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  • Oracle 多表关联更新

     
    
    drop table course;
    create table course (
    id integer,
    teacherNo integer,
    teacherDesc varchar2(100),
    teacherName varchar2(50),
    courseName varchar2(50)
    );
    insert into course values(1,100,'Mr.zhang','ZhangSan','English');
    insert into course values(2,101,'Mr.wang','WangWu','History');
    insert into course values(2,101,'Mr.wang','WangWu','Chinese');
    
    update course set teacherDesc='Good Teacher' where teacherNo=101;
    commit;
    drop table teacher;
    create table teacher(
    id integer,
    teacherDesc varchar2(100),
    teacherName varchar2(50)
    );
    insert into teacher values(100,'Mr.zhang','ZhangSan');
    insert into teacher values(101,'Mr.wang','WangWu');
    update teacher set teacherDesc='Excellent Teacher' where id=101;
    commit;
    select c.teacherdesc, c.teachername, t.teacherdesc, t.teachername
      from course c, teacher t
     where c.teacherno = t.id
       and (c.teacherdesc != t.teacherdesc or c.teachername != t.teachername);
    
    
    update course c
       set (c.teacherdesc, c.teachername) =
           (select t.teacherdesc, t.teachername
              from teacher t
             where c.teacherno = t.id
               and (c.teacherdesc != t.teacherdesc or
                   c.teachername != t.teachername))
     where exists (select 1
              from teacher t
             where c.teacherno = t.id
               and (c.teacherdesc != t.teacherdesc or
                   c.teachername != t.teachername));
    commit;
    select c.teacherdesc, c.teachername, t.teacherdesc, t.teachername
      from course c, teacher t
     where c.teacherno = t.id
       and (c.teacherdesc != t.teacherdesc or c.teachername != t.teachername);
    
    select c.teacherdesc, c.teachername, t.teacherdesc, t.teachername
      from course c, teacher t
     where c.teacherno = t.id;
    
       


    update时报ORA-01779:
    数据准备:

    CREATE TABLE test1 ( id integer primary key, num integer );    
    INSERT INTO test1 VALUES (1,0);   
    INSERT INTO test1 VALUES (2,0);   
    INSERT INTO test1 VALUES (3,0);   
    INSERT INTO test1 VALUES (4,0);  
    
    CREATE TABLE test2 ( id integer, num integer, upd integer ); 
    INSERT INTO test2 VALUES (1,10, 0);   
    INSERT INTO test2 VALUES (2,20, 1); 
    
    commit;  
    

    执行如下更新语句会报错:ORA-01779: 无法修改与非键值保存表对应的列
    01779, 00000, "cannot modify a column which maps to a non key-preserved table"
    // *Cause: An attempt was made to insert or update columns of a join view which
    //        map to a non-key-preserved table.
    // *Action: Modify the underlying base tables directly.

    UPDATE (SELECT T1.ID ID1, T1.NUM NUM1, T2.ID ID2, T2.NUM NUM2   
              FROM TEST1 T1, TEST2 T2   
             WHERE T1.ID = T2.ID   
               AND T2.UPD = 1)   
       SET NUM1 = NUM2;   
    

    这个错误的意思是,子查询的结果中,更新数据源(test2)的内容不唯一,导致被更新对象(test1)中的一行可能对应数据源(test2)中的多行。
    本例中,test2表的id不唯一,因此test2表中可能存在id相同但是num不相同的数据,这种数据是无法用来更新 test1 的。【这个报错属于事前检查,没有通过校验】

    解决方法就是保证数据源的唯一性,例如本例中可以为test2.id创建一个唯一索引:

    CREATE UNIQUE INDEX test2_idx_001 ON test2 (id);

    之后上面的更新就可以执行了。
    也可以使用如下命令来使用test2中记录具有唯一性

    alter table test2 modify id unique; 
    

    另外也可以强制 Oracle 执行,方法是加上 BYPASS_UJVC 注释。

    UPDATE (SELECT /*+ BYPASS_UJVC */  
             T1.ID ID1, T1.NUM NUM1, T2.ID ID2, T2.NUM NUM2   
              FROM TEST1 T1, TEST2 T2   
             WHERE T1.ID = T2.ID   
               AND T2.UPD = 1)   
       SET NUM1 = NUM2;  
    

    BYPASS_UJVC的作用是跳过Oracle的键检查。
    这样虽然能够执行了,但是如果test2中存在不唯一的数据,test1就会被更新多次而导致意想不到的结果。【有风险,不建议使用】  

    http://www.linuxidc.com/Linux/2012-08/69089.htm

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  • 原文地址:https://www.cnblogs.com/softidea/p/4891213.html
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