关于ORACLE TOP 1 的问题 高手解答！ 关于显卡的问题.求高手解答

oracle \u8bed\u53e5\u95ee\u9898\uff0c\u8bf7\u9ad8\u624b\u56de\u7b54\u4e00\u4e0b\uff0c\u8c22\u8c22

\u7b2c\u4e00\u79cd\u5199\u6cd5:
with a as(select t.user_name,sum(t.movie_price) price from moivehistory t group by t.user_name )
select a.user_name,a.price from a where a.price = (select max(price) from a)
\u4e00\u697c\u53ea\u5b9e\u73b0\u5206\u7ec4\u7684\u76ee\u7684,\u6ca1\u6709\u6c42\u51fa\u6700\u5927
\u4e8c\u697coracle\u91cc\u6ca1\u6709top\u8fd9\u6837\u7684\u7528\u6cd5,\u8fd9\u4e2a\u662fSQLSERVER\u7684
\u4e09\u697c\u7684SQL\u90a3\u91cc\u5c11\u7a7a\u683c,\u8dd1\u4e0d\u901a
\u8fd8\u6709\u5176\u5b83\u7684\u5199\u6cd5,
\u7b2c\u4e8c\u79cd\u5199\u6cd5:
select *
from (select t.user_name, sum(t.movie_price) price
from moivehistory t
group by t.user_name)
where price = all (select max(price)
from (select t.user_name, sum(t.movie_price) price
from moivehistory t
group by t.user_name))
\u7b2c\u4e09\u79cd\u5199\u6cd5:
select *
from (select t.user_name, sum(t.movie_price) price
from moivehistory t
group by t.user_name) c
where (select count(*)
from (select t.user_name, sum(t.movie_price) price
from moivehistory t
group by t.user_name)
where price > c.price) = 0

\u60f3\u95ee\u4e2a\u95ee\u9898,\u60a8\u7684\u8fd9\u4e2a\u663e\u5b58\u7684\u5927\u5c0f\u60a8\u662f\u7528\u4ec0\u4e48\u770b\u7684,\u5c1d\u8bd5\u7528GPU-Z\u770b\u770b,\u6709\u7684\u65f6\u5019\u8fd9\u4e9b\u53ef\u80fd\u88ab\u4e00\u4e9b\u5176\u4ed6\u7684\u8f6f\u4ef6\u5360\u7528\u7684.

Top 1 就是 返回 第一条 数据

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