在Oracle上使用内部联接更新语句

Question

我有一个在MySQL中工作正常的查询,但是当我在Oracle上运行时,我收到以下错误:

SQL错误:ORA-00933:SQL命令未正确结束
00933. 00000 - "SQL命令未正确结束"

查询是:

UPDATE table1
INNER JOIN table2 ON table1.value = table2.DESC
SET table1.value = table2.CODE
WHERE table1.UPDATETYPE='blah';

Answer 1

该语法在Oracle中无效.你可以这样做:

UPDATE table1 SET table1.value = (SELECT table2.CODE
                                  FROM table2 
                                  WHERE table1.value = table2.DESC)
WHERE table1.UPDATETYPE='blah'
AND EXISTS (SELECT table2.CODE
            FROM table2 
            WHERE table1.value = table2.DESC);

或者您可以这样做:

UPDATE 
(SELECT table1.value as OLD, table2.CODE as NEW
 FROM table1
 INNER JOIN table2
 ON table1.value = table2.DESC
 WHERE table1.UPDATETYPE='blah'
) t
SET t.OLD = t.NEW

(这取决于内联视图是否被Oracle认为是可更新的).

Answer 2

用这个:

MERGE
INTO    table1 trg
USING   (
        SELECT  t1.rowid AS rid, t2.code
        FROM    table1 t1
        JOIN    table2 t2
        ON      table1.value = table2.DESC
        WHERE   table1.UPDATETYPE='blah'
        ) src
ON      (trg.rowid = src.rid)
WHEN MATCHED THEN UPDATE
    SET trg.value = code;

Answer 3

合并where子句为我工作:

MERGE into table1
USING table2
ON (table1.id = table2.id)
WHEN MATCHED THEN UPDATE SET table1.startdate = table2.start_date
WHERE table1.startdate > table2.start_date;

您需要该MERGE子句,因为WHERE无法更新子句中引用的列.

Answer 4

 UPDATE ( SELECT t1.value, t2.CODE
          FROM table1 t1
          INNER JOIN table2 t2 ON t1.Value = t2.DESC
          WHERE t1.UPDATETYPE='blah')
 SET t1.Value= t2.CODE

Answer 5

不要使用上面的一些答案.

有人建议使用嵌套的SELECT,不要这样做,这是极其缓慢的.如果要更新许多记录,请使用join,如下所示:

update (select bonus 
        from employee_bonus b 
        inner join employees e on b.employee_id = e.employee_id 
        where e.bonus_eligible = 'N') t
set t.bonus = 0;

有关详细信息,请参阅此链接. http://geekswithblogs.net/WillSmith/archive/2008/06/18/oracle-update-with-join-again.aspx.

此外,请确保您要加入的所有表都有主键.

Answer 6

如图所示这里,第一个解决方案由托尼·安德鲁斯提出的一般语法是:

update some_table s
set   (s.col1, s.col2) = (select x.col1, x.col2
                          from   other_table x
                          where  x.key_value = s.key_value
                         )
where exists             (select 1
                          from   other_table x
                          where  x.key_value = s.key_value
                         )

我认为这很有趣,特别是如果你想要更新多个字段.

Answer 7

甲骨文运行良好

merge into table1 t1
using (select * from table2) t2
on (t1.empid = t2.empid)
when matched then update set t1.salary = t2.salary