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);

或者你可以这样做:

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Oracle是否认为内联视图是可更新的 (第二个语句是否可更新取决于列出的一些规则 这里)。

 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

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;

对table2使用description而不是desc，

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'
;

下面的语法适合我。

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;

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

用WHERE子句合并:

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WHERE子句，因为ON子句中引用的列不能更新。

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

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

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`

如这里所示，Tony Andrews提出的第一个解决方案的通用语法是:

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，特别是当你想要更新多个字段时。

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。

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，不需要指向表。

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;
/

甲骨文基地在这方面做得很好。

https://oracle-base.com/articles/misc/updates-based-on-queries

从这个链接-我使用了上述查询的修改，这对我来说没有作用(答案来自mathguy，它使用rowid)

MERGE /*+ APPEND PARALLEL(8) */ INTO dest_table tt
USING source_table st
ON (tt.identifier = st.identifier)
WHEN MATCHED THEN
  UPDATE SET tt.number = st.number;

这里我有两个表:source和dest，它们都有一个共同的varchar字段，我将源标识字段(PK)添加到dest表中。