与内Update语句加入甲骨文

Question

我有在MySQL工作正常查询，但是当我在Oracle运行它，我得到以下错误：

  

SQL错误：ORA-00933：SQL命令未正确结束，点击   00933. 00000 - “SQL命令不能正常结束”

在查询是：

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

Answer 1

这语法不是Oracle中有效。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);

或者你的可能的能够做到这一点：

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

（这取决于如果内嵌视图由Oracle视为更新）。

Answer 2

使用这样的：

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

MERGE与WHERE子句：

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;

您需要的WHERE条款，因为ON子句中引用的列不能被更新。

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

不要使用上述一些问题的答案。

一些建议使用嵌套的SELECT的，不这样做，这是极为缓慢。如果你有大量的记录更新，使用加入，所以是这样的：

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;

请参阅此链接的更多细节。 http://geekswithblogs.net/WillSmith /archive/2008/06/18/oracle-update-with-join-again.aspx 。

此外，确保有对所有要加入的表的主键。

Answer 6

正如所指出的此处，对于第一溶液的一般语法提出由Tony安德鲁斯是：

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
                         )

我觉得这很有趣，特别是如果你想更新多个字段。

Answer 7

这下面的语法为我工作。

UPDATE
(SELECT A.utl_id,
    b.utl1_id
    FROM trb_pi_joint A
    JOIN trb_tpr B
    ON A.tp_id=B.tp_id Where A.pij_type=2 and a.utl_id is null
)
SET utl_id=utl1_id;

Answer 8

它工作正常预言

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

Answer 9

使用的描述代替递减为表2，

update
  table1
set
  value = (select code from table2 where description = table1.value)
where
  exists (select 1 from table2 where description = table1.value)
  and
  table1.updatetype = 'blah'
;

Answer 10

UPDATE table1 t1
SET t1.value = 
    (select t2.CODE from table2 t2 
     where t1.value = t2.DESC) 
WHERE t1.UPDATETYPE='blah';

Answer 11

UPDATE IP_ADMISSION_REQUEST ip1
SET IP1.WRIST_BAND_PRINT_STATUS=0
WHERE IP1.IP_ADM_REQ_ID        =
  (SELECT IP.IP_ADM_REQ_ID
  FROM IP_ADMISSION_REQUEST ip
  INNER JOIN VISIT v
  ON ip.ip_visit_id=v.visit_id
  AND v.pat_id     =3702
  ); `enter code here`

Answer 12

就像完整性的问题，因为我们正在谈论甲骨文，这可能做得一样好：

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;
/

Answer 13

UPDATE (SELECT T.FIELD A, S.FIELD B
FROM TABLE_T T INNER JOIN TABLE_S S
ON T.ID = S.ID)
SET B = A;

A和B是别名字段，就不需要以指向该表。

Answer 14

update table1  a 
   set a.col1='Y' 
 where exists(select 1 
                from table2 b
               where a.col1=b.col1 
                 and a.col2=b.col2
             )