我在页面上有ext:FileUploadField。文件上传后,我需要显示该文件的链接。 我动态创建一个LinkButton,将它添加到Panel1上,我看不到LinkButton!我不知道为什么!
<ext:Panel ID="Panel1" runat="server">
<Content>
<ext:FileUploadField ID="FileUploadField1" runat="server" EmptyText="Choose a file" FieldLabel="File" Icon="ImageAdd" />
</Content>
<Buttons>
<ext:Button ID="SaveButton2" runat="server" Text="Upload">
<DirectEvents>
<Click OnEvent="UploadClick"></Click>
</DirectEvents>
</ext:Button>
</Buttons>
</ext:Panel>
protected void UploadClick(object sender, DirectEventArgs e)
{
if (this.FileUploadField1.HasFile)
{
var attachment = new Attachment { ............ };
if (UploadAttachment(attachment))
{
X.Msg.Show( ...... );
var linkButton = new LinkButton();
linkButton.ID = "fdsfdsfds";
linkButton.Text = attachment.Name;
linkButton.NavigateUrl = "#";
linkButton.Render();
Panel1.Add(linkButton);
// Panel1.Render(true);
Panel1.DoLayout(true,true);
}
else
{
//................
}
}
else
{
//................
}
}
答案 0 :(得分:1)
我猜你需要将它添加到面板的按钮列表而不是面板本身。如果您有合适的布局并添加第二个项目,那么您可能还会遇到布局问题。
答案 1 :(得分:0)
尝试使用此代码:
X.Msg.Show( ...... );
var linkButton = new LinkButton();
linkButton.ID = "fdsfdsfds";
linkButton.Text = attachment.Name;
linkButton.NavigateUrl = "#";
linkButton.Render(Panel1, RenderMode.AddTo);
这将直接向Panel1添加链接按钮