Question

这是我的查询

SELECT dia
FROM CRES
WHERE pro_id = 2
AND 8103434563 LIKE
  ( SELECT dial_pattern||'%' FROM CDIVN WHERE dial_id = 1
  );

现在select dial_pattern||'%' from CDIVN where dial_id = 1可以提供multiple个结果。因此，我的主要查询失败，原因是"sub query returns more than one row"。这是因为我提到了like。

但我的逻辑需要like，因为我希望8103434563具有来自表CDIVN的模式匹配条件。

如何修改我的查询。请帮忙。

=======

 CREATE TABLE CDIVN
   (    DIAL_PATTERN_ID NUMBER NOT NULL ENABLE, 
    DIAL_PATTERN VARCHAR2(30 BYTE) NOT NULL ENABLE, 
    OTHERS VARCHAR2(64 BYTE), 
     CONSTRAINT "CDIVN_PK" PRIMARY KEY ("DIAL_PATTERN_ID", "DIAL_PATTERN")

   );


Insert into CDIVN (DIAL_PATTERN_ID,DIAL_PATTERN,OTHERS) values (1,'810','abc');
Insert into CDIVN (DIAL_PATTERN_ID,DIAL_PATTERN,OTHERS) values (1,'811','xyz');
Insert into CDIVN (DIAL_PATTERN_ID,DIAL_PATTERN,OTHERS) values (1,'812','aaa');
Insert into CDIVN (DIAL_PATTERN_ID,DIAL_PATTERN,OTHERS) values (5,'999','www');
Insert into CDIVN (DIAL_PATTERN_ID,DIAL_PATTERN,OTHERS) values (9,'333','ewe');


  CREATE TABLE CRES
   (    PROFILE_ID NUMBER NOT NULL ENABLE, 
    PROFILE_NAME VARCHAR2(50 BYTE) NOT NULL ENABLE, 
    DIALLED_PATTERN VARCHAR2(15 BYTE), 
     CONSTRAINT "CRES_PK" PRIMARY KEY ("PROFILE_ID")

   ) ;


Insert into CRES (PROFILE_ID,PROFILE_NAME,DIALLED_PATTERN) values (1,'A','1');
Insert into CRES (PROFILE_ID,PROFILE_NAME,DIALLED_PATTERN) values (2,'B','5');
Insert into CRES (PROFILE_ID,PROFILE_NAME,DIALLED_PATTERN) values (3,'C','9');

我有其他来源的CRES-profile id和一个号码，分别为1和81034345。

现在select DIALLED_PATTERN from CRES where PROFILE_ID=1;

这会给我DIALLED_PATTERN为1。

现在select DIAL_PATTERN from CDIVN where DIALLED_PATTERN_ID = 1 ( DIALLED_PATTERN )

这将使DIAL_PATTERN为810,811,812。

现在，如果81034345与任何一个810％或811％或812％匹配。然后我需要810％作为我的答案。

Answer 1

如果我可以这么说，我也对sql有一个 IN LIKE 功能的疯狂痴迷。要执行此操作，您可以将其作为JOIN写在表CRES上，而不是将子查询放在您的条件中。

我看到的方式是子查询

( SELECT dial_pattern||'%' FROM CDIVN WHERE dial_id = 1
  );

条件中的

不会直接影响表CRES中的行，但会用作所有结果的过滤器。意思是，如果子查询没有返回（a）值，则整个查询也不应返回任何内容。

Answer 2

收集like-values并构建一个regexp-expression，例如

-- Test Data
with CDIVN as
 (select 1 as dial_id, 'ABC' as val
    from dual
  union all
  select 1 as dial_id, 'ACD' as val
    from dual
  union all
  select 1 as dial_id, 'XXA' as val from dual),

CRES as
 (select 2 as proc_id, 'ABCD' as val
    from dual
  union all
  select 2 as proc_id, 'DABCD' as val
    from dual
  union all
  select 2 as proc_id, 'ACF' as val
    from dual
  union all
  select 2 as proc_id, 'XXAF' as val from dual)

,
-- Build regexp expression: 1, 'ABC|ACD|XXA'
CDIVN_PATTERN as
 (select dial_id,
         listagg(val, '|') within group(order by dial_id) as val_pattern
    from CDIVN
   group by dial_id)

-- Use this expression by regexp_like
select *
  from CRES c
 where regexp_like(c.val,
                   (select '^' || p.val_pattern
                      from cdivn_pattern p
                     where p.dial_id = 1));

Answer 3

你正朝着正确的方向前进，只需要加入两个表，而不需要子查询，你可以直接使用LIKE dial_pattern||'%'。

SQL> SELECT a.*, b.*
  2  FROM CRES A
  3  JOIN CDIVN b
  4  ON (b.DIAL_PATTERN_ID = A.profile_id)
  5  WHERE TO_CHAR(8103434563) LIKE dial_pattern||'%';

PROFILE_ID PROFILE_NAME DIALLED_PATTERN DIAL_PATTERN_ID DIAL_PATTER OTHERS
---------- ------------ --------------- --------------- ----------- ------
         1 A            1                             1 810         abc

Answer 4

如果我理解正确，我会选择：

SELECT dia
FROM   CRES
WHERE  pro_id = 2
AND    exists (
         SELECT null
         FROM   CDIVN
         WHERE  dial_id = 1 and
                TO_CHAR(8103434563) like dial_pattern||'%');

Oracle DB：与多个值一样

4 个答案: