How do I force full post-back from a button within an UpdatePanel?
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
You can use the Triggers property of the UpdatePanel to register actions that trigger a full postback.
Add a PostBackTrigger object to that property, containig the ControlID of the control which needs to trigger a full postback.
<asp:UpdatePanel ID="UpdatePanel1" UpdateMode="Conditional" runat="server">
<ContentTemplate>
...
</ContentTemplate>
<Triggers>
<asp:PostBackTrigger ControlID="myFullPostBackControlID" />
</Triggers>
</asp:UpdatePanel>
Method 2
Just adding this because nobody else has. It is possible to do this in code-behind in one line of code without any of the above methods. Just put this in page_load:
Visual Basic
ScriptManager.GetCurrent(Me).RegisterPostBackControl(myButtonID)
C#
ScriptManager.GetCurrent(this).RegisterPostBackControl(myButtonID);
Method 3
From here:
Use the PostBackTrigger control to enable controls inside an UpdatePanel to cause a postback instead of performing an asynchronous postback.
<Triggers>
<asp:PostBackTrigger ControlID="controlID" />
</Triggers>
Method 4
I had the same problem using an ImageButton inside a WebDataGrid.
I put the Line from EvilDr inside the preRender for the webdatagrid instead. Worked great!
for (int i = 0; i < wdgMyData.Rows.Count; i++)
{
ScriptManager.GetCurrent(this).RegisterPostBackControl((ImageButton)wdgMyData.Rows[i].Items[3].FindControl("btnDownloadExcel"));
}
Method 5
Its an old question there is bit tricky approach as well, when everything is in update panels and you want to do full post back so that your Document.Ready code works on click.
1. Create a dummy button outside the update panel and click it from code behind like this
ScriptManager.RegisterStartupScript(Page, this.GetType(), "ResetDoc", "ResetDocumentReady();", true);
2. Define the function on front end like this
function ResetDocumentReady() {
$("[id$=DummyButton]").click();
}
NOTE: But the best way is using trigger, the answer posted by @Thibault Falise 🙂
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0