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'
;

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`

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。

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'
;

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。

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