Question

我正在寻找在Oracle中进行哈希和替换算法的方法。

我有一张客户表，里面有客户ID和客户电话号码。我有另一张表，其中列出了参考电话号码值。我想使用这些参考值并在主表中一致地替换它们。

应用此替换逻辑时应该有一致的屏蔽，即同一客户出现在不同表中的任何地方，电话号码替换应该相同。

主表名1： - CUS_PHONE_NUM
主表名称2： - CUS_PHONE_NUM_2
参考表名称： - REF_PHONE_NUMBER

我曾尝试过以下方法，但性能非常糟糕 - 对于20000名客户来说需要18分钟。你能否提出一个更好的方法来满足这个要求。

    DROP TABLE CUS_PHONE_NUM  ;
    DROP TABLE CUS_PHONE_NUM2 ;
    DROP TABLE REF_PHONE_NUMBER ;

    --  Main table 1 
    CREATE TABLE CUS_PHONE_NUM  (
            CUS_ID VARCHAR2(09) ,
      PHONE_NUMBER NUMBER(12)    ); 

    --  Main table 2
    CREATE TABLE CUS_PHONE_NUM_2  (
            CUS_ID VARCHAR2(09) ,
      PHONE_NUMBER NUMBER(12)    ); 

    --  reference table 
    CREATE TABLE REF_PHONE_NUMBER  ( REF_PHONE_NUMBER_VALUE NUMBER(12)); 


    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134321) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134322) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134323) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134324) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134325) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134326) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134327) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134328) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134329) ; 
    INSERT INTO REF_PHONE_NUMBER VALUES ( 0470134320) ; 

    TRUNCATE TABLE CUS_PHONE_NUM ;
    COMMIT; 

    INSERT INTO CUS_PHONE_NUM VALUES ( '401795401' , 0426881030  ) ; 
    INSERT INTO CUS_PHONE_NUM VALUES ( '401795402' , 0426881031  ) ; 
    INSERT INTO CUS_PHONE_NUM VALUES ( '401795403' , 0426881032  ) ; 
    INSERT INTO CUS_PHONE_NUM VALUES ( '401795404' , 0426881033  ) ; 
    INSERT INTO CUS_PHONE_NUM VALUES ( '401795405' , 0426881034  ) ; 
    INSERT INTO CUS_PHONE_NUM VALUES ( '401795406' , 0426881035  ) ; 
    INSERT INTO CUS_PHONE_NUM VALUES ( '401795407' , 0426881036  ) ; 

    TRUNCATE TABLE CUS_PHONE_NUM_2 ;
    COMMIT; 

    INSERT INTO CUS_PHONE_NUM_2 VALUES ( '401795401' , 0426881030  ) ; 
    INSERT INTO CUS_PHONE_NUM_2 VALUES ( '401795403' , 0426881032  ) ; 
    INSERT INTO CUS_PHONE_NUM_2 VALUES ( '401795405' , 0426881034  ) ; 
    INSERT INTO CUS_PHONE_NUM_2 VALUES ( '401795407' , 0426881036  ) ; 
    COMMIT ; 


    -- SQL1 to replace reference values for Main table 1 - CUS_PHONE_NUM

    WITH REF_PHONE_NUMBER_HASH AS
      (
       SELECT REF_PHONE_NUMBER_VALUE ,
              row_number() over(order by REF_PHONE_NUMBER_VALUE) - 1   REF_PHONE_NUMBER_VALUE_HASH,
              count(*) over() max_buckets
         FROM REF_PHONE_NUMBER
       ),
       CUS_PHONE_NUM_HASH  AS
       (select CUS_ID,
               PHONE_NUMBER
         from CUS_PHONE_NUM )

     SELECT c.*, n.* ,ora_hash(PHONE_NUMBER, n.max_buckets)   
       FROM CUS_PHONE_NUM_HASH c
       JOIN REF_PHONE_NUMBER_HASH n
         ON ora_hash(PHONE_NUMBER, n.max_buckets) = n.REF_PHONE_NUMBER_VALUE_HASH;



    -- SQL1 to replace reference values for Main table 2 - CUS_PHONE_NUM_2

    WITH REF_PHONE_NUMBER_HASH AS
      (
       SELECT REF_PHONE_NUMBER_VALUE ,
              row_number() over(order by REF_PHONE_NUMBER_VALUE) - 1   REF_PHONE_NUMBER_VALUE_HASH,
              count(*) over() max_buckets
         FROM REF_PHONE_NUMBER
       ),
       CUS_PHONE_NUM_HASH  AS
       (select CUS_ID,
               PHONE_NUMBER
         from CUS_PHONE_NUM_2 )

     SELECT c.*, n.* ,ora_hash(PHONE_NUMBER, n.max_buckets)   
       FROM CUS_PHONE_NUM_HASH c
       JOIN REF_PHONE_NUMBER_HASH n
         ON ora_hash(PHONE_NUMBER, n.max_buckets) = n.REF_PHONE_NUMBER_VALUE_HASH;

Answer 1

尝试将max_bucket移到外面。它改变了对测试数据的解释计划。但我不知道它对你真实的如何有效。我也不知道你的索引。

 WITH REF_PHONE_NUMBER_HASH AS
      (
       SELECT REF_PHONE_NUMBER_VALUE ,
              row_number() over(order by REF_PHONE_NUMBER_VALUE) - 1   REF_PHONE_NUMBER_VALUE_HASH
         FROM REF_PHONE_NUMBER
       ),
       CUS_PHONE_NUM_HASH  AS
       (select CUS_ID,
               PHONE_NUMBER
         from CUS_PHONE_NUM )
   , BUCKET as
     (select count(*) max_buckets from  REF_PHONE_NUMBER) 
     SELECT c.*, n.* ,ora_hash(PHONE_NUMBER, b.max_buckets)   
       FROM CUS_PHONE_NUM_HASH c
       cross join BUCKET b
       JOIN REF_PHONE_NUMBER_HASH n
         ON ora_hash(PHONE_NUMBER, b.max_buckets) = n.REF_PHONE_NUMBER_VALUE_HASH

也许你也可以考虑硬编码max_bucket值。在编码之后，您可以为此创建索引：

create index idx_CUS_PHONE_NUM_01 on CUS_PHONE_NUM
( ora_hash(Phone_NUmber, 255) )

Oracle中的引用查找替换

1 个答案: