oracle - 如何将数据插入到架构中具有多个引用的对象关系表中

Question

我有一个表，其架构为 Table(number, ref, ref, varchar2, varchar2,...)。

我如何将一行数据插入到这个表中？

当我做：

insert into table 
values (1, select ref(p), ref(d), '239 F.3d 1343', '35 USC § 283', ... 
           from plaintiff p, defendant d
            where p.name='name1' and d.name='name2');

我收到“缺少表达式”错误。

如果我做：

insert into table
1, select ref(p), ref(d), ... 
from plaintiff p, defendant 
where p.name=...;

我收到“缺少关键字值”错误。

Answer 1

您的插入语法已关闭。尝试：

insert into table  
(select 1, ref(p), ref(d), '239 F.3d 1343', '35 USC § 283', ... 
   from plaintiff p, defendant d where p.name='name1' and d.name='name2');

一般来说，明确提及您要插入的列也是一种很好的做法，以避免列顺序更改后出现问题，以及自行记录代码：

insert into table (col1, col2, col3, ...) 
(select 1, ref(p), ref(d), '239 F.3d 1343', '35 USC § 283', ... 
   from plaintiff p, defendant d where p.name='name1' and d.name='name2');

Answer 2

给定这样的表...

SQL> create table cases
  2      (case_no number
  3       , plaintiff_ref   REF person_t SCOPE IS plaintiffs
  4       , defendant_ref   REF person_t SCOPE IS defendants
  5       , col1 varchar2(30)
  6       , col2 varchar2(30)
  7       )
  8  /

Table created.

SQL>

我们可以像这样填充它...

SQL> insert into cases
  2  select 1, ref(p), ref(d), '239 F.3d 1343', '35 USC § 283'
  3  from plaintiffs p, defendants d
  4  where p.id = 1000
  5  and d.id=2000
  6  /

1 row created.

SQL>

...或者像这样...

SQL> declare
  2      p_ref REF person_t;
  3      d_ref REF person_t;
  4  begin
  5      select ref(p) into p_ref
  6      from plaintiffs p
  7      where p.id = 1000;
  8      select ref(d) into d_ref
  9      from defendants d
 10      where d.id = 2000;
 11
 12      insert into cases
 13      values
 14      (2, p_ref, d_ref, 'YYT A.2e 789', '26 FTW § 169');
 15  end;
 16  /

PL/SQL procedure successfully completed.

SQL>

REF 长得令人瞠目结舌：

SQL> 从案例 2 中选择 * /

   CASE_NO
----------
PLAINTIFF_REF
--------------------------------------------------------------------------
DEFENDANT_REF
--------------------------------------------------------------------------
COL1                           COL2
------------------------------ ------------------------------
         1
0000220208771EFF0FAD71409F85A448C831C0C7B041CAA1874D514FDC9D18EF12DA22C12D
0000220208981D65F90A004146A1A390DC1048858777ECAC51136743B39A75F37D22DC1379
239 F.3d 1343                  35 USC § 283

         2
0000220208771EFF0FAD71409F85A448C831C0C7B041CAA1874D514FDC9D18EF12DA22C12D
0000220208981D65F90A004146A1A390DC1048858777ECAC51136743B39A75F37D22DC1379
YYT A.2e 789                   26 FTW § 169


SQL>