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`

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

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。

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 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`

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。