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
Oracle: s_emp s_dept s_region 行:Row(tuple) 列:Column(attribute) conn:改变用户 Drop:删除用户 drop user briup cascade; 改变语言:alter session set nls_language='american'; alter session set nls_language='simplified chinese; 显示当前用户:show user; 用@符号导入sql:@F:\Linux笔
查询用户数据表,需要根据时间正序排序,然后获取时间最早的前三条数据,是不是第一印象想这么写: select * from users where rownum<4 order by datatime asc 哈哈,这就错了,因为oracle会先去前三条数据,然后再按照时间排序,你是得不到正确结果的,应该先排序,再获取数据 select * from (select y.* from users y order by y.datatime asc) where rownum<4
1>找出相应的发票; SELECT DISTINCT ai.invoice_id, ai.invoice_num invoice_num, pv.segment1 vendor_num, pv.vendor_name vendor_name, 'The fully Paid invoice should be closed!' status FROM ap.ap_invoices_all ai, po.po_vendors pv, ap.ap_payment_schedules_all aps
比如sql: select zyxdm from table where bindid=2265254 查询结果为:1|4|8|9|10 将这个查询结果转成多行,结果如下: ID 1 4 8 9 10 转换的sql : with CTE1 as (select zyxdm from table where bindid=2265254) SELECT REGEXP_SUBSTR((select * from CTE1), '[^|]+', 1, ROWNUM) id FROM DUAL CONN
for update模拟锁表 --session 1 SQL> select * from tt for update; --session 2 SQL> update tt set id=1 where id=2; 按用户查询锁的情况 set line 300 col OBJECT_NAME format a30 col event format a30 col type format a10 col object_name format a15 col object_type format