Oracle 查询出来的数据取第一条 --------------------------------------------------------------------------- 转载自:http://www.itpub.net/thread-246442-1-1.html select * from (select * from <table> order by <key>) where rownum=1; select * from (select * from &l
oracle 分组取第一行数据 SELECT * FROM ( SELECT ROW_NUMBER() OVER(PARTITION BY x ORDER BY y DESC) rn, t.* FROM test1 t ) WHERE rn = 1; 查找oracle 执行的语句 select t.*from v$sqlarea t where t.FIRST_LOAD_TIME like '2018-11-05%' order by t.FIRST_LOAD_TIME desc
检查重复记录 -- 检查重复code1 select count(identity) num, identity from event_log where code='code1' order by num desc 删除重复记录 DELETE FROM event_log WHERE `code`='code1' AND identity IN ( SELECT identity from ( ) a ) AND id NOT IN ( SELECT keepId FROM ( ) b ) 其
MySQL分组排序(取第一或最后) 方法一:速度非常慢,跑了30分钟 SELECT custid, apply_date, rejectrule FROM ( SELECT *, IF ( , ) AS rank, @pkey := custid FROM ( SELECT custid, createTime, SUBSTR( createTime, , ) AS apply_date, rejectRule, STATUS FROM ( SELECT * FROM credit.`apply
a = [1, 2, 3, 4, 5, 6, 7, 8]l=[i**2 for i in a if i**2>=16] #列表推导式+if判断print(l)print(type(l)) b={1:1,2:2,3:3}d={i**2 for i in b }print(d) #输出的是一个集合print(type(d)) c=(1,2,3,4,5) #元组,输出的是生成器t=(i**2 for i in c)print(t)print(type(t)) d1={1,2,3,4,5,6} #集合s