I feel really silly having to ask this question as I know I should not be having so much trouble with this simple task....but I am trying to launch my .msi when a user pushes a button of a form. I am certain this is a one liner but I cannot for the life of me figure this out. I have the .MSI file on my desktop so I want the button to also be able to have the user select where the msi file is. If anyone could help me that would be grand...
Look at Process.Start.
Process.Start("path to msi");
To get the path to the file, you can use the FileDialog class (assuming winforms).
OpenFileDialog openFileDialog1 = new OpenFileDialog();
if(openFileDialog1.ShowDialog() == DialogResult.OK)
{
Process.Start(openFileDialog1.FileName);
}
Look at using this to get the file:
FileDialog dialog = new FileDialog();
Related
I'm creating an application to automate some processes. One of them is creating docx file from the dotx template.
Steps are quite easy: app opens MS Word with test.dotx file and SaveAs it to c:\temp as a test.docx. It should be as close to user's actions as possible. When the file is opened (from dotx so it is docx already) all I need is to open SaveAs dialog and push "save" (or just "enter", because focus is set on "save" button).
The problem is how to "hit" the save/enter. I tried SendKeys but I am in ShowDialog() which is waiting for the result and cannot perform SendKeys at the moment. Of course if I press enter from keyboard or cklick on "Save" all works perfectly, but this one "press" I'd like to do from the code. Could you please point me how to solve this (if it is possible at all)? Thank you.
Here is the part of the code I'm strugglig with:
SaveFileDialog saveFileDialog1 = new SaveFileDialog
{
InitialDirectory = #"C:\Temp\",
DefaultExt = "docx",
FileName = "test"
};
if (saveFileDialog1.ShowDialog() == DialogResult.OK)
{
object FileName = saveFileDialog1.FileName;
doc.SaveAs(ref FileName);
}
If hit the save/enter is your requirement, the recommendation is use Spy++ (this Tool can be installed by Visual Studio Installer) to capture both: "the dialog with save button" and "save button" alias/class; then, programming with C# PInvoke (for example, using FindWindow and FindWindowEx) to send/post message to simulate this "save" button.
I've been trying to follow tutorials and various Stack Overflow posts, etc. to implement an OpenFileDialog to select a file. The problem is, it seems I can't get my program to continue with the rest of the logic. Not entirely sure if it has something to do with the fact I'm trying to open the file dialog inside my main window or what. Consider the following snippet:
public MainWindow()
{
InitializeComponent();
string file = "";
// Displays an OpenFileDialog so the user can select a file.
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "Files|*.txt;*.out";
openFileDialog1.Title = "Select a File";
openFileDialog1.ShowHelp = true;
// Show the Dialog.
// If the user clicked OK in the dialog and
// a file was selected, open it.
if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
{
//file = openFileDialog1.FileName;
//file = openFileDialog1.OpenFile().ToString();
//openFileDialog1.Dispose();
}
openFileDialog1 = null;
Console.WriteLine("File path is: " + file);
As you can see, I've even tried setting the "Help" value to true before the dialog finishes. I've tried to select both the file name for the file string, etc. but to no avail - the program seems to simply wait after the file is selected from the dialog. Would anyone here have a suggestion for a solution?
Previously I had the same problem with WPF. When you are working with WPF, System.Windows.Form Namespace is not included to your Project references;
and in the other hand actually, there are two OpenFileDialog, the first is System.Windows.Forms.OpenFileDialog (this is what you have) and the second is Microsoft.Win32.OpenFileDialog. if you want to get your code to work you must add System.Windows.Forms into your references:
Solution Explorer -> YourProject -> References (Right Click and Add Reference...) -> Assembly -> Framework -> Find And Select System.Windows.Forms -> OK
and the Next Solution is to use Microsoft.Win32, it's pretty easy. just add this Namespace into your code file and Change your code like this:
string file = "";
// Displays an OpenFileDialog so the user can select a file.
OpenFileDialog openFileDialog1 = new OpenFileDialog();
openFileDialog1.Filter = "Files|*.txt;*.out";
openFileDialog1.Title = "Select a File";
// Show the Dialog.
// If the user clicked OK in the dialog and
// a file was selected, open it.
if (openFileDialog1.ShowDialog() == true)
{
file = openFileDialog1.FileName;
//file = openFileDialog1.OpenFile().ToString();
//openFileDialog1.Dispose();
}
openFileDialog1 = null;
Console.WriteLine("File path is: " + file);
OpenFileDialog.ShowDialog() is a modal method:
FileDialog is a modal dialog box; therefore, when shown, it blocks the
rest of the application until the user has chosen a file. When a
dialog box is displayed modally, no input (keyboard or mouse click)
can occur except to objects on the dialog box. The program must hide
or close the dialog box (usually in response to some user action)
before input to the calling program can occur.
This means that invoking this method will block your main thread until the dialog box is closed. A few of the options you have are:
Invoke the FileDialog after the main window is initialized.
Invoke the FileDialog from a thread. In this case, be careful to synchronize.
I am working on winforms application in C#. What I want to achieve is to get a file from user for which I am using the following code:
OpenFileDialog dlg = new OpenFileDialog();
if (dlg.ShowDialog() == DialogResult.OK)
{
string sFileName = dlg.FileName;
//my code goes here
}
Now, everything is working fine but I want to put 3 radio buttons in the same dialog box, meaning I would now get two things from this dialog box
string sFileName = dlg.FileName; //same as in case of traditional dialog box
//some thing like this which tells which radio button is selected:
dlg.rbTypes.Selected
How do I achieve this?
Yes, that's possible, I did the same kind of customization with SaveFileDialog successfully and it's pretty interesting.
Follow the following links:
http://www.codeproject.com/KB/dialog/OpenFileDialogEx.aspx
http://www.codeproject.com/KB/cs/getsavefilename.aspx
http://www.codeproject.com/KB/dialog/CustomizeFileDialog.aspx
Also my own questions too will help you:
Change default arrangement of Save and Cancel buttons in SaveFileDialog
How to stop overwriteprompt when creating SaveFileDialog using GetSaveFileName
You have to use the WinAPI for this and you need to write the ShowDialog method in your own calling the GetOpenFileName windows function inside it, instead of calling .net's OpenFileDialog. The GetOpenFileName will create the windows OpenFileDialog. (Refer to http://msdn.microsoft.com/en-us/library/ms646927%28v=vs.85%29.aspx). This together with writing the HookProc procedure and catching events such as WM_INITDIALOG, CDN_INITDONE inside it will help you do what you want.
To add radio buttons etc., you have to call the windows functions such as CreateWindowEx and SendMessage....
The 2nd link has the exact direction to the customization...
Ask for any clarifications...
On XP you need to use the hook procedure method and the GetOpenFileName API. On Vista and later this will result in a horrid looking file dialog with limited utility, e.g. no search. On Vista you should use IFileDialog and to customise the dialog you need the IFileDialogCustomize interface. Because the new Vista dialogs are exposed as COM interfaces they are quite easy to consume in .net.
I asked this question a minute ago and was not specific enough so let me try again.
I am trying to generate a report of inventory information that is already made and have it update from the user input into text boxes on the form and then have a button to make the .txt file of the report show to the screen and have the updated information on it.
I have the GUI created and have the button created and the .txt file is created. I just need to know how to make it where I can click the button and have the .txt file appear to the screen.
Using System.Diagnostics;
...
String filename = "C:\\....\data.txt"; \\ File Created With Information
Process.Start(filename); \\ Will open file with default program
The above code can be used to open an external program to display your text file.
As usual I recommend using try/catch since you are dealing with external I/O (files).
You can just start the notepad process with your *.txt file as the argument and start the process can't you?
Found this link that might help you: http://www.csharp-station.com/HowTo/ProcessStart.aspx
Assign a click event to your button (in your class constructor for instance):
button.Click += new EventHandler(button_Click);
In the event, start notepad.exe in a new process:
void button_Click(Object sender, EventArgs e) {
ProcessStartInfo startInfo = new ProcessStartInfo();
startInfo.FileName = "notepad.exe";
startInfo.Arguments = "C:\Path\To\My\file.txt";
Process.Start(startInfo);
}
OpenFileDialog does not show complete filename in Windows 7. The problem is also reported connect.microsoft.com.
There is also a work around by setting openFileDialog.AutoUpgradeEnabled = false. But that causes old xp style dialog. Is there any way to fix it without doing the above workaround. There is no window handle in the dialog so i cannot figure out how to add custom window message handler to do sendmessage() to fix it.
using (OpenFileDialog openFileDialog = new OpenFileDialog())
{
openFileDialog.FileName = "abcdefghijklmnopqrstuvwxyz";
openFileDialog.ShowDialog();
}
On the Connect there is another workaround listed as:
Posted by Robert Breitenhofer on 10/09/2010 at 01:52 Add:
openFileDialog.ShowHelp = true;
before calling ShowDialog().
I only have Win XP so I cannot test this, hope this helps.