How to visible next fileupload control on click on linkbutton ?

Question

0.00/5 (No votes)

See more:

Hi to all,
I am working on project which has based as ticket submitting which having "submit_ticket.aspx" page with C# code behind and sql server 2008 r2 database!
For this i have taken 5 file upload controls and one link button has "id=lnkaddmore" !
The first file upload control has visible by default and remaining aren't visible !
I wanna to show adding next file upload control should visible on click of lnkaddmore but previous shouldn't postback !
I tried lots !
but I didn't get success yet !
Anybody Help Me Please ?
Thanks in advance !

Posted 18-Apr-14 20:00pm

prakashvidwans

Add a Solution

Comments

Ganesh KP 19-Apr-14 6:12am

I think because of postback the page is getting refreshed and set the state of a page to the initial first time load. So, can u please post the code of where you are trying to do set the visibility of the control.

Ganesh KP 19-Apr-14 6:12am

Try these links also

http://stackoverflow.com/questions/9360791/visible-property-of-asp-net-control-not-working

http://stackoverflow.com/questions/11309171/asp-net-update-panels-and-postback

ZurdoDev 19-Apr-14 8:22am

FileUpload does an ajax postback I believe so if you are setting visible in C# it won't work without a full postback.

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Karteek Panakani · Answer 1 · 2014-04-19T02:59:00

XML

<div id="divFile">
    <div>
        <asp:FileUpload ID="FileUpload1" runat="server" Style="width: 200px;" />
        <asp:LinkButton ID="lnkAddFiles" runat="server" OnClientClick="return AddFile()">Add</asp:LinkButton>
    </div>
</div>

XML

<script type="text/javascript">
       var i = 1;
       function AddFile() {
           i++;
           var div = document.createElement('DIV');
           div.innerHTML = '<input id="file' + i + '" name="file' + i + '" type="file" style="margin-bottom:10px;width:200px;" /><a href="#" style="font-weight:bold; margin-left:10px;" onclick="return RemoveFile(this)">Remove</a>';
           document.getElementById("divFile").appendChild(div);
           return false;
       }
       function RemoveFile(file) {
           document.getElementById("divFile").removeChild(file.parentNode);
           return false;
       }
   </script>

Use the above design and javascript. it may use full to you.