SQL Server 2019安装后出现连接失败

Posted 2023-04-04 Jennifer_糖糖

解决SQL Server 连接失败的问题

最近因学习数据库系统原理，下载安装了一个2019版本的，启动服务后，发现使用Aqua Data Studio根本连不上，打开设备管理器 一看1433端口没有监听；

于是在网站上搜索解决方案，基本上都是说在 SQLServer的管理配置界面设置一下，TCP/IP的端口即可。也的确如此；但是我配置完端口之后，依旧无法正常连接。

因为我遇到的情况相对特殊一些，我的计算机上安装了虚拟机，虚拟了很多的网卡以及IP，所以需要设置所有IP的port并enabled，然后重启服务即可：

 一般解决方法：

1、打开配置管理器，检查微软SQL服务器(MSSQLServer)的TCP/IP协议是否开启。

 确保上面设置正确后，进行最后的SQL服务器防火墙设置，(问题大多出在这里)

1、可以直接关闭防火墙，但是这样不安全，不建议采用。

2、找到微软SQL服务器(MSSQLServer)的TCP/IP协议。在SQL服务器防火墙已经处于运行状态下，点击“TCP/IP协议”选择属性

 从上图中我们可以看出，这台SQL服务器上支持TCP/IP协议的是1433端口。下一步要做的是在防火墙的配置中允许1433端口支持TCP/IP协议即可。打开“控制面板”选择“Windows防火墙”——“高级设置”。右击“入站规则(Inboud Rules)”选项，选择“新建规则(NewRule)”：

 

 

打开“新建规则”选项后，利用“新建入站规则向导”为1433端口配置“内部绑定协议”配置为适用于TCP/IP协议即可。

配置过程：

 点击下一步

 下一步

 点击完成

 完成上述步骤后，SQL服务器即可支持远程链接了。赶紧试试吧！

 

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#??????????????????

1???sql

 1 Create table GroupInfo([Id] int,[GroupName] nvarchar(50),[ParentGroupId] int)
 2 
 3 Insert GroupInfo
 4 
 5 select 0,??????????????????,null union all
 6 
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 8 select 2,??????????????????,1 union all
 9 select 3,??????????????????,1 union all
10 select 4,????????????????????????,2 union all
11 select 5,????????????????????????,2 union all
12 select 6,????????????????????????,3 union all
13 select 7,????????????????????????,3 union all
14 
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16 select 9, ?????????????????????,8 union all
17 select 10,????????????????????????,8 union all
18 select 11,???????????????????????????,9 union all
19 select 12,???????????????????????????,9 union all
20 select 13,??????????????????????????????,10 union all
21 select 14,??????????????????????????????,10 

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 1 --????????????????????????????????????????????????
 2 with
 3 CTE
 4 as
 5 (
 6     select * from GroupInfo where Id=1
 7     union all
 8     select G.* from CTE inner join GroupInfo as G
 9     on CTE.Id=G.ParentGroupId
10 )
11 select * from CTE order by Id

 

 1 --????????????????????????????????????????????????
 2 with
 3 CTE
 4 as
 5 (
 6     select * from GroupInfo where Id=14
 7     union all
 8     select G.* from CTE inner join GroupInfo as G
 9     on CTE.ParentGroupId=G.Id
10 )
11 select * from CTE order by Id

2?????????????????????

 1 --??????????????????
 2 with
 3 CTE
 4 as
 5 (
 6     select Id,GroupName,ParentGroupId,GroupPath=CAST( GroupName as nvarchar(max)) from GroupInfo where Id=1
 7     union all
 8     select G.*,CAST(CTE.GroupPath+???//???+G.GroupName as nvarchar(max)) as GroupPath from CTE 
 9     inner join GroupInfo as G
10     on CTE.Id=G.ParentGroupId
11 )
12 select * from CTE

 3?????????????????????????????????????????????????????????

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 2 WITH    
 3 CTE
 4 AS 
 5 ( 
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 7     WHERE ParentGroupId = 0
 8     UNION ALL
 9     SELECT   GroupInfo.* ,CAST(sort + RIGHT(???000??? + CAST(GroupInfo.[Id] AS VARCHAR),3) AS VARCHAR(MAX)) AS sort
10     FROM CTE
11     INNER JOIN GroupInfo ON CTE.Id = GroupInfo.ParentGroupId
12 )
13 SELECT * FROM CTE ORDER BY sort 

 

4?????????????????????????????????????????????????????????

1 --??????????????????
2 WITH CTE AS (
3     SELECT *,1 AS [Level] FROM GroupInfo WHERE ParentGroupId=0
4     UNION ALL
5     SELECT G.*,CTE.Level+1 FROM GroupInfo as G 
6     JOIN CTE ON CTE.Id =G.ParentGroupId
7 )
8 SELECT * FROM CTE