Oracle 18c - REGEXP_REPLACE 的替代品
Oracle 18c - Alternative to REGEXP_REPLACE
迁移到Oracle 18c企业版后，基于函数的索引创建失败
这是我的索引 DDL：
CREATE INDEX my_index ON my_table
(UPPER( REGEXP_REPLACE ("DEPT_NUM",'[^[:alnum:]]',NULL,1,0)))
TABLESPACE my_tbspace
PCTFREE 10
INITRANS 2
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS UNLIMITED
PCTINCREASE 0
BUFFER_POOL DEFAULT
);
我收到以下错误：
ORA-01743: only pure functions can be indexed
01743. 00000 - "only pure functions can be indexed"
*Cause: The indexed function uses SYSDATE or the user environment.
*Action: PL/SQL functions must be pure (RNDS, RNPS, WNDS, WNPS). SQL
expressions must not use SYSDATE, USER, USERENV(), or anything
else dependent on the session state. NLS-dependent functions
are OK.
这是 18c 中的已知错误吗？如果不再支持这个基于函数的索引，还有什么写这个函数的方法？
很可能是 REGEXP_REPLACE 导致了问题，请参阅 。您可以使用用户定义的函数绕过限制（感谢 Bob Jarvis）
CREATE OR REPLACE FUNCTION KEEP_ALNUM(strIn IN VARCHAR2)
RETURN VARCHAR2
DETERMINISTIC
AS
BEGIN
RETURN UPPER(REGEXP_REPLACE(strIn, '[^[:alnum:]]', NULL, 1, 0));
END KEEP_ALNUM;
/
CREATE INDEX DEPTS_1 ON DEPTS(KEEP_ALNUM(DEPT_NUM));
只需确保函数具有关键字 DETERMINISTIC，然后您甚至可以像下面这样定义无用的函数并在其上创建函数索引
CREATE OR REPLACE FUNCTION SillyValue RETURN VARCHAR2 DETERMINISTIC
AS
BEGIN
RETURN DBMS_RANDOM.STRING('p', 20);
END;
/
有几个解决方法。
第一个是 hack。
您可能知道，当您创建 FBI 时，Oracle 会在其上创建隐藏列和索引。
此外，您甚至可以指定该列的名称而不是 FBI 表达式，Oracle 将使用索引。
set lines 70 pages 70
column column_name format a15
column data_type format a15
drop table my_table;
create table my_table(dept_num, dept_descr) as select rownum||'*', 'dummy' from dual connect by level <= 1e6;
create index my_index
on my_table(upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0)));
select column_name, data_type from user_tab_cols where table_name = 'MY_TABLE';
explain plan for
select * from my_table where upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0)) = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
explain plan for
select * from my_table where SYS_NC00003$ = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
输出
Table dropped.
Table created.
Index created.
COLUMN_NAME DATA_TYPE
--------------- ---------------
DEPT_NUM VARCHAR2
DEPT_DESCR CHAR
SYS_NC00003$ VARCHAR2
3 rows selected.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
因此，为了模仿 FBI，您可以创建一个隐藏列并在其上创建一个索引。
这可以在 Oracle 11g 中使用 dbms_stats.create_extended_stats.
完成
drop index my_index;
begin
for i in (select dbms_stats.create_extended_stats
(user, 'my_table', '(upper(regexp_replace("DEPT_NUM", ''[^[:alnum:]]'', null, 1, 0)))') as col_name
from dual)
loop
execute immediate(utl_lms.format_message('alter table %s rename column "%s" to my_hidden_col','my_table', i.col_name));
end loop;
end;
/
select column_name, data_type from user_tab_cols where table_name = 'MY_TABLE';
create index my_index on my_table(my_hidden_col);
explain plan for
select * from my_table where upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0)) = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
explain plan for
select * from my_table where MY_HIDDEN_COL = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
输出
Index dropped.
PL/SQL procedure successfully completed.
COLUMN_NAME DATA_TYPE
--------------- ---------------
DEPT_NUM VARCHAR2
DEPT_DESCR CHAR
MY_HIDDEN_COL VARCHAR2
3 rows selected.
Index created.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
从 Oracle 12c 开始记录了隐藏的列，因此它变得更加简单。
alter table my_table add (my_hidden_col invisible as
(upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0))) virtual);
create index my_index on my_table(my_hidden_col);
另一种方法是在没有正则表达式的情况下实现相同的逻辑。
create index my_index on my_table(
translate(upper(dept_num, '_'||translate(dept_num, '_ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789', '_'), '_')));
但在这种情况下，您必须确保谓词中所有带有正则表达式的表达式都替换为新表达式。
我发现最简单的解决方法是使用 NLS_UPPER 而不是 UPPER:
创建索引
CREATE INDEX my_index ON my_table
( REGEXP_REPLACE (NLS_UPPER("DEPT_NUM"),'[^[:alnum:]]',NULL,1,0)))
TABLESPACE my_tbspace
PCTFREE 10
INITRANS 2
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS UNLIMITED
PCTINCREASE 0
BUFFER_POOL DEFAULT
);
问题是 regexp_replace 不是确定性的。更改 NLS 设置时出现问题：
alter session set nls_language = english;
with rws as (
select 'STÜFF' v
from dual
)
select regexp_replace ( v, '[A-Z]+', '#' )
from rws;
REGEXP_REPLACE(V,'[A-Z]+','#')
#Ü#
alter session set nls_language = german;
with rws as (
select 'STÜFF' v
from dual
)
select regexp_replace ( v, '[A-Z]+', '#' )
from rws;
REGEXP_REPLACE(V,'[A-Z]+','#')
#
U-umlaut 位于英语字母表的末尾。但是在德语中的 U 之后。所以第一个语句 不会 替换它。第二个。
在 Oracle 数据库 12.1 和更早版本中，regexp_replace 错误地 标记为确定性。 12.2 通过使其成为非确定性来修复此问题。
仔细考虑是否有任何解决方法可以正确管理变音符号。
MOS 说明 2592779.1 对此进行了进一步讨论。
迁移到Oracle 18c企业版后，基于函数的索引创建失败
这是我的索引 DDL：
CREATE INDEX my_index ON my_table
(UPPER( REGEXP_REPLACE ("DEPT_NUM",'[^[:alnum:]]',NULL,1,0)))
TABLESPACE my_tbspace
PCTFREE 10
INITRANS 2
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS UNLIMITED
PCTINCREASE 0
BUFFER_POOL DEFAULT
);
我收到以下错误：
ORA-01743: only pure functions can be indexed
01743. 00000 - "only pure functions can be indexed"
*Cause: The indexed function uses SYSDATE or the user environment.
*Action: PL/SQL functions must be pure (RNDS, RNPS, WNDS, WNPS). SQL
expressions must not use SYSDATE, USER, USERENV(), or anything
else dependent on the session state. NLS-dependent functions
are OK.
这是 18c 中的已知错误吗？如果不再支持这个基于函数的索引，还有什么写这个函数的方法？
很可能是 REGEXP_REPLACE 导致了问题，请参阅
CREATE OR REPLACE FUNCTION KEEP_ALNUM(strIn IN VARCHAR2)
RETURN VARCHAR2
DETERMINISTIC
AS
BEGIN
RETURN UPPER(REGEXP_REPLACE(strIn, '[^[:alnum:]]', NULL, 1, 0));
END KEEP_ALNUM;
/
CREATE INDEX DEPTS_1 ON DEPTS(KEEP_ALNUM(DEPT_NUM));
只需确保函数具有关键字 DETERMINISTIC，然后您甚至可以像下面这样定义无用的函数并在其上创建函数索引
CREATE OR REPLACE FUNCTION SillyValue RETURN VARCHAR2 DETERMINISTIC
AS
BEGIN
RETURN DBMS_RANDOM.STRING('p', 20);
END;
/
有几个解决方法。
第一个是 hack。 您可能知道，当您创建 FBI 时，Oracle 会在其上创建隐藏列和索引。 此外，您甚至可以指定该列的名称而不是 FBI 表达式，Oracle 将使用索引。
set lines 70 pages 70
column column_name format a15
column data_type format a15
drop table my_table;
create table my_table(dept_num, dept_descr) as select rownum||'*', 'dummy' from dual connect by level <= 1e6;
create index my_index
on my_table(upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0)));
select column_name, data_type from user_tab_cols where table_name = 'MY_TABLE';
explain plan for
select * from my_table where upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0)) = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
explain plan for
select * from my_table where SYS_NC00003$ = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
输出
Table dropped.
Table created.
Index created.
COLUMN_NAME DATA_TYPE
--------------- ---------------
DEPT_NUM VARCHAR2
DEPT_DESCR CHAR
SYS_NC00003$ VARCHAR2
3 rows selected.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
因此，为了模仿 FBI，您可以创建一个隐藏列并在其上创建一个索引。
这可以在 Oracle 11g 中使用 dbms_stats.create_extended_stats.
drop index my_index;
begin
for i in (select dbms_stats.create_extended_stats
(user, 'my_table', '(upper(regexp_replace("DEPT_NUM", ''[^[:alnum:]]'', null, 1, 0)))') as col_name
from dual)
loop
execute immediate(utl_lms.format_message('alter table %s rename column "%s" to my_hidden_col','my_table', i.col_name));
end loop;
end;
/
select column_name, data_type from user_tab_cols where table_name = 'MY_TABLE';
create index my_index on my_table(my_hidden_col);
explain plan for
select * from my_table where upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0)) = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
explain plan for
select * from my_table where MY_HIDDEN_COL = '666';
select * from table(dbms_xplan.display(format => 'BASIC'));
输出
Index dropped.
PL/SQL procedure successfully completed.
COLUMN_NAME DATA_TYPE
--------------- ---------------
DEPT_NUM VARCHAR2
DEPT_DESCR CHAR
MY_HIDDEN_COL VARCHAR2
3 rows selected.
Index created.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
Explain complete.
PLAN_TABLE_OUTPUT
----------------------------------------------------------------------
Plan hash value: 2234884270
--------------------------------------------------------
| Id | Operation | Name |
--------------------------------------------------------
| 0 | SELECT STATEMENT | |
| 1 | TABLE ACCESS BY INDEX ROWID BATCHED| MY_TABLE |
| 2 | INDEX RANGE SCAN | MY_INDEX |
--------------------------------------------------------
9 rows selected.
从 Oracle 12c 开始记录了隐藏的列，因此它变得更加简单。
alter table my_table add (my_hidden_col invisible as
(upper(regexp_replace(dept_num, '[^[:alnum:]]', null, 1, 0))) virtual);
create index my_index on my_table(my_hidden_col);
另一种方法是在没有正则表达式的情况下实现相同的逻辑。
create index my_index on my_table(
translate(upper(dept_num, '_'||translate(dept_num, '_ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789', '_'), '_')));
但在这种情况下，您必须确保谓词中所有带有正则表达式的表达式都替换为新表达式。
我发现最简单的解决方法是使用 NLS_UPPER 而不是 UPPER:
创建索引CREATE INDEX my_index ON my_table
( REGEXP_REPLACE (NLS_UPPER("DEPT_NUM"),'[^[:alnum:]]',NULL,1,0)))
TABLESPACE my_tbspace
PCTFREE 10
INITRANS 2
MAXTRANS 255
STORAGE (
INITIAL 64K
MINEXTENTS 1
MAXEXTENTS UNLIMITED
PCTINCREASE 0
BUFFER_POOL DEFAULT
);
问题是 regexp_replace 不是确定性的。更改 NLS 设置时出现问题：
alter session set nls_language = english;
with rws as (
select 'STÜFF' v
from dual
)
select regexp_replace ( v, '[A-Z]+', '#' )
from rws;
REGEXP_REPLACE(V,'[A-Z]+','#')
#Ü#
alter session set nls_language = german;
with rws as (
select 'STÜFF' v
from dual
)
select regexp_replace ( v, '[A-Z]+', '#' )
from rws;
REGEXP_REPLACE(V,'[A-Z]+','#')
#
U-umlaut 位于英语字母表的末尾。但是在德语中的 U 之后。所以第一个语句 不会 替换它。第二个。
在 Oracle 数据库 12.1 和更早版本中，regexp_replace 错误地 标记为确定性。 12.2 通过使其成为非确定性来修复此问题。
仔细考虑是否有任何解决方法可以正确管理变音符号。
MOS 说明 2592779.1 对此进行了进一步讨论。