项目中有一个业务需求是:默认载入15条历史记录(按时间顺序从早到晚). 以下是我构造的sql逻辑,mark一下,亲測可行. SELECT * FROM (SELECT *FROM group_chatmsg_v WHERE ((group_Id=46 AND send_user_id=28 AND receive_user_id=70) OR (group_Id=46 AND receive_user_id=28 AND STATUS=1)) AND is_delete =0 ORDER BY…
<?php mysql_connect("localhost","root","root");mysql_select_db("test");//保留最新的1000条记录$limit=1000;$query="select `id` from `news`";$result=mysql_query($query);$num=mysql_num_rows($result);if($num>$lim…
1.sqlite3数据库select * from QG order by random() limit 6 以下显示前10条记录 2.SQL Server数据库select top 10 * from table_name; 3.DB2数据库select * from table_name fetch first 10 rows only; 4.Oracle数据库select * from table_name where rownum <=10; 5.MySQL数据库select * f…
继上篇<SQL 列转行 合并多条记录>后,有网友反馈新的需求还是不太会用. 现举例说明 一,网友需要如下的效果: 其实,这个需求依然可以我上篇的方法进行解答,但为了实现分组,需要distinct group1,同时,为了根据key1,key2,key3是否相同进行分组,所以要用where进行连接. 语法格式:select .....from t1 where key1=a.key1 and key2=a.key2... for xml path('') 二,进入正题,代码如下: select…
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…
写一条存储过程,实现往User中插入一条记录并返回当前UserId(自增长id) --推荐写法 if(Exists(select * from sys.objects where name=N'Usp_InsertedID')) drop proc Usp_InsertedID go create proc Usp_InsertedID as insert into [User] output inserted.UserID values(N'张三蛋',3) --另一种写法(SCOPE_IDEN…
CREATE TABLE [TestTable] ( ) NOT NULL , ) NOT NULL , ) ))) GO ALTER TABLE [TestTable] ADD PRIMARY KEY ([uid], [key]) GO ) GO ) GO ) GO ) GO ) GO ) GO WITH cte AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY uid ORDER BY [key] DESC) AS rn FROM vtable )…