Oracle中实现MySQL show index from table命令SQL脚本分享

数据库  /  管理员 发布于 6年前   192

实验数据初始化:

复制代码 代码如下:
create table t as select * from hr.employees;
create index inx_t1 on t(employee_id,first_name desc,last_name);
create index inx_t2 on t(job_id,hire_date);

显示该表所有索引的信息。

以dba登录

复制代码 代码如下:
set linesize 300;
set pagesize 100;
col c1 format a20;
col c2 format a20;
col c3 format a20;
col c4 format a20;
col c5 format a20;
col INDEX_NAME format a20;
select INDEX_NAME,
max(decode(COLUMN_POSITION,1,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c1,
max(decode(COLUMN_POSITION,2,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c2,
max(decode(COLUMN_POSITION,3,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c3,
max(decode(COLUMN_POSITION,4,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c4,
max(decode(COLUMN_POSITION,5,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c5
from (
select INDEX_NAME,COLUMN_NAME,COLUMN_LENGTH,COLUMN_POSITION,DESCEND
from dba_ind_columns
where table_owner='LIHUILIN'
AND table_name='T'
order by INDEX_NAME,column_position
) group by INDEX_NAME;

以普通用户登录

复制代码 代码如下:
set linesize 300;
set pagesize 100;
col c1 format a20;
col c2 format a20;
col c3 format a20;
col c4 format a20;
col c5 format a20;
col INDEX_NAME format a20;
select INDEX_NAME,
max(decode(COLUMN_POSITION,1,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c1,
max(decode(COLUMN_POSITION,2,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c2,
max(decode(COLUMN_POSITION,3,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c3,
max(decode(COLUMN_POSITION,4,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c4,
max(decode(COLUMN_POSITION,5,COLUMN_NAME||','||COLUMN_LENGTH||','||DESCEND,null)) c5
from (
select INDEX_NAME,COLUMN_NAME,COLUMN_LENGTH,COLUMN_POSITION,DESCEND
from user_ind_columns
where table_name='T'
order by INDEX_NAME,column_position
) group by INDEX_NAME;

但是可以看到,以倒序创建的索引字段，都是以SYS等命名。

Oracle把这种倒序创建的索引字段看成函数索引。

它的信息保存在user_ind_expressions视图。

user_ind_expressions视图的COLUMN_EXPRESSION字段类型是long型。

王工的版本可以解决这个问题

复制代码 代码如下:
CREATE OR REPLACE FUNCTION long_2_varchar (
   p_index_name IN user_ind_expressions.index_name%TYPE,
   p_table_name IN user_ind_expressions.table_name%TYPE,
   p_COLUMN_POSITION IN user_ind_expressions.table_name%TYPE)
   RETURN VARCHAR2
AS
   l_COLUMN_EXPRESSION LONG;
BEGIN
   SELECT COLUMN_EXPRESSION
     INTO l_COLUMN_EXPRESSION
     FROM user_ind_expressions
    WHERE index_name = p_index_name
          AND table_name = p_table_name
          AND COLUMN_POSITION = p_COLUMN_POSITION;

   RETURN SUBSTR (l_COLUMN_EXPRESSION, 1, 4000);
END;
/

复制代码 代码如下:
set linesize 300;
set pagesize 100;
col c1 format a20;
col c2 format a20;
col c3 format a20;
col c4 format a20;
col c5 format a20;
col INDEX_NAME format a20;
SELECT INDEX_NAME,
         MAX (DECODE (COLUMN_POSITION, 1, COLUMN_NAME || ' ' || DESCEND, NULL))
            c1,
         MAX (DECODE (COLUMN_POSITION, 2, COLUMN_NAME || ' ' || DESCEND, NULL))
            c2,
         MAX (DECODE (COLUMN_POSITION, 3, COLUMN_NAME || ' ' || DESCEND, NULL))
            c3,
         MAX (DECODE (COLUMN_POSITION, 4, COLUMN_NAME || ' ' || DESCEND, NULL))
            c4,
         MAX (DECODE (COLUMN_POSITION, 5, COLUMN_NAME || ' ' || DESCEND, NULL))
            c5
    FROM ( SELECT a.INDEX_NAME,
                   REPLACE (
                      DECODE (
                         descend,
                         'DESC', long_2_varchar (b.index_name,
                                                 b.table_NAME,
                                                 b.COLUMN_POSITION),
                         a.column_name),
                      '"',
                      '')
                      COLUMN_NAME,
                   a.COLUMN_LENGTH,
                   a.COLUMN_POSITION,
                   DESCEND
              FROM user_ind_columns a
                   LEFT JOIN
                   user_ind_expressions b
                      ON a.index_name = b.index_name
                         AND a.table_name = b.table_name
             WHERE a.table_name = 'T'
          ORDER BY INDEX_NAME, column_position)
GROUP BY INDEX_NAME;