SQL查询回车和换行SQL 2008

时间:2018-07-09 16:51:40

标签: sql sql-server sql-server-2008

SQL查询仅返回1行#1

select  * from email_table  
where emailAddress like '%'+char(13)+'%'  

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在SQL 2008中需要带回车符返回前两行。

2 个答案:

答案 0 :(得分:0)

您应该能够在SQL中直接添加一个新行:

select *
from @email_table
where emailAddress like '%
%'

答案 1 :(得分:0)

您可以执行以下操作来排除故障:

/*DECLARE @email_table TABLE (emailaddress varchar(2000))
INSERT INTO @email_table(emailaddress) SELECT '#1gmail.com  ' + CHAR(13) + CHAR(10)
INSERT INTO @email_table(emailaddress) SELECT '#2gmail.com' + CHAR(13) + CHAR(10)
INSERT INTO @email_table(emailaddress) SELECT '#3gmail.com'
*/

SELECT *,CONVERT(VARBINARY(MAX),emailaddress) AS [Binary Representation]
FROM @email_table
--WHERE emailAddress LIKE '%' + CHAR(13) + '%'

[Binary Representation]行的最后一个字符应为0D0A(如果EmailAddress字段为NVARCHAR,则为0D000A00),如果不是,则look up它们实际上是什么字符。