我有一个查询,在MySQL工作得很好,但当我在Oracle上运行它时,我得到以下错误:

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

查询为:

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

当前回答

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;

其他回答

不要使用上面的一些答案。

有些人建议使用嵌套SELECT,不要这样做,它非常慢。如果你有很多记录要更新,使用join,就像这样:

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;

更多细节请参见此链接。 http://geekswithblogs.net/WillSmith/archive/2008/06/18/oracle-update-with-join-again.aspx。

另外,确保所有要连接的表上都有主键。

它工作得很好

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

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;

只是作为一个完整的问题,因为我们谈论的是Oracle,这也可以做到:

declare
begin
  for sel in (
    select table2.code, table2.desc
    from table1
    join table2 on table1.value = table2.desc
    where table1.updatetype = 'blah'
  ) loop
    update table1 
    set table1.value = sel.code
    where table1.updatetype = 'blah' and table1.value = sel.desc;    
  end loop;
end;
/
UPDATE table1 t1
SET t1.value = 
    (select t2.CODE from table2 t2 
     where t1.value = t2.DESC) 
WHERE t1.UPDATETYPE='blah';