我使用的是T-sql(Sql server 2008)存储过程,它根据许多业务条件返回一个custmized值。 (这必须在数据库上完成,因为我对C#
代码中的许多存储过程进行了泛型调用)。
我的问题在于换行符。我在我的代码Char(13);Char(13) + char(10); char(10)...
中使用但它不起作用。
注意:如果我测试了一个 INSERT 语句,但是在 SELECT 语句中没有换行符,它会生成一个白色空间(完成一个简单的查询)。 贝娄我的代码
ALTER PROCEDURE [dbo].[getDebtorAddress] (
@idFolder int
)
AS
BEGIN
DECLARE
@type bit ,
@firstName varchar(100) ,
@familyName varchar(100) ,
@raisonSocial varchar(100),
@jeuneFille varchar(100) ,
@saleName varchar (100) ,
@enseigne varchar (100) ,
@sigle varchar (100) ,
@address varchar (max),
@lieuDit varchar(100) ,
@postalCode varchar (100),
@city varchar (100),
@country varchar (100),
@finalAddress nvarchar(max)
SET NOCOUNT ON;
SELECT DISTINCT
@firstName = Actor.M_EN_NOM,
@familyName = Actor.M_EN_PRENOM1,
@raisonSocial = Actor.M_EN_RAISONSOCIALE,
@jeuneFille = Actor.M_EN_NOMJEUNEFILLE,
@saleName = Actor.M_EN_NOMCOMMERCIAL,
@enseigne = Actor.M_EN_ENSEIGNE,
@sigle = Actor.M_EN_SIGLE,
@address = ActorCP.M_PR_CP_ADRESSE,
@lieuDit = ActorCP.M_PR_CP_LIEUDIT,
@postalCode = PostalCode.C_PA_CP_CODEPOSTALE,
@city = PostalCode.C_PA_CP_VILLE,
@country = Country.C_PA_PA_LIBELLE,
@type = Actor.M_EN_TYPE
FROM M_EN Actor
LEFT OUTER JOIN M_DE Debtor ON Actor.M_EN_ID =Debtor.M_EN_ID
LEFT OUTER JOIN M_DO Folder ON Debtor.M_DE_ID= Folder.M_DE_ID
LEFT OUTER JOIN M_PR_CP ActorCP ON Actor.M_EN_ID = ActorCP.M_EN_ID
LEFT OUTER JOIN C_PA_CP PostalCode ON ActorCP.C_PA_CP_ID = PostalCode.C_PA_CP_ID
LEFT OUTER JOIN C_PA_PA Country ON ActorCP.C_PA_PA_ID = Country.C_PA_PA_ID
WHERE Folder.M_DO_ID = @idFolder
if(@type=0)
begin
if (@firstName is not Null and @firstName !='')
set @finalAddress= @firstName
if(@familyName is not Null and @familyName != '')
set @finalAddress = @finalAddress + ' ' + @familyName +CHAR(13)
if(@jeuneFille is not null and @jeuneFille !='' )
set @finalAddress=@finalAddress + ' ' + @jeuneFille + CHAR(13)
if(@address is not null and @address != '')
set @finalAddress=@finalAddress + REPLACE(@address,'|',char(13))
if(@lieuDit is not null and @lieuDit != '')
set @finalAddress = @finalAddress + @lieuDit + CHAR(13)
if(@postalCode is not null and @postalCode != '')
set @finalAddress = @finalAddress + @postalCode + CHAR(13)
if(@country is not null and @country != '')
set @finalAddress = @finalAddress + @country + CHAR(13)
end
else
begin
if (@raisonSocial is Null or @raisonSocial = '')
Begin
if(@firstName is not null and @firstName != '')
set @finalAddress = @finalAddress + @firstName + CHAR(13)
if(@familyName is not null and @familyName != '')
set @finalAddress = @finalAddress + @familyName + CHAR(13)
end
else
set @finalAddress = @finalAddress + @raisonSocial + CHAR(13)
if(@jeuneFille is not null and @jeuneFille !='' )
set @finalAddress=CHAR(13) + @finalAddress + ' ' + @jeuneFille + CHAR(13)
if(@saleName is not null and @saleName != '')
set @finalAddress=@finalAddress + @saleName + CHAR(13)
if(@enseigne is not null and @enseigne != '')
set @finalAddress=@finalAddress + @enseigne + CHAR(13)
if(@sigle is not null and @sigle != '')
set @finalAddress=@finalAddress + @sigle + CHAR(13)
if(@address is not null and @address != '')
set @finalAddress=@finalAddress + REPLACE(@address,'|',char(13))
end
Create Table #TempAddress
(
TempAddress_ID int ,
adresse nvarchar(max)
)
INSERT INTO #TempAddress (#TempAddress.adresse) VALUES(@finalAddress)
SELECT #TempAddress.adresse from #TempAddress
DROP TABLE #TempAddress
END
目前的结果是
aaaaa bbbbb ccccc dddddd
但我的预期结果是
aaaaa
bbbbb
ccccc
ddddd
有任何建议或帮助吗? 感谢
答案 0 :(得分:3)
正如我在comment中解释的那样,在Windows上,换行符为CHAR(13)+CHAR(10)
,但是否会将其呈现为换行符取决于您的打印方式。
在SELECT语句中没有换行符,它会生成一个空格[...]在简单的SELECT中不起作用
“简单的SELECT”并没有解释你想要做什么。请记住,您是唯一一个可以看到您的屏幕的人。如果您希望其他人了解您的问题,您需要明确。
如果您的意思是SQL Server Management Studio中的网格视图不显示换行符,那么您是正确的:它不会,直到您单击工具栏中的“结果到文本”按钮为suggested in the first comment。
答案 1 :(得分:2)
你的T-SQL工作正常,它插入了char(13),但是根据你的客户端你需要添加一个额外的char(10)。
char(13) + char(10)
在客户端而不是在sql查询中进行这样的格式化可能是个更好的主意。
答案 2 :(得分:0)
您还可以使用以下语法:
......请注意第2行后的最后一个单引号...
| -> press enter and begin a new line with another...
SELECT 'use ' + DB_NAME() + ';' + '
' + 'CREATE USER ' + users.name + ' for ' + users.name + ';'
| -> continued quote in line 3.
这将产生以下输出:
use MyDatabase;
CREATE USER User1 for User1;