这称为相关更新

UPDATE table1 t1
   SET (name, desc) = (SELECT t2.name, t2.desc
                         FROM table2 t2
                        WHERE t1.id = t2.id)
 WHERE EXISTS (
    SELECT 1
      FROM table2 t2
     WHERE t1.id = t2.id )

假设连接产生保留键的视图，您还可以

UPDATE (SELECT t1.id, 
               t1.name name1,
               t1.desc desc1,
               t2.name name2,
               t2.desc desc2
          FROM table1 t1,
               table2 t2
         WHERE t1.id = t2.id)
   SET name1 = name2,
       desc1 = desc2

试试这个：

MERGE INTO table1 t1
USING
(
-- For more complicated queries you can use WITH clause here
SELECT * FROM table2
)t2
ON(t1.id = t2.id)
WHEN MATCHED THEN UPDATE SET
t1.name = t2.name,
t1.desc = t2.desc;

尝试

UPDATE Table1 T1 SET
T1.name = (SELECT T2.name FROM Table2 T2 WHERE T2.id = T1.id),
T1.desc = (SELECT T2.desc FROM Table2 T2 WHERE T2.id = T1.id)
WHERE T1.id IN (SELECT T2.id FROM Table2 T2 WHERE T2.id = T1.id);

Update table set column = (select...)

从来没有为我工作，因为集只需要1个值- SQL错误：ORA-01427：单行子查询返回多行。
解决方案如下：

BEGIN
For i in (select id, name, desc from table1) 
LOOP
Update table2 set name = i.name, desc = i.desc where id = i.id;
END LOOP;
END;

这就是在SQLDeveloper工作表上运行它的确切方式。他们说这很慢，但这是唯一的解决办法，为我工作的情况下。

这里似乎是一个更好的答案，使用'in'子句，允许连接多个键：

update fp_active set STATE='E', 
   LAST_DATE_MAJ = sysdate where (client,code) in (select (client,code) from fp_detail
  where valid = 1) ...

完整的示例如下：http://forums.devshed.com/oracle-development-96/how-to-update-from-two-tables-195893.html-从Web存档，因为链接已失效。
问题在于，在where子句中，要用作键的列放在'in'前面的括号中，而select语句的列名放在括号中。where（column1，column2）in（select（column1，column2）from table where“the set I want”）;

Oracle Database 23c为update和delete添加了直接联接：

create table t1 (
  c1 int, c2 int
);
create table t2 (
  c1 int, c2 int
);

insert into t1 values ( 1, 1 ), ( 2, 2 );
insert into t2 values ( 1, 42 );

select * from t1;

        C1         C2
---------- ----------
         1          1
         2          2

update t1
set    t1.c2 = t2.c2
from   t2
where  t1.c1 = t2.c1;

select * from t1;

        C1         C2
---------- ----------
         1         42
         2          2

delete t1
from   t2
where  t1.c1 = t2.c1;

select * from t1;

        C1         C2
---------- ----------
         2          2

BEGIN
For i in (select id, name, desc from table2) 
LOOP
Update table1 set name = i.name, desc = i.desc where id = i.id and (name is null or desc is null);
END LOOP;
END;

如果你的表t1和它的备份t2有很多列，这里有一个紧凑的方法。
此外，我的相关问题是，只有一些列被修改，许多行没有对这些列进行编辑，所以我想不去管它们--基本上是从整个表的备份中恢复列的子集。如果只想恢复所有行，请跳过where子句。
当然，更简单的方法是删除和插入为选择，但在我的情况下，我需要一个解决方案，只是更新。
诀窍是，当你从一对具有重复列名的表中选择 * 时，第二个表将被命名为_1。所以我的想法是

update (
    select * from t1 join t2 on t2.id = t1.id
    where id in (
      select id from (
        select id, col1, col2, ... from t2
        minus select id, col1, col2, ... from t1
      )
    )
  ) set col1=col1_1, col2=col2_1, ...