2013-10-03 133 views
-1

我有一個下拉列表是leavetype在此sickleave,緊急請假類型,當我選擇數據庫值緊急請假去病假即在數據庫下拉列表在asp.net

leasid leavedescription leavetypeid: 
1  abcdefe   1 

那裏,因爲我在選擇2表格

錯誤在哪裏?

這是

cc.leavee(Txt_leaveValue, Convert.ToInt32(leavedrop.SelectedValue)); 

LMS.leavetype le = new LMS.leavetype(); 
     DropDownList2.DataSource = le.getdep(); 
     DropDownList2.DataValueField = "leaveID"; 
     DropDownList2.DataTextField = "leavetypeID"; 
     DropDownList2.DataBind(); 

回答

0

你爲的IsPostBack在你的Page_Load()事件檢查的代碼?

if (!Page.IsPostBack) 
{ 
    // init dropdown here 
    DMSLIB.leavetype le = new DMSLIB.leavetype(); 
    DropDownList2.DataSource = le.getdep(); 
    DropDownList2.DataValueField = "leaveID"; 
    DropDownList2.DataTextField = "leavetypeID"; 
    DropDownList2.DataBind(); 
} 

如果您忘記該測試,則在讀取該值之前重新初始化該測試。

+0

是的,我忘了添加此代碼如果(!page.ispostback) –

+0

太棒了!現在「接受」我的答案,我們都會很高興:) – egrunin

+0

haha​​hahha ....:D –