG_ACCOUNT_2 AS

SELECTA.ACCT_ID,

A.SEQ_NBR,

A.CUST_ID,

A.EFF_DATE,

A.EXP_DATE,

LAG(A.CUST_ID)OVER(PARTITIONBYA.ACCT_IDORDERBYA.SEQ_NBR)PREV_CUST_ID,

LEAD(A.CUST_ID)OVER(PARTITIONBYA.ACCT_IDORDERBYA.SEQ_NBR)NEXT_CUST_ID

FROM  TMP_MIG_ACCOUNT   A

ORDERBYA.ACCT_ID,A.SEQ_NBRASC;

SELECT*FROM  MIG_ACCOUNT_2  ;

构建level2 临时表

DROP TABLETMP_MIG_ACCOUNT_1;

CREATE TABLE TMP_MIG_ACCOUNT_1 AS

SELECTA.*,row_number()OVER(PARTITIONBYa.acct_idORDERBYa.seq_nbrASC)RN

FROMMIG_ACCOUNT_2 A

WHEREA.CUST_ID!=NVL(A.PREV_CUST_ID,0);

DROP TABLETMP_MIG_ACCOUNT_2;

CREATE TABLE TMP_MIG_ACCOUNT_2 AS

SELECT A.*,row_number()OVER(PARTITIONBYa.acct_idORDERBYa.seq_nbrASC)RN

FROM MIG_ACCOUNT_2 A

WHEREA.CUST_ID!=NVL(A.NEXT_CUST_ID,0);

SELECT*FROMTMP_MIG_ACCOUNT_1;

SELECT*FROMTMP_MIG_ACCOUNT_2;

组装数据

SELECTT1.ACCT_ID,

T1.SEQ_NBR,

T1.CUST_ID,

T1.EFF_DATE,

T1.EXP_DATE,

T1.PREV_CUST_ID,

T1.NEXT_CUST_ID,

T1.RN,

       T2.EXP_DATE LAST_EXP_DATE

FROMTMP_MIG_ACCOUNT_1 T1,TMP_MIG_ACCOUNT_2 T2

WHERET1.ACCT_ID=T2.ACCT_ID

ANDT1.CUST_ID=T2.CUST_ID

ANDT1.RN=T2.RN;

结果集为

这篇关于记一次oracle 连续记录合并处理 （LAG，LEAD，row_number()）的文章就介绍到这儿，希望我们推荐的文章对大家有所帮助，也希望大家多多支持为之网！