2016-07-26 89 views
0

嘗試插入一行數據時出現以下錯誤。它指出該字符串未被識別爲有效的日期時間。轉換爲日期時間時字符串未被識別爲有效日期時間

C#

protected void saveExceptionAdd(object sender, System.EventArgs e) 
    { 
     //Default value secruity 
     if (ddlTimeFromAdd.SelectedIndex == 0 || ddlTimeToAdd.SelectedIndex == 0) 
     { 
      lblAddExcept.Visible = true; 
      lblAddExcept.Text = "Fields Required."; 
      divExceptionAdd.Focus(); 
     } 

     else 
     { 
      string EFTVFROM = txtDatefromAdd.Text.ToString() + ddlTimeFromAdd.SelectedValue.ToString(); 
      string EFTVTO = txtDatetoAdd.ToString() + ddlTimeToAdd.SelectedValue.ToString(); 

      DateTime eftvfromdt = Convert.ToDateTime(EFTVFROM); 
      DateTime eftvtodt = Convert.ToDateTime(EFTVTO); 

      //Update WeekDay restriction 
      CDSSQLConnections.RunStoredProcedureWithNParams("connDataStore", "sp_AB_BULLETIN_EXCEPTION_INSERT", 
      new Dictionary<string, object> { { "EFTVFROM", eftvfromdt }, { "EFTVTO", eftvtodt }, { "ABSTATUS", ddlStatus.SelectedValue }, { "LASTMODBY", CDSSecurity.CurrentUserID } }); 
      divExceptionAdd.Visible = false; 
      lblAddExcept.Visible = false; 
      repException.DataBind(); 
     } 
    } 
+3

那麼,什麼是你試圖轉換到'DateTime'字符串? – Siyual

+0

@Siyual EFTVFROM和EFTVTO變量 –

+0

您可能需要使用'DateTime.ParseExact',但它確實取決於您期望得到的格式或格式。 – juharr

回答

3

你可以告訴解析器日期時間格式應使用解析字符串

DateTime dt=DateTime.ParseExact("24/01/2013", "dd/MM/yyyy", CultureInfo.InvariantCulture); 
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