使用新添加的 STRING_AGG 函数(在 SQL Server 2017 中),如以下查询所示,我可以获得下面的结果集。 SELECT ProjectID, STRING_AGG( newID.value, ',') WITHIN GROUP (ORDER BY newID.value) AS NewField FROM [dbo].[Data] WITH(NOLOCK) CROSS APPLY STRING_SPLIT([bID],';') AS newID WHERE newID.val...
SQL 分析端點 能串連字串運算式的值,並在這些值之間放置分隔符號值。 系統不會在字串結尾處新增分隔符號。 Transact-SQL 語法慣例 語法 syntaxsql STRING_AGG( expression, separator ) [<order_clause>]<order_clause>::=WITHINGROUP(ORDERBY<order_by_expression_list>[ASC|DESC] ) ...
However, I would like to apply an order by clause do the Agg function like i wrote in line 10. When I run que query in SQL Server it works just fine and I get the expected output, however when I test the same exact query in service studio I get an error and the same ...
SELECT STRING_AGG([value], ',') WITHIN GROUP (ORDER BY[value])FROM (SELECT[value]FROM STRING_SPLIT(@str,',')) source Source Code 或者使⽤⼀般资料表运算式CTE (Common Table Expression);WITH dump_data AS (SELECT[value]FROM STRING_SPLIT(@str,','))SELECT STRING_AGG([value], ',')...
使用CREATE ASSEMBLY 语句在 SQL Server 中注册程序集 搜索 程序集(数据库引擎 CREATE ASSEMBLY StringAgg FROM 'c:\StringAgg.dll' WITH PERMISSION_SET = SAFE; 启用SQL Server 运行 CLR 代码的功能 exec sp_configure 'clr enabled',1 RECONFIGURE WITH OVERRIDE; ...
但偶尔也会碰到一些后端为SQL Server的应用，并且其只允许来自预定义的主机名或应用程序列表的连接。这些...
The following example replaces null values with 'N/A' and returns the names separated by commas in a single result cell. SQL USEAdventureWorks2022; GOSELECTSTRING_AGG(CONVERT(NVARCHAR(max),ISNULL(FirstName,'N/A')),',')AScsvFROMPerson.Person; GO ...
SELECTSTRING_AGG([value],',') WITHINGROUP(ORDERBY[value])FROM(SELECT[value]FROMSTRING_SPLIT(@str,',')) source 1. 2. Source Code 或者使用一般资料表运算式CTE (Common Table Expression) ;WITHdump_dataAS(SELECT[value]FROMSTRING_SPLIT(@str,',') ...
repo #安装 SQL Server yum install -y mssql-server #选择 SQL Server 的一个版本: 1) ...
The following example replaces null values with 'N/A' and returns the names separated by commas in a single result cell.SQL Kopyala USE AdventureWorks2022; GO SELECT STRING_AGG(CONVERT(NVARCHAR(max), ISNULL(FirstName,'N/A')), ',') AS csv FROM Person.Person; GO ...