一.背景 一张person表,有id和name的两个字段,id是唯一的不允许重复,id相同则认为是重复的记录. 二.解决 select id from group by id having count(*) > 1 按照id分组并计数,某个id号那一组的数量超过1条则认为重复. http://blog.163.com/ability_money/blog/static/185339259201221443031331/ http://blog.163.com/aner_rui/blog/stat
select id from group by id having count(*) > 1 按照id分组并计数,某个id号那一组的数量超过1条则认为重复. 如何查询重复的数据 select 字段1,字段2,count(*) from 表名 group by 字段1,字段2 having count(*) > 1 PS:将上面的>号改为=号就可以查询出没有重复的数据了. Oracle删除重复数据的SQL(删除所有): 删除重复数据的基本结构写法: 想要删除这些重复的数据,可以使用下面语句
下面两中方式都是将 srcTbl 的数据插入到 destTbl,但两句又有区别的: 方式一 (select into from)要求目标表(destTbl)不存在,因为在插入时会自动创建. select * into destTbl from srcTbl 方式二 (insert into select from)要求目标表(destTbl)存在,由于目标表已经存在,所以我们除了插入源表(srcTbl)的字段外,还可以插入常量,如例中的:5.特别注意的是:插入的字段顺序要和查询出的字段顺序一致
SELECT 'list.add("' || t.dummy || '");' as listFROM dual t where rownum < 600; 执行结果: SELECT '''' || t.dummy || ''',' as 查询出的字段加引号FROM dual t where rownum < 600; 执行结果:
select B.enterprise_code, B.enterprise_name, sum(B.h0_overnum) AS over00, sum(B.h1_overnum) AS over01, sum(B.h2_overnum) AS over02, sum(B.h3_overnum) AS over03, sum(B.h4_overnum) AS over04, sum(B.h5_overnum) AS over05, sum(B.h6_overnum) AS over06, su
在Oracle中实现select top N:由于Oracle不支持select top 语句,所以在Oracle中经常是用order by 跟rownum的组合来实现select top n的查询.简单地说,实现方法如下所示:select 列名1 ...列名n from(select 列名1 ...列名n from 表名 order by 列名1)where rownum <=N(抽出记录数)order by rownum asc 如:select id,name from (selec
对数据库某些意外情况,引起的重复数据,如何处理呢? ----------------查重复: select * from satisfaction_survey s and s.project_no in (select ss.project_no from satisfaction_survey ss group by ss.project_no ) order by s.project_no, s.submit_time 思路:从预先选出的数据中(s.as_side = 0),根据特定字段(
查询包含小写的所有数据: select oper_no from info_oper where regexp_like(oper_no,'[[:lower:]]'); select oper_no from info_oper where regexp_like(oper_no,'[a-z]'); 查询包含大写的所有数据: select oper_no from info_oper where regexp_like(oper_no,'[[:upper:]]'); select oper_no
1,mysql concat('a','b','c')和concat_ws('a','b','c')的区别:前者如果有某个值为空,结果为空;后者如果有某个值为空,可以忽略这个控制 SELECT concat_ws('',FORE_TWO,THIRD_POSITION,FOURTH_POSITION,FIFTH_POSITION) AS code FROM TABLE_A WHERE ID= #{id} 2,oracle SELECT FORE_TWO||THIRD_POSITION||FOURT
1. 新增一个表,通过另一个表的结构和数据 create table XTHAME.tab1 as select * from DSKNOW.COMBDVERSION 2. 如果表存在: insert into tab1 select * from tab2; 3.同一个表中,将A字段的指赋给B字段: update table_name set B = A; 4. 将一个表的字段数据插入到另一个表的字段数据中 insert into XTHAME.tab1(pk_bdversion,vbdco