表1:
id name desc
-----------------------
1 a abc
2 b def
3 c adf
表2:
id name desc
-----------------------
1 x 123
2 y 345
在oracle SQL中，我如何运行一个SQL更新查询，可以更新表1与表2的名称和desc使用相同的id?所以最终的结果是
表1:
id name desc
-----------------------
1 x 123
2 y 345
3 c adf
问题是从更新一个表从另一个数据，但特别为oracle SQL。
当前回答
这就是所谓的相关更新
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 )
假设连接的结果是键保留视图，您也可以这样做
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
其他回答
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;
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;
Update table set column = (select...)
从未为我工作，因为设置只期望1个值- SQL错误:ORA-01427:单行子查询返回多行。
下面是解决方案:
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，但这是我处理这个案子的唯一办法。
下面似乎是一个更好的答案，使用'in'子句，允许多个键的连接:
update fp_active set STATE='E',
LAST_DATE_MAJ = sysdate where (client,code) in (select (client,code) from fp_detail
where valid = 1) ...
完整的例子如下: http://forums.devshed.com/oracle-development-96/how-to-update-from-two-tables-195893.html -从网络档案，因为链接已死。
问题在于，在'in'之前的where子句的括号中有你想用作键的列，在括号中有相同的列名的select语句。 where (columnn1,column2) in (select (columnn1,column2) from table where I want的set);
try
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);
