我有一个查询，在MySQL工作得很好，但当我在Oracle上运行它时，我得到以下错误:
SQL错误:ORA-00933: SQL命令未正确结束 00933. 00000 - "SQL命令未正确结束"
查询为:
UPDATE table1
INNER JOIN table2 ON table1.value = table2.DESC
SET table1.value = table2.CODE
WHERE table1.UPDATETYPE='blah';
当前回答
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;
/
其他回答
UPDATE table1 t1
SET t1.value =
(select t2.CODE from table2 t2
where t1.value = t2.DESC)
WHERE t1.UPDATETYPE='blah';
UPDATE ( SELECT t1.value, t2.CODE
FROM table1 t1
INNER JOIN table2 t2 ON t1.Value = t2.DESC
WHERE t1.UPDATETYPE='blah')
SET t1.Value= t2.CODE
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
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。
