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HI ALL,
I am trying to run my application as a Operating system(Directly from Boot menu).is any one help for about this topic please help.
Thanks
If you can think then I Can.
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First, you need to have written an operating system...
Have you written an operating system?
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No i have not written an Operating System. But i wanna try for it. Give me some idea for written an Operating System.
If you can think then I Can.
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Before you learn how to bake pie, you must learn how to create a universe.
Go Cosmos[^] 
I are troll
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A bootloader is 512 bytes long, and has a specific two-byte signature at the end of the file. It gets loaded at 0x7c00:0x0. You get put in real mode. BIOS interrupts are available. But before you even start writing an operating system, make sure you know C++ and assembly language (especially pointers). If you need to get any more knowledge then go to the OSDev Wiki and read anything you can find. Write some code, and follow tutorials to start with
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Thanks Dear,
It's really halpful for me. Can i write for OS Development in MFC VC++.
If you can think then I Can.
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Using MFC?? No, you can't since MFC relies on Windows, which, of course, won't exist under your operating system.
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No. If you want a function, then write it. You have no malloc, strlen or strcat. You're on your own in terms of libraries. If you have the base infrastructure, you could port an existing one, but that's a lot further than a bootloader. That said, GrUB does a rather nice job of loading your kernel for you - just don't expect it to be able to load a VB.Net program - that uses the .net framework, which runs on Windows, which you can't use to boot from
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You want to run a VB.NET app as the boot program?? Never going to happen, since the .NET Framework relies on the existance of Windows, which won't be loaded if you attempt to do what you want.
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Hello Didi,
Thanks for helping. I think it's really helpful for me. It's really very big project so it's take time for learning and Implementing.
I will contect u after some time.
Thanks & Regard
Anubhava
If you can think then I Can.
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Hi to all,
I have this strange behavior in my report for a couple of months. Can anyone have idea, Please help me.
I have >7 reportViewer on my form. Retrieving data from ms access database and putting it on the report.
Now all reports works fine except 2.
WHENEVR I TRY TO VIEW THESE 2 REPORTS IN RUNTIME, ONE DIALOG BOX APPEAR BY NAME "DATABASE LOGIN"....AND ASKING ME TO ENTER
1. SERVER NAME (DISABLE FIELD)
2. DATABASE NAME (DISABLE FIELD)
3. USERNAME (VALUE IS ADMIN) (DISABLE FIELD)
4. PASSWORD
Please help me with it.
Thanks in advance
-----------------------------
I am a beginner
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Hello Himangshu,
I think when you are showing second report at this time Connection was already open and It generate problem for creating another connection.
So show one report at a time or Create Connection With Crystal Report by using Code and pass user name and password.
If you can think then I Can.
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thanks for your reply Anubhava
I am showing only one report at a time...I have show button for each report....all reports and its show button are in different tab-page.
And I have done this through code only.
In this case what could be the problem?
Regards,
Himangshu S
-----------------------------
I am a beginner
modified on Wednesday, May 27, 2009 2:26 AM
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Hello Himangshu,
Can u create connection by using DSN.i think it is better way to find result.Because i am using it.
If you can think then I Can.
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Hi Anu
But I have absolutely no idea about it.
And I have very less time too...
Anyway could you give brief ..if u don't mind
Regards,
Himangshu
-----------------------------
I am a beginner
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I had a similar problem with a project and was never able to find a very nice solution. Basically I designed my reports off an access database that I was never really going to create or have a copy of on the user's machine. I was planning to just create an in memory dataset that matched the schema and load that with data when passing the data into the report. However, when the app runs the crystal report goes out and looks for the database at whatever file location the design database was at on the development machine. When it doesn't find it displayed the login dialog box. To work around the issue, I was forced to always include a blank copy of the design database with my program, and dynamically set a connection between my report and the design database before loading the viewer (because the path is always going to be different between user machines and the development machine). So typically you would just set the report's data source like this:
ReportDocument1.SetDataSource(dsMyDataSet)
I had to do this (I store my design database in the user application data area:
strDesignDBLocation = My.Computer.FileSystem.SpecialDirectories.CurrentUserApplicationData & System.IO.Path.DirectorySeparatorChar & "MyDesignDB.mdb"
ReportDocument1.DataSourceConnections.Item(0).SetConnection(strDesignDBLocation, "", False)
ReportDocument1.SetDataSource(dsMyDataSet)
I'm not sure if you have the same issue, but maybe this is similar and it gives you a place to start looking. If it IS the same problem, please let me know if you find an easier fix.
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thnaks for your reply.
It is not exactly, but ya almost similar problem..I will surely let you know if I get any solution...
My problem is....Application will be installed in an LAN environment. There will be one server wherer the database will be stored....from all other client machine, the db should be access like "\\server\c\folder\db1.mdb"....this connection string is hardcoded..
thats it....
Now problem is for all other reports it works fine except two,.....which are showing the database login
dialpg box....I am going mad to solve it.....need it by today...Donno wat to do...
Anyway, thanks
-----------------------------
I am a beginner
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Creating Connection Using DSN is very easy.
Go to Control Panel > Administrator Tool > ODBC Data Source then use select System DSN or USER DSN.
then click on add button then a dialog will be open and select Microsoft Access Driver(*.mdb) and finish.
then a window will be open then following information must be enter.
Data source name : Name of DSN
Database Path : path of data base
Login : name of admin
Password : password of access database if required
then Select Crystal report and Create connection via DSN.
If you can think then I Can.
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thanks a lot anu...
I will surely try this and come to you back....right now suddenly I have to be busy with job hunting...
so May be i will get back to u by next week....Hope thats ok....
Thanks a lot anyway
-----------------------------
I am a beginner
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Hello
Can any one tell me why?
Warning 1 Function 'ReturnMultipleValue' doesn't return a value on all code paths. A null reference exception could occur at run time when the result is used.
Public Function ReturnMultipleValue(ByVal MyQuery As String) As DataSet
Try
Dim ds As DataSet
con = New OleDbConnection(ConnString)
da = New OleDbDataAdapter(MyQuery, con)
ds = New DataSet
con.Open()
da.Fill(ds)
con.Close()
Return ds
Catch ex As Exception
mdChecking.Prompt(ex.Message.ToString)
If con.State = ConnectionState.Open Then
con.Close()
End If
End Try
End Function
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Because if your code breaks and enters the Exception handler, there is nothing in there returning a DataSet value, nor is there anything after the End Try statement that returns a value.
A better practice would be to declare your return variable before the Try statement, set it's default return value, then go through the code in your Try block WITHOUT returning a value, then after the End Try statement, put your Return statement.
You should really be using a Try/Catch/Finally setup here. Something more like (this is NOT entirely best practice!):
Public Function ReturnMultipleValue(ByVal MyQuery As String) As DataSet
' Default return value.
Dim ds As DataSet = Nothing
Try
' Setup and retrieve your data...
con = New OleDbConnection(ConnString)
da = New OleDbDataAdapter(MyQuery, con)
ds = New DataSet
con.Open()
da.Fill(ds)
Catch ex As Exception
' This will only execute when something goes wrong in the above Try block...
mdChecking.Prompt(ex.Message.ToString)
Finally
' This block will execute no matter what happens in the Try and/or Catch blocks...
If con.State <> ConnectionState.Closed Then
con.Close()
End If
End Try
' Returns either your default value (if something went wrong), or the new value from
' the Try block.
Return ds
End Function
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Thank You
That removed the Warning
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hi there,
I am in a little trouble here. I have a database with employee names details etc.. and I also have a child sub table for their absence. what I want my program to do:
I have a list box that loads full employee names and selects the record
once I click on it I want it to display the info from a child sub table according to the name selected.
I already know how to write to the child sub table but do not know how to display it.
any help would be appreciated.
the code I have written so far fills up the list box with all absence disregarding individual record.
thanks
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What you are trying to do goes under the name of Master-Detail. Describing how to do it would take too much space, so it is best if you Google for master detail vb.net or master-detail vb.net you will find lots of examples/tutorials. Choose one that seems to fit your situation most, and have a play.
Henry Minute
Do not read medical books! You could die of a misprint. - Mark Twain
Girl: (staring) "Why do you need an icy cucumber?"
“I want to report a fraud. The government is lying to us all.”
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