如何在SQL表中创建多列拼接的新列？解决语法报错问题

需要将表dbo.equi2022t中的6个字符列拼接为一个名为censusbloc的char(15)列，涉及列信息如下：

censusstate char(2)
censuscounty  char(3)
censustrac   char(6)
censusblocgroup char(1)
censusbloccode char(2)
censusbloccode2 char(1)

尝试执行以下SQL语句时触发语法错误：

alter table dbo.equi2022t
add censusbloc char(15) AS 
(Concat(CensusState,CensusCounty,Censustrac,Censusblocgroup,Censusbloccode,Censusbloccode2));

错误提示：

Msg 156, Level 15, State 1, Line 364
Incorrect syntax near the keyword 'AS'.

针对MariaDB的修正语法 #

MariaDB创建计算列时，需在ADD后明确指定COLUMN关键字，修正后的语句如下：

ALTER TABLE dbo.equi2022t
ADD COLUMN censusbloc CHAR(15) AS (
    CONCAT(censusstate, censuscounty, censustrac, censusblocgroup, censusbloccode, censusbloccode2)
);

若使用SQL Server（根据错误提示推测） #

如果实际使用的是SQL Server，基础语法无需COLUMN关键字，语句如下：

ALTER TABLE dbo.equi2022t
ADD censusbloc CHAR(15) AS (
    CONCAT(censusstate, censuscounty, censustrac, censusblocgroup, censusbloccode, censusbloccode2)
);

若需要将计算列持久化存储（而非实时计算），可添加PERSISTED关键字：

ALTER TABLE dbo.equi2022t
ADD censusbloc CHAR(15) AS (
    CONCAT(censusstate, censuscounty, censustrac, censusblocgroup, censusbloccode, censusbloccode2)
) PERSISTED;

内容的提问来源于stack exchange，提问作者Tim Wilcox