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

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。

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。

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