进参问题!请高手来帮这看看!不够+分!顶这有分!
源码:
public class Autokan
{
public string KA3;
}
public class AutoJg
{
public string Jg;
}
public class LookedClass
{
public static Autokan Saves( string pid )
{
SqlConnection con = showclass.datacon.AutoCon();
SqlCommand cmd = new SqlCommand( "Kan_GetList " + pid,con);
con.Open();
Autokan Kan = new Autokan();
SqlDataReader sdr = cmd.ExecuteReader();
if( sdr.Read() )
{
Kan.KA3 = sdr[ "auto_price "].ToString();
}
sdr.Close();
con.Close();
return Kan;
}
public static string Jg()
{
SqlConnection con = showclass.datacon.AutoCon();
SqlCommand cmd = new SqlCommand( "AutoPrice_JiaGe " + JiaGe,con);
con.Open();
System.Text.StringBuilder Jg = new System.Text.StringBuilder();
SqlDataReader sdr = cmd.ExecuteReader();
while(sdr.Read())
{
Jg.Append( " <table cellpadding=0 cellspacing=0 class=kkkkkkkk> ");
Jg.Append( " <tr> ");
Jg.Append( " <td width=25 align=center> <font color=#FF8600> "+sdr[ "auto_price "].ToString()+ " </font> </td> ");
Jg.Append( " <td width=140 align=left> "+sdr[ "autoBrand "].ToString()+sdr[ "autoType "].ToString()+ " </td> ");
Jg.Append( " </tr> ");
Jg.Append( " </table> ");
}
sdr.Close();
con.Close();
return Jg.ToString();
}
}
存储过程:
CREATE PROCEDURE AutoPrice_JiaGe
@JiaGe int
AS
select a.[PriceID],a.[AutoID],a.[auto_price],b.[autoBrand],b.[autoType] from AutoPrice a,AutoInfo b where auto_price > @JiaGe-5 and auto_price <@JiaGe+5 and a.AutoID=b.AutoID
GO
======================
问:我怎样才能吧Kan.KA3 = sdr[ "auto_price "].ToString();得来的数进参到
public static string Jg()类中让他等于JiaGe它??
请大伙帮我看看吧!
------解决方案--------------------没看懂……
------解决方案--------------------public static string Jg()类中已经有方法取结果
JiaGe又是什么
看半天代码真痛苦,表述清楚一点嘛
------解决方案--------------------waiting for csasp
------解决方案--------------------帮顶了
------解决方案--------------------帮顶了,在线关注
------解决方案--------------------混乱啊
------解决方案--------------------不大明白~
在Jg()中加?
JiaGe = LookedClass.Saves(pid)
------解决方案--------------------看不懂 lz太高了
------解决方案--------------------用拼音起的名还真不知道是什么意思,真佩服楼主