[英]Updating a key on table from another table in Oracle
當鍵值為 (abc) 時,我試圖通過從表 (t2) 獲取值來更新表 (t1) 上的鍵。
當我將其限制為特定人員時,它按預期工作
update table_a t1
set t1.u_key = (select t2.u_key
from table_b t2
where t2.name_f=t1.name_f
and t2.name_l=t1.name_l
and rownum<=1
and t2='NEVADA')
where t1.u_key = 'abc'
and e.name_f='Lori'
and e.name_l='U'
;
我最初嘗試不使用 rownum,它說返回的行太多。
為了使用 t1.u_key='abc' 運行所有數據並取出特定名稱,我嘗試了這個一直運行到超時的方法。
update table_a t1
set t1.u_key = (select t2.u_key
from table_b t2
where t2.name_f=t1.name_f
and t2.name_l=t1.name_l
and rownum<=1
and t2='NEVADA')
where t1.u_key = 'abc'
;
你能看看它並建議我錯過什么嗎?
您應該首先查看單獨運行內部 SELECT 語句時返回的內容:
SELECT t2.u_key FROM table_b t2
WHERE t2.name_f IN (SELECT name_f FROM table_a WHERE u_key = 'abc')
AND t2.name_l IN (SELECT name_l FROM table_a WHERE u_key = 'abc')
AND t2='NEVADA'
檢查結果,您將看到返回的行不止一行。
如果每個鍵應該只有匹配的行,您還需要將鍵添加到內部 SELECT 中,但我無法告訴您如果沒有額外的表描述以及可能來自table_a
和table_b
一些示例條目,它應該是什么樣子。
用這個:
update (
SELECT t2.u_key t2key,
t1.ukey t1key
FROM table_b t2,
table_a t1
where t2.name_f=t1.name_f
and t2.name_l=t1.name_l
and t2='NEVADA'
and rownum<=1 )
SET t1key = t2key
where t1key = 'abc';
merge into table_a t1
using(
select name_f, name_l, max(u_key) as new_key
from table_b t2
where t2='NEVADA'
group by name_f, name_l
) t2
on (t1.name_f=t2.name_f and t1.name_l=t2.name_l and t1.u_key='abc')
when matched then
update set t1.u_key=t2.new_key
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