ASP.NET

16-Jul-07 20:25

Pleae help me to acess Menu Control's Items. How Can I manipulate Menu Items from code (C#)?

Thanks,
I'm using Master Pages if it makes some differance.

sidbaruah17-Jul-07 1:47

Re: Accessing Menu Items (C#) [modified]

17-Jul-07 1:47

Use the following!!!

menu_name.Items[i]."properties....

You can get and set them as you like!!!

I was born dumb!!
Laugh | :laugh:

Programming made me laugh Laugh | :laugh:

!!!
--sid--

totocaster17-Jul-07 3:49

17-Jul-07 3:49

Thansk for response, I knew that but It doesn't work Frown | :(

That's Why I asked it here. Here is my situation:
I have a master page where the menu is located, and in Page_Load i'm tring to write this:

if (!Context.User.IsInRole("Administrator"))
{
// here i'm tring to disable some menu items
}

but When I type Menu1 (which is Menu's ID) nothing heppens. Frown | :(

and i do not know why... Frown | :(

Thanks

sidbaruah17-Jul-07 18:24

17-Jul-07 18:24

To access the objects in the master page you need to create a masterpage object in the page you are referring!!!

Before pageload write
master_page_class_name masterpage_objectname;

and write this is pageload:
masterpage_objectname= (master_page_class_name)Page.Master;

Now use the masterpage object to alter the menu
hope that helps!!

I was born dumb!!
Laugh | :laugh:

Programming made me laugh Laugh | :laugh:

!!!
--sid--

totocaster17-Jul-07 20:19

17-Jul-07 20:19

but I'm trying to reach menu1 from *.master.cs file. Frown | :(

Thanks for info anyways.

sidbaruah17-Jul-07 20:26

17-Jul-07 20:26

Thats quite weird!!
You should be able to do that...anyways..try to figure out something im not getting any ideas Smile | :)

I was born dumb!!
Laugh | :laugh:

Programming made me laugh Laugh | :laugh:

!!!
--sid--

totocaster17-Jul-07 20:32

17-Jul-07 20:32

maybe that's because menu is in big LoginView?

sidbaruah17-Jul-07 20:37

17-Jul-07 20:37

wats a big login view??

I was born dumb!!
Laugh | :laugh:

Programming made me laugh Laugh | :laugh:

!!!
--sid--

totocaster17-Jul-07 20:43

Description field binding

17-Jul-07 20:43

LoginView is a standart control which comes with VS. I have a master page with this login view like so:

<%@ Master Language="C#" AutoEventWireup="true" CodeBehind="Environment.master.cs" Inherits="Optical_Box_Manamgemt_System.Environment" %>

<title>Title

<asp:loginview id="LoginView1" runat="server">
<loggedintemplate>
//Everything Goes Here, Including Menu and Content Place Holder

if you are skype user you can add me, my skype name is same as my nickname here.
Thanks for Help Wink | ;)

samerh16-Jul-07 20:23

samerh

16-Jul-07 20:23

Hi all,
iam binding a description field varchar(500) to a grid view. i want to display in the grid view the only first 20 or 30 characters only and not all the characters in this field. Is this possible? Thanks

Error in invoking upload_serverclick method

Hari_101016-Jul-07 19:55

16-Jul-07 19:55

Hi all,

I want to upload multiple files in ASP.Net 2.0 with the file upload control. So I have download the source code "Upload multiple files Hotmail style -
Kwame Hubbard" from the following site.

http://www.codeproject.com/aspnet/multiuploads.asp

It seems that everytime I invoke this method, it will tell me that "Error - a file name must be specified" no matter what I do. The add and remove methods work just fine.

In anycase, in the upload method, it seems to always think that filecount or ListBox1.Items.Count is equal to 0, and therefore won't proceed with the uploading.

Is it because the buttons themselves are webcontrols, while the upload is a htmlcontrol? Somehow the html control is not able to access the existing value of filecount.

So could you please help me out to solve this problem. Please its very urgent.

Thanks and Regards,
Hariharan C

N a v a n e e t h16-Jul-07 20:12

16-Jul-07 20:12

Hello

I am not getting why you are going behind the way what the article explained ? Uploading multiple files will be such a simple task if you use gridview and normal HTML upload file control. What you need to do is

1. Place a GridView
2. Add template column
3. Add File upload control to template column and make it runat=server
4. Codebehind, use gridviews FindControl() method to find the upload control.
5. Use uploadcontrolObj.PostedFile.SaveAs(filename)

If you are uploading very huge files, you need to change something in Web.Config.

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Hari_101016-Jul-07 20:21

16-Jul-07 20:21

Hi Navaneeth,

I am not able to follow your mehtod. So could you please explain me little bit clearly with the code.

Thanks and Regards,
Hariharan C

N a v a n e e t h16-Jul-07 20:42

16-Jul-07 20:42

Hari_1010 wrote:
I am not able to follow your mehtod. So could you please explain me little bit clearly with the code.

Please let me know, Are you familiar with gridview ?

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Hari_101016-Jul-07 20:45

16-Jul-07 20:45

Yes Navaneeth, I am familiar with gridview.

N a v a n e e t h16-Jul-07 20:59

16-Jul-07 20:59

First you need to place a gridview. Add a template column. In ItemTemplate add
<input type="file" id="MyFileUpload" runat="server">

Assume you need to upload 5 file. So you need 5 rows in grid view. For that create a dummy datatable with 5 records and bind to gridview. Now you will see gridview binding with 5 records and in each row you will get each file upload control.

Down to gridview place a button. Inside button click write

for ( int i=0;i<GridviewRows;i++)
{
    HtmlInputFile fileUpload = (HtmlInputFile)GridView.Rows[i].Cells[FileUploadCell].
FindControl("MyFileUpload");
    fileUpload.PostedFile.SaveAs("YourServerLocation");//Need to use server.mappath
}

FindControl method will find the control from gridview and give you the object. Using that object you can do upload. You need to handle exception efficiently. This will raise NullReferenceException if FindControl() is not getting the object. So make sure that you will specify the cell where FileUploadControl resides.

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Hari_101016-Jul-07 21:07

16-Jul-07 21:07

Hi Navaneeth,

Thanks for your code. But my problem is, my client dont want that much file upload control to be seen in the UI. The requirement of my client is, he needs only one file upload control and in one click he wants to select more than one file at time.

Can we create any web user control. Do you have any suggestion for that.

Thanks and Regards,
Hariharan C.

N a v a n e e t h16-Jul-07 21:32

16-Jul-07 21:32

Hari_1010 wrote:
my client dont want that much file upload control to be seen in the UI. The requirement of my client is, he needs only one file upload control and in one click he wants to select more than one file at time.

Why don't you tell this first. So whatever code I written for you in previous posts became unusable Mad | :mad:

. No issues, let's start again from the scratch.

Now for this requirement we can follow the method what that article say's. But you can't reuse the article code since it won't work perfectly. You need to do some small changes in his code. You can try with this

1. Put one file upload control.( As your client said )
2. Put list box just to show the files selected. No much role in this movie
3. Place two buttons with "Add to List", and "Upload" texts.
4. OnClick of "Add to List" button, get the postedfile object and store that in one arraylist.
5. Now move arraylist to one session. ( Used to get the value across page postbaccks )
6. On Upload button click cast session to arraylist. Then cast array list object to HttpPostedFile object. Now use SaveAs() method in that object

In that article author is not using sessions. Your Upload button code would be

ArrayList ObjArr = (ArrayList)Session["files"];
HttpPostedFile postFile = (HttpPostedFile)ObjArr[0];
Response.Write(postFile.FileName);
Response.Write(postFile.ContentLength.ToString());
postFile.SaveAs(Server.MapPath("/filename.zip"));

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Hari_101016-Jul-07 20:58