2012-01-14 2 views
1

나는 asp.net VB에서 gridview를 만들었습니다. 그리고 난 ID를 열에 새 페이지 등 http://localhost/defualt.aspx?id= (IDrow에서 datarow)에 착륙 할 수있는 링크를 만들고 싶어 링크를 추가하고 싶습니다. gridview로 가능합니까? 코드는 다음과 같습니다gridview (asp.net)에 링크를 추가하는 방법은 무엇입니까?

<asp:GridView ID="GridView1" runat="server" AutoGenerateColumns="False" 
    DataKeyNames="EmployeeID" DataSourceID="SqlDataSource1"> 
    <Columns> 
     <asp:BoundField DataField="EmployeeID" HeaderText="EmployeeID" 
      InsertVisible="False" ReadOnly="True" SortExpression="EmployeeID" /> 
     <asp:BoundField DataField="LastName" HeaderText="LastName" 
      SortExpression="LastName" /> 
     <asp:BoundField DataField="FirstName" HeaderText="FirstName" 
      SortExpression="FirstName" /> 
    </Columns> 
</asp:GridView> 
<asp:SqlDataSource ID="SqlDataSource1" runat="server" 
    ConnectionString="<%$ ConnectionStrings:NwindConnectionString %>" 
    ProviderName="<%$ ConnectionStrings:NwindConnectionString.ProviderName %>" 
    SelectCommand="SELECT [EmployeeID], [LastName], [FirstName] FROM [Employees]"> 
</asp:SqlDataSource> 

답변

2

당신은

 <asp:HyperLinkField DataNavigateUrlFields="EmployeeID" 
DataNavigateUrlFormatString="/default.aspx?id={0}" 
          DataTextField="EmployeeID" HeaderText="EmployeeID" /> 
HyperLinkField를 사용할 수있다
0

EmployeeID의 BoundField를 삭제하고 TemplateField를 대체하십시오. 이 시도 :

<asp:TemplateField HeaderText="EmployeeID" InsertVisible="False" 
       ShowHeader="False" SortExpression="EmployeeID"> 
       <ItemTemplate> 
        <asp:LinkButton ID="LinkButton1" runat="server" CausesValidation="false" 
         CommandName="" PostBackUrl='<%# "http://localhost/defualt.aspx?id=" & Eval("EmployeeID") %>' Text='<%# Eval("EmployeeID") %>'></asp:LinkButton> 
       </ItemTemplate> 
      </asp:TemplateField> 
관련 문제