C# , ASP.Net 中 关于 like in 实现参数化查询的问题。2008-09-18 18:17对于 普通的 select等sql语句, 正常的参数化 语句 格式:
select * from profile where EmployeeID= @EmployeeID
for example:
string loginString = "select * from profile where EmployeeID= @EmployeeID";
but please attention to the like sql sentence:
select * from profile where EmployeeID Like ‘%’ + @EmployeeID + ‘%’;
The accurate search format is :
Select * from profile where EmployeeID like +@EmployeeID ;
So the
String = "SELECT * from Box WHERE BoxID like '%' + @subString + '%'"
对本文提供了有价值的文章有:
c# sql like 参数
2008-08-08 10:09
参数化的意义在于把对应的值从参数中提供,对于like语句,like后面的值则包括了单引号中的所有部分,包括百分号(%),因此在参数化like对应的值时,应该把百分号移到参数值中提供,像这样:
Cmd.Parameters["@KeyWord"].Value = "%" + StrKeyWord + "%";
可别奢想在sql语句中像这样的样子:
Select * From [TableName] Where [Column1] like '%@KeyWord%'
不会报错,不过你不可能查询到想要的结果.
select * from profile where EmployeeID= @EmployeeID
for example:
string loginString = "select * from profile where EmployeeID= @EmployeeID";
but please attention to the like sql sentence:
select * from profile where EmployeeID Like ‘%’ + @EmployeeID + ‘%’;
The accurate search format is :
Select * from profile where EmployeeID like +@EmployeeID ;
So the
String = "SELECT * from Box WHERE BoxID like '%' + @subString + '%'"
对本文提供了有价值的文章有:
c# sql like 参数
2008-08-08 10:09
参数化的意义在于把对应的值从参数中提供,对于like语句,like后面的值则包括了单引号中的所有部分,包括百分号(%),因此在参数化like对应的值时,应该把百分号移到参数值中提供,像这样:
Cmd.Parameters["@KeyWord"].Value = "%" + StrKeyWord + "%";
可别奢想在sql语句中像这样的样子:
Select * From [TableName] Where [Column1] like '%@KeyWord%'
不会报错,不过你不可能查询到想要的结果.