代碼實現AJAX非同步回傳(彈出javascript指令碼)、非同步更新的錯誤處理

最後更新：2018-12-04 來源：互聯網

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--------------------------------代碼實現非同步回傳--------------------------------------------
頁面關鍵代碼

<body>
    <form id="form1" runat="server">
  <asp:ScriptManager ID="ScriptManager1" runat="server">
  </asp:ScriptManager>

  <asp:UpdatePanel ID="UpdatePanel1" runat="server" UpdateMode="Conditional">
   <ContentTemplate>
    <%= DateTime.Now %>
    <asp:Button ID="Button2" runat="server" Text="Button" />
   </ContentTemplate>
  </asp:UpdatePanel>
  <asp:Button ID="Button1" runat="server" Text="Button" OnClick="Button1_Click" />
    </form>
　</body>
　後台代碼
public partial class Demo2_API : System.Web.UI.Page
{
    protected void Page_Load(object sender, EventArgs e)
    {

　　　//讓button1能夠引發非同步回傳(且button1不在UpdatePanel中)
ScriptManager.GetCurrent(this).RegisterAsyncPostBackControl(this.Button1);

　　　//讓button2能夠引發同步回傳(且button2在UpdatePanel中)
ScriptManager.GetCurrent(this).RegisterPostBackControl(this.Button2);

}

protected void Button1_Click(object sender, EventArgs e)
{
this.UpdatePanel1.Update();//更新UpdatePanel1中的內容
}
}

----------------------非同步回傳時彈出javascript指令碼-----------------------
頁面關鍵代碼

  <asp:UpdatePanel ID="UpdatePanel1" runat="server">
   <ContentTemplate>
    <%= DateTime.Now %>
    <asp:Button ID="Button1" runat="server" Text="Button" OnClick="Button1_Click1" />
   </ContentTemplate>
  </asp:UpdatePanel>　　　
  <asp:UpdatePanel ID="UpdatePanel2" runat="server">
   <ContentTemplate>
    <%= DateTime.Now %>
   </ContentTemplate>
  </asp:UpdatePanel>
    </form>
</body>

後台代碼：

public partial class Demo5_ScriptManagerReg : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{

}

protected void Button1_Click1(object sender, EventArgs e)
{

　　　//在UpdatePanel1更新時彈出javascript指令碼
ScriptManager.RegisterStartupScript(this.UpdatePanel1, this.GetType(), "UpdatePanel1", "alert(1)", true);

　　　//在UpdatePanel2更新時彈出javascript指令碼
ScriptManager.RegisterStartupScript(this.UpdatePanel2, this.GetType(), "UpdatePanel2", "alert(2)", true);

}
}

-----------------------利用代碼（建立UpdatePanel）實現AJAX效果-----------------------------
頁面代碼：
<body>
    <form id="form1" runat="server">
  <asp:ScriptManager ID="ScriptManager1" runat="server">
  </asp:ScriptManager>

</form>
</body>

後台代碼：

public partial class Demo7_ProgramicUpdatePanel : System.Web.UI.Page
{
    protected void Page_Load(object sender, EventArgs e)
    {
  UpdatePanel up = new UpdatePanel();
  up.ID = "UpdatePanel1";
  this.Form.Controls.Add(up);

LiteralControl lc = new LiteralControl(DateTime.Now.ToString());
up.ContentTemplateContainer.Controls.Add(lc);

  Button btn = new Button();
  btn.Text = "Refresh";
  up.ContentTemplateContainer.Controls.Add(btn);
    }
}

--------------------AJAX非同步更新的錯誤處理-----------------------------------------------------

可以通過ScriptManager控制項的OnAsyncPostBackError事件和AsyncPostBackErrorMessage屬性捕捉和設定回傳時的錯誤訊息

樣本：

頁面代碼：

後台代碼：

public partial class UpdatePanelErrorMessage : System.Web.UI.Page
{
    protected void Page_Load(object sender, EventArgs e)
    {

    }
    protected void Button1_Click(object sender, EventArgs e)
    {
        try
        {
            double a = Convert.ToDouble(TextBox1.Text.Trim());
            double b = Convert.ToDouble(TextBox2.Text.Trim());
            double res = a * b;
            Label2.Text = res.ToString("f3");//輸出字串格式的控制
        }
        catch (Exception ex)
        {
            if (TextBox1.Text.Length > 1 && TextBox2.Text.Length > 1)
            {
                ex.Data["cuo"] = "這兩個數無法相乘!";
            }
            throw ex;
        }
    }
    protected void ScriptManager1_AsyncPostBackError(object sender, AsyncPostBackErrorEventArgs e)
    {
        if (e.Exception .Data ["cuo"]!=null )
        {
            ScriptManager1.AsyncPostBackErrorMessage = e.Exception.Data["cuo"].ToString();
        }
        else
        {
            ScriptManager1.AsyncPostBackErrorMessage = e.Exception.Message;
        }
    }
}

本文章原先以中文撰寫並發佈於 aliyun.com，亦設英文版本，僅作資訊用途。本網站不對文章的準確性，完整性或可靠性或其任何翻譯作出任何明示或暗示的陳述或保證。如對該文章有任何疑慮或投訴，請傳送電郵至 info-contact@alibabacloud.com 並提供相關疑慮或投訴的詳細說明。職員會於 5 個工作天內與您聯絡，一經驗證之後，即會刪除該侵權內容。

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