where filename includes the path. It should open the mdb file in Microsoft Access.
This works fine on Windows 7 (my system), but when I try it on Windows XP (client machine) it comes up with the file download security warning dialog asking to Open, Save or Cancel. If I click Open it presents the same dialog but now with Save and Cancel only.
I am sure there is a quick and easy way to open a file in its proper program through VB.NET. I know I can open the database using Interop, but I don't want to go down that route.
Years ago when coding with VB6 and earlier, I used a file open command to create, read, and write a file that made it easy to store useful information.I believe it was binary and you wrote/read the data to a specific line in the file. When you opened the file with something like notepad, it just showed random ascii chars. When I needed a line of data, I could specify what line to load it from.
I have been trying to open a solution file downloaded from my class for the past two days. When I open the assignment file on my destop to access the solution file, it's not there. I have the variables design, obj, bin, and others but not the .sln file which is the one that I need. How can I fix that problem?
I try to map to the network drives and i am able to read the number of folders in the drives and access right for FULL CONTROL ( READ , write etc. ) is given to me and in fact "everyone in the same domain group that i am in. However i encounter the following code error where FileOpen(1, ReadString, OpenMode.Binary) --> It prompts that access to the path of \xxxxxxxxxfolder is denied.
I am trying to write a section of Access VB code that opens the Windows dialog box and allows my user to select and open a file to verify it is the correct one. (Then I use the file path and name to copy, rename and move the file). I have been trying to use the OpenFileName structure and have succeeded in capturing the file path and name my user selected... My main problem in all of this is getting the file to actually open so the user can verify it before I commit the actions noted. This probably can be done as a part of the OpenFileName dialog if I understood it better, but but isn't there also a trivial way to open a file if you have the path and name?
i use this command to open my ms access database file, but it is not working, what is the problem.... i learn this command from internet tutorials. [code]
I made a program in Visual Basic 6.0 and am trying to convert it to vb.net.so i am at the stage where the program needs to load the Access file and populate the listbox according to which radio button i chose.
I am trying to open an Access file but so far I am not succeeding. [Code] I receive following error: Cannot start your application. The workgroup information file is missing or opened exclusively by another user. I am sure the file is not open and I tried already to add in the connection string: ;User ID=User and / or Jet OLEDB:System database=C:SYSTEM.MDW; at the and but without result.
Possible Duplicate: Access Denied in SQL 2008 EXPRESS?I have create my code in order to write images in a remote sql server All the details of accessing and writing are fine until now, including the system account right now i'm in the command of:
SqlFileStream = New SqlFileStream(filePathName, fileToken, FileAccess.Write)
and when i'm trying to execute it the Server returns the error 'Access denied' I have try all the posible ( those which i know) combinations to overcome this error but nothing Please give me the best assistance you may have I've put a sniffer in my PC to lookup the packages between Server and my PC, so here what i got: the first addres is the Server address and the second is my PC address.
I dare to say that this error comes from the Windows program when the SQL 2008 tries to write some DATA to the filies which creates on C:sqlRemData.....
I am new to this and have almost 0 programming knowledge. I want to automate transferring data from excel spreadsheet to access. i surfed around on the net to see if there were examples of codes that i can copy.this is what i currently have.
This is probably a pretty novice design question. I'm trying to work my way through a number of requirements and give the users the experience they're looking for.I've written a tool that does big calcluation-type things. It currently consists of a class library and command line tool (separate .NET projects.) We're using an Access database format as the file type because it can keep all the various tables together in one file. A few other items about the application: There are not many users. There are no concerns with scalability. There are not great concerns with updates. Desktop is desired. Not web.Using VB and .NET 3.5 SP1
I now need to develop a GUI front end that will allow typical File/Open and File/Save type operations.Users expect that they can open a file, edit it some, then either choose to save it or close it unsaved without any changes being written back to the file. Saving it would obviously save all changes affecting all tables back to the file.
Does it then make sense to use a temp file for something like a proxy then? To, when a user "opens" a file, copy the source Access file to a local temp file and then use that for the editing session? Then, if the user "saves", copy the local temp file back to the source path?
Update: [tagged with ms-access tag too] Also, I omitted the fact that users would expect typical File / Save As functionality too. I think the design I've put in question in this post is what is traditionally called the Proxy design pattern. Has anyone tried this (successfully!) with Access database files before? Words of caution or advice?
I am writing a program to calculate Pi to several hundred billion decimal places and this will require lots of GB of memory. I wrote a test program in VB2008 that saved the first 16 digits of Pi (without the decimal point) to both a text file and a random access file, just to be sure it was outputting the numbers properly. For reference the first 16 digits of Pi are:
I've gone through about 16 hours and two packs of cigarettes trying to figure this out. First a little background. I was using 6.0 up until 2004 when I went to prison. I'm out now, and trying to relearn the trade, using VS 2005. I'm currently porting some 6.0 code from another project, SpyCast Webcam Studio, into VB 8.0. It's disheartening, to say the least. None of the old built-in subs/functions work anymore, so I have to scour the forums to relearn each and every function.The section I'm doing now takes a snapshot from the webcam (Video API --> PictureBox --> Save as Jpeg), then upens the file to upload it to the server via HTTP POST. I've been using this code in SpyCast for years with no trouble, but I spent many hours trying to piece together the right code to open the binary file to read its contents. I pieced together two methods I found around the forums, one using FileStream() though the code I found wasn't for binary files, even though it said it was, so that code doesn't really work. Method two uses Microsoft.VisualBasic.FileOpen() and works better.
Here's the kicker. By the time I run through the rest of the rigamarole of uploading the file, by the time I read it on the webserver, it's *slightly* corrupted. It's a valid Jpeg, no errors, but the picture looks like when I use to watch the Playboy Channel when I was a kid scrambled with weird colors and whatnot. [code] Each "chunk" is basically one "line" of the file. It looks like a single LineInput() return is the text between two carriage returns. Am I correct? I tested this with a flat text file, and it looks true. However, That one input line returns the text or data with the carriage returns *stripped*! ***?!? =( Fine, I have no problem adding my own vbCrLf to each LineInput(), if I were opening text. but this's binary. A character could be Chr(10) or Chr(13), both of which are removed from the original file contents.So I could very well need to use something other than LineInput(), but I haven't found any other examples on the web using this method.
I need a code for the open file dialog that if a file is not supported then a message box appears stating the error. This is my open file dialog code. [code] The message box pops up like it is suppose to, when you click ok the open file comes back up like it is suppose to. When you put in the correct file the message box comes back up again. Also the message pops up even if the correct file is put in the open file. but if you close the file dialog the image is where it is suppose to be.I believe the code for the public sub wrongfile is what is the problem. can some one look it over and let me know what changes are needed.
It is saved on my desktop for easy access.I can close the program. I can click the solution to open it right back up, no problems.I then zip my file (also to the desktop)
My problem: I moved the original file (Project A) to another folder and extracted the unzipped Project A onto the desktop. When I open the new Project A's solution file, nothing happens besides VB starting up. It doesn't show me anything.
I need to open an excel file from vb.net and then search it for specific data. I then need to take those data and insert them into text boxes on a form that i have created. This is all controlled by a button click. I already have some code that will open a file dialog box and let me navigate to the correct file, but I am having trouble with the search portion. I have tried the Find function but I am not sure of the proper syntax. I am using Visual Studio 2008 and Excel 2003.
Well im making a program that will Add the Safe File Name of the File opened in the Open File Dialog to the first column of the listview and the File name and path to the second. What i want it so when i double click on the Safe file name it will open the file specified in the path in the second column.i already got the add file to it.
code snippet that would allow a progress bar to track the input of a text file? Normally I would not bother with this, but the text files are > 10,000 lines long, which is noticable even on a fast machine. The number of lines is variable, so I would assume that one would not use a fixed value to calculate when the progress bar goes 100%.
Is there a better way of finding the path of a file that was opened with the open file dialogue? This is what I did. It works, but it seems like there should be a way to get the path through one of the open dialogue options.
Code: 'm_PicSource = OpenFileDialog1.FileName 'm_PicSource is a global var Dim strCnt As Integer = m_PicSource.Length - 1
what im trying to do and what i have done so far is probably a little backwards not sure sure how to id it any other way.when the user generates an error because a particular file doenst exist ive got my dialoge box to generate an .txt file for each file it errors on, so that the administrator can check and update missing files as an when - shouldnt be many files missing in the first place, should just be new product ranges or item just launched.what im stuck on is:
1. is there a better way
2. how do i open a .txt file and add the extra lines to it, so i re use the file but keeping the previous data.
