Question

我们有一个名为asamembr的表，其中包含两个字段：cust_code和mbrcode。

还有另一个表成员消息与外键具有相同的字段但是当我使用以下查询来创建约束时：

alter table 'informix'.messageclubmembership add constraint foreign key 
            (membership_number, member_code)
            references 'informix'.asamembr
            (cust_code, mbr_code) 
            on delete cascade 
            constraint fk_messageclubm926;

我收到此错误：

 Cannot find unique constraint or primary key on referenced table (informix.asamembr)

请告诉我如何在两个字段cust_code和mbr_code上查询表asamembr上是否存在主键？

Answer 1

首先查找PK的索引名称（pk_idx列）

select c.constrname, c.constrtype as tp , c.idxname as pk_idx , t2.tabname, c2.idxname
from sysconstraints c, systables t, outer (sysreferences r, systables t2, sysconstraints c2)
where t.tabname = "asamembr"
  and t.tabid = c.tabid
  and r.constrid = c.constrid
  and t2.tabid = r.ptabid
  and c2.constrid = r.constrid

其中constrtype：

constrtype CHAR（1）识别约束类型的代码：
  C =检查约束
  N =非NULL
  P =主键
  R =参考
  T =表
  U =唯一

然后，检查索引列（查找PK约束的相同索引名称）：

   select unique
        t.tabname
      , i.idxname
      , i.idxtype
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part1 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part2 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part3 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part4 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part5 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part6 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part7 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part8 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part9 )
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part10)
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part11)
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part12)
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part13)
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part14)
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part15)
      , (select c.colname from syscolumns c where c.tabid = i.tabid and c.colno = i.part16)
      from sysindexes i , systables t
      where i.tabid = t.tabid
        and t.tabname = "asamembr";

其中idxtype：

idxtype CHAR（1）索引类型：
  U =独特的   D =允许重复   G = Nonbitmap generali
  g =位图一般化   u =唯一，位图
  d =非唯一，位图

在Informix online manuals搜索“sysconstraints”或“sysindexes”

Answer 2

要查看表asamembr的表模式，可以在命令行中使用dbschema：

dbschema –d yourdbname –t asamembr

有关示例，请参阅here

或者您可以转到dbaccess yourdbname > Table > Info > asamembr并在那里查看表信息，但我更喜欢使用dbschema，因为它会在一个地方显示所有内容。

关于您的错误，引用的列（或使用多列约束格式时的列集，这是您的情况）必须具有唯一或主键约束。在你的情况下看起来并非如此。查看更多信息here

查询以检查informix中的表是否存在主键

2 个答案: