有时,我们需要对比两张表的数据,找到在其中一张表,不在另一张表中的数据 hql 如下: SELECT * FROM (SELECT id FROM a WHERE dt = '2019-03-17' ) a LEFT JOIN (SELECT id FROM b ) b ON a.id = b.id WHERE b.id IS NULL;
1.在面试的时候碰到一个 问题,就是让写一张表中有id和name 两个字段,查询出name重复的所有数据,现在列下: select * from xi a where (a.username) in (select username from xi group by username having count(*) > 1) 2.查询出所有数据进行分组之后,和重复数据的重复次数的查询数据,先列下: select count(username) as '重复次数',username from
表名:student 表结构及数据: +----+--------+---------+------+------------+--------------+---------+ | id | name | english | math | birthday | native_place | chinese | +----+--------+---------+------+------------+--------------+---------+ | 1 | 潘怡茹 |
MySql : 有N张表,N未知,每张表都有一个字段(id),每张表的字段结构不完全一样,如何查询所有表里面所有id的最大值?如下图所示: 对上面三张表进行操作的话,结果应该为:9 SQL语句: select greatest( (select max(id) from table_1), (select max(id) from table_2), (select max(id) from table_3) )
在A表中存在一个字段“AccountId”,而且不同记录之间的“AccountId”值有可能会相同,现在就是需要查询出在该表中的各记录之间,“AccountId”值存在重复的项,这里count记录该字段重复的记录数(此处只针对该字段是否重复进行查询):Select AccountId,Count(*) From A Group By AccountId Having Count(*) > 1 如果查询多个字段也相同大则如下:Select AccountId,Name,Count(*) From
.查找表中多余的重复记录,重复记录是根据单个字段(peopleId)来判断 select*from people where peopleIdin (select peopleIdfrom peoplegroupby peopleIdhaving ) .删除表中多余的重复记录,重复记录是根据单个字段(peopleId)来判断,只留有rowid最小的记录 deletefrom people where peopleIdin (select peopleIdfrom peoplegroupby pe
将表T_wz_wz中的部分数据插入到表t_wz_kc: insert into t_wz_kc(wzid,jldwid,kcsl,yfpkcsl,cshwcbz) select wzid,jldwid,0,0,'Y' from t_wz_wz where yxbz='Y' --去重复-- and wzid not in (select wzid from t_wz_kc)
今天在QQ群众讨论到一个问题,记录下下来,一边以后用的时候可以翻阅 总结除了三种方法 --方法1,,这一种方法不行,,错误的认识了,@@ROWCOUNT,,,唉,,学艺不精,,丢人啊 SELECT TOP 1 @@ROWCOUNT FROM dbo.tb_CommonDataDict --方法2if exists(SELECT TOP 1 * FROM dbo.tb_CommonDataDict )--方法3if (select Count(*) from a)>0
方法一: select t1.num1,t2.num2,t3.num3 from (select count(*) num1 from table1) t1, (select count(*) num2 from table2) t2, (select count(*) num3 from table3) t3 方法二: select sum(t.num1),sum(t.num2),sum(t.num3) from ( as num3 from table1 union as num3 from
语法: update table1 set table1.列=table2.列 from table2 where table2.列=table1.列update dbo.PATIENT setPATIENT_ADDUSERID=OLD.PATIENT_ADDUSERID,PATIENT_UPDATEUSERID=OLD.PATIENT_UPDATEUSERID,PATIENT_UPDATEUSERNAME=OLD.PATIENT_UPDATEUSERNAME,PATIENT_ADDUS
select * from DB_PATCH awhere lower(a.db_name) in (select lower(db_name) from DB_PATCH group by lower(db_name) having count(*) > 1) delete from DB_PATCH awhere lower(a.db_name) in (select lower(db_name) from DB_PATCH group by lower(db_name) having co
with a as ( select id,name,parentid from categories where id=53 union all select x.id,x.name,x.parentid from categories x,a where x.parentid=a.id ) select * from a
select consumer_id,user_name,mobile,invite_code from csr_consumer where invite_count<(select count(1) from csr_invite_picture) select invite_picture_id,blank_file_store_id,logo_file_store_id from csr_invite_picture t1 where t1.invite_picture_id not i
--创建新增本地数据库的存储过程create or replaceprocedure pro_electric_record as begin insert into electric_meter_record(id,basestation_id,name,meter_number,createtime,electric_meter_id) select sys_guid(),substr(s.sname,0,36),s.sname,s.svalue,sysdate,(select
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,vbd
将一张表中的数据插入另外一张表 1.两张表结构相同 insert into 表1名称 select * from 表2名称 2.两张结构不相同的表 insert into 表1名称(列名1,列名2,列名3···) select 列1,列2,列3 ··· from 表2名称 3.从另一张表获取某几个字段插入另一张表 insert into 表1名称(列名1,列名2,列名3···)values(数据1,数据2,(select 列名3 from 表2名称 )) 三张结构相同的表查询一条记录 SELEC
/*查询某张表被哪些存储过程或者视图用到的sql语句*/select distinct object_name(id) from syscomments where id in (select id from sysobjects where type in('V','P')) and text like '%表名%'
注:本文来源于<oracle查询某张表的外键(最终解决办法)> 一:几个查询表外键的脚本 select b.table_name, b.column_name from user_constraints a inner join user_cons_columns b on a.constraint_name = b.constraint_name where a.r_constraint_name in ( select e.constraint_name from user_constra