说明: MBGL_GZJH jh_id 对应 mbgl_gzjh_fkmx jh_id mbgl_gzjh_fkmx jh_id 有重复多条,但是 FKRQ 不一样,我们去 FKRQ 最新的一条. select * FROM MBGL_GZJH A LEFT JOIN (select BC.* from (select max(FKRQ) as maxrq,JH_ID from mbgl_gzjh_fkmx GROUP BY JH_ID ) AC LEFT JOIN mbgl_gzjh_fkmx
sql 查询某个条件下多条数据中最新的一条数据或最老的一条数据 test_user表结构如下: 需求:查询李四.王五.李二创建的最初时间或者最新时间 1:查询最初的创建时间: SELECT * FROM( SELECT * FROM test_user ) AS tu WHERE NOT EXISTS ( SELECT * FROM( SELECT * FROM test_user ) AS tu2 WHERE tu2.user_name=tu.user_name AND DATE(tu.tim
一.问题 groupBY分组后取最新一条记录的SQL的解决方案. 二.解决方案 select Message,EventTime from PT_ChildSysAlarms as a where EventTime = (select max(b.EventTime) from PT_ChildSysAlarms as b where a.PtName = b.PtName ) group by Message,EventTime order by EventTime desc
如题,解决思路如下: 1.首先我们需要找出拥有重复数据的记录 ---以name字段分组 select Name,COUNT(Name) as [count] from Permission group by Name-- having COUNT(Name)>1 ---以name字段分组,筛选出有重复出现的字段 having 用于条件筛选 作用相当于where 区别where用于group by 之前,having用于之后 查找拥有重复记录的数据 执行结果: 2.然后我们需要知道要要保留的记录的
主表: 辅表: 一个app对应多个apk,现在要取上线(Apk_Status最大的)的应用 select * from [dbo].[tbl_APP] as app join ( * from tbl_Apk as BB where AA.Apk_APPId=BB.Apk_APPId and AA.Apk_Status < BB.Apk_Status)) as apk on app.APP_Id = apk.Apk_APPId 重点是取关联的右表: select * from [dbo].[tb
delete from people where peopleName in (select peopleName from people group by peopleName having count(peopleName) > 1) and peopleId not in (select min(peopleId) from people group by peopleName having count(peopleName)>1)
方法一: 使用TOP SELECT TOP 1 * FROM user; SELECT TOP 1 * FROM user order by id desc; 方法二: 使用LIMIT SELECT * FROM user LIMIT 1; SELECT * FROM user ORDER BY id ASC LIMIT 1; SELECT * FROM user ORDER BY id DESC LIMIT 1; --------------------- 原文:https://blog.cs
mysql: SELECT * from (SELECT H_TEMPERATURE,TH_TIME FROM wenshidu WHERE TH_TIME <= STR_TO_DATE('2016-03-16 10:04:52','%Y-%m-%d %H:%i:%s') order by TH_TIME desc) as total limit 0,8; oracle: SELECT * from (SELECT H_HUMIDITY FROM HOME_MONITOR WHE
方法一: 使用TOP SELECT TOP 1 * FROM user; SELECT TOP 1 * FROM user order by id desc; 方法二: 使用LIMIT SELECT * FROM user LIMIT 1; SELECT * FROM user ORDER BY id ASC LIMIT 1; SELECT * FROM user ORDER BY id DESC LIMIT 1; 转载至:https://www.cnblogs.com/weimingxin/p
一.条件字段为数值的情况 select * from tb where id=@id; --当前记录 select top 1 * from tb where id>@id order by id; --下一条记录 select top 1 * from tb where id<@id order by id desc --上一条记录 二.以排序的思路出发的一种方案 ;WITH TUsers AS ( SELECT *, ROW_NUMBER() OVER (
用到的是这个函数: group_concat() select group_buying_id, group_concat(app_user_ids) from org_user_group group by group_buying_id 输出的结果: select group_buying_id, group_concat(app_user_ids) from org_user_group where group_buying_id=10 这是查询一条
数据库中二张表,用户表和奖金记录表,奖金记录表中一个用户有多条信息,有一个生效时间,现在要查询: 奖金生效时间在三天前,每个用户取最新一条奖金记录,且用户末锁定 以前用的方法是直接写在C#代码中的: for(所有末锁定用户) 查询奖金记录表 top 1 where uid=??? order by 生效时间 desc if(上面查询的记录生效时间在三天前) 输出 今天花了大部分时间研究了一下,终于找到一条SQL语句就可以把结果查出来的方法: 那个row_number函数在MSSQL2008以上才