[20180926]神奇的规避ORA-01795方法.txt

[20180926]神奇的规避ora-01795方法.txt

–//大家知道in里面的值限制1000个值,如果超出报ora-01795错误.

d:\> ooerr 01795
01795, 00000, “maximum number of expressions in a list is 1000”

d:\> oerr ora 01795
01795, 00000, “maximum number of expressions in a list is 1000”
// *cause:    number of expressions in the query exceeded than 1000.
//            note that unused column/expressions are also counted
//            maximum number of expressions that are allowed are 1000.
// *action:   reduce the number of expressions in the list and resubmit.

–//别人的awr报表我仔细查看,发现对方规避ora-01795的方法,使用集合的概念,仔细一想难道集合就没有1000个的限制吗?
–//我仔细看对方程序是写死的,2000个集合,明显这样是没有问题的.
–//有时候还是给佩服开发的想象力,我还是做一些例子说明问题.

1.环境:
scott@test01p> @ ver1

port_string                    version        banner                                                                               con_id
—————————— ————– ——————————————————————————– ———-
ibmpc/win_nt64-9.1.0           12.1.0.1.0     oracle database 12c enterprise edition release 12.1.0.1.0 – 64bit production              0

scott@test01p> create table t as select * from all_objects;
table created.

scott@test01p> create unique index i_t_object_id on t(object_id);
index created.

–//分析表略.

2.测试1:
variable b1 number;
variable b2 number;
exec :b1 :=42;
exec :b2 :=47;

scott@test01p> select * from t where (1,object_id) in ((1,:b1),(1,:b2));
owner object_name subobject_name object_id data_object_id object_type          created             last_ddl_time       timestamp
—– ———– ————– ——— ————– ——————– ——————- ——————- ——————-
sys   i_icol1                           42             42 index                2013-06-28 09:03:29 2013-06-28 09:03:29 2013-06-28:09:03:29
sys   i_user2                           47             47 index                2013-06-28 09:03:29 2013-06-28 09:03:29 2013-06-28:09:03:29

–//ok没有问题,里面仅仅包含2个集合,如果更多呢?

3.测试2:
select ‘variable b’||to_char(level)||’ number;’ txt from dual connect by level<=1001
union all
select ‘exec :b’||to_char(level)||’ :=’|| to_char(level) txt from dual connect by level<=1001
union all
select ‘select * from t where (1,object_id) in (‘ txt from dual
union all
select ‘(1,:b’||to_char(level)||’),’ txt from dual connect by level<=1000
union all
select ‘(1,:b’||to_char(1001)||’));’ txt from dual ;

–//将输出保存1个脚本,适当编辑执行.ok,没有任何问题.执行计划如下:
plan hash value: 3525592940

—————————————————————————————————————————————-
| id  | operation                    | name          | starts | e-rows |e-bytes| cost (%cpu)| e-time   | a-rows |   a-time   | buffers |
—————————————————————————————————————————————-
|   0 | select statement             |               |      1 |        |       |   196 (100)|          |    891 |00:00:00.01 |      40 |
|   1 |  inlist iterator             |               |      1 |        |       |            |          |    891 |00:00:00.01 |      40 |
|   2 |   table access by index rowid| t             |   1001 |   1001 |   112k|   196   (0)| 00:00:01 |    891 |00:00:00.01 |      40 |
|*  3 |    index unique scan         | i_t_object_id |   1001 |   1001 |       |   178   (0)| 00:00:01 |    891 |00:00:00.01 |      22 |
—————————————————————————————————————————————-

–//1001个集合没有任何问题.
–//修改b200-b999变量等于0,测试看看:
plan hash value: 3525592940

—————————————————————————————————————————————-
| id  | operation                    | name          | starts | e-rows |e-bytes| cost (%cpu)| e-time   | a-rows |   a-time   | buffers |
—————————————————————————————————————————————-
|   0 | select statement             |               |      1 |        |       |   196 (100)|          |    193 |00:00:00.01 |      16 |
|   1 |  inlist iterator             |               |      1 |        |       |            |          |    193 |00:00:00.01 |      16 |
|   2 |   table access by index rowid| t             |    203 |   1001 |   112k|   196   (0)| 00:00:01 |    193 |00:00:00.01 |      16 |
|*  3 |    index unique scan         | i_t_object_id |    203 |   1001 |       |   178   (0)| 00:00:01 |    193 |00:00:00.01 |      12 |
—————————————————————————————————————————————-

–//可以发现starts=203,重复值不会扫描的.

4.我的估计:
–//对方2000个集合写死的,我估计可能是一个数组,初始赋值0(或者查询不到的值),然后赋值.
–//通过这样的方式实现绑定变量.
–//我个人还是趋向这样的方式使用临时表更加合理简单一些.

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