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Requirement: I want to create an application, which should be able to create a file, with the file extension be specified by me. And, when i double click to open that file, it should open it's content in that application. How can this be achieved? ( like word,excel..etc) Code so far: For example.. I have created a application with a Richtextbox control to display data,buttons to create a new file, openfile,savefile and to exit. And let my file extension be ".gun". Time to Answer: Now, what happens is i create new file, save(savefiledialog) it with .gun extension. This files gets opened when i open it using 'open(openfiledialog) button' which is present in my application. But when i double this file (which is placed on desktop) it opens only the application and not with the content.? How can i get it with the content when the file is double clicked?
You will need to associate the file extension *.gun with your application - the article http://support.microsoft.com/kb/185453 describes for VB but its basicallyde sticking some info in the registry...
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Look for method "Main" in your application. If you have no parameters in it "Main()" then change it to "Main(string[] args)". If you double-click .gun file, then it is like calling using command prompt: MyApplication.exe myFile.gun Just take your .gun file in "args" variable.
Or just call
Environment.GetCommandLineArgs() -
Requirement: I want to create an application, which should be able to create a file, with the file extension be specified by me. And, when i double click to open that file, it should open it's content in that application. How can this be achieved? ( like word,excel..etc) Code so far: For example.. I have created a application with a Richtextbox control to display data,buttons to create a new file, openfile,savefile and to exit. And let my file extension be ".gun". Time to Answer: Now, what happens is i create new file, save(savefiledialog) it with .gun extension. This files gets opened when i open it using 'open(openfiledialog) button' which is present in my application. But when i double this file (which is placed on desktop) it opens only the application and not with the content.? How can i get it with the content when the file is double clicked?
Hi, When i use Main(string[] args) or Environment.GetCommandLineArgs; it shows me the following error. Error 1 'string' does not contain a definition for 'rtbFile' (Richtextbox instance)
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Hi, When i use Main(string[] args) or Environment.GetCommandLineArgs; it shows me the following error. Error 1 'string' does not contain a definition for 'rtbFile' (Richtextbox instance)
show us the code.
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show us the code.
The error is ok now. and i have done main(string[] args) or Environment.GetCommandLineargs..... But still when i double click the created file(file with .gun extension),, it still doesn't contain the content.
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The error is ok now. and i have done main(string[] args) or Environment.GetCommandLineargs..... But still when i double click the created file(file with .gun extension),, it still doesn't contain the content.
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Just for example :
static void Main(string[] args)
{
string fileToOpen = args[0]; // it is the same like openFileDialog.FileName// Now you MUST open the file "fileToOpen"
}I dont get what u try to say. This is my code. Can u suggest with this code....(ofdFile=Openfiledialog,rtbFile=richtextbox) void btnOpen_Click(object sender, EventArgs e) { ofdFile = new OpenFileDialog(); ofdFile.DefaultExt = @"*.gun"; ofdFile.Filter = @"Gun Files|*.gun"; if (ofdFile.ShowDialog() == System.Windows.Forms.DialogResult.OK && ofdFile.FileName.Length > 0) { //ofdFile.FileName. rtbFile.LoadFile(ofdFile.FileName, RichTextBoxStreamType.PlainText); } }
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I dont get what u try to say. This is my code. Can u suggest with this code....(ofdFile=Openfiledialog,rtbFile=richtextbox) void btnOpen_Click(object sender, EventArgs e) { ofdFile = new OpenFileDialog(); ofdFile.DefaultExt = @"*.gun"; ofdFile.Filter = @"Gun Files|*.gun"; if (ofdFile.ShowDialog() == System.Windows.Forms.DialogResult.OK && ofdFile.FileName.Length > 0) { //ofdFile.FileName. rtbFile.LoadFile(ofdFile.FileName, RichTextBoxStreamType.PlainText); } }
And my main function is in Program.cs file
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I dont get what u try to say. This is my code. Can u suggest with this code....(ofdFile=Openfiledialog,rtbFile=richtextbox) void btnOpen_Click(object sender, EventArgs e) { ofdFile = new OpenFileDialog(); ofdFile.DefaultExt = @"*.gun"; ofdFile.Filter = @"Gun Files|*.gun"; if (ofdFile.ShowDialog() == System.Windows.Forms.DialogResult.OK && ofdFile.FileName.Length > 0) { //ofdFile.FileName. rtbFile.LoadFile(ofdFile.FileName, RichTextBoxStreamType.PlainText); } }
pass the args[0] to your form class.
static void Main(string[] args)
{
Application.Run(new Form1(args[0]));
}For example your form class is Form1. Then change the constructor to
public Form1(string fileToBeOpened)
{
if(fileToBeOpened != string.Empty)
{
// rtbFile.LoadFile(fileToBeOpened , RichTextBoxStreamType.PlainText);
}
} -
pass the args[0] to your form class.
static void Main(string[] args)
{
Application.Run(new Form1(args[0]));
}For example your form class is Form1. Then change the constructor to
public Form1(string fileToBeOpened)
{
if(fileToBeOpened != string.Empty)
{
// rtbFile.LoadFile(fileToBeOpened , RichTextBoxStreamType.PlainText);
}
}It shows me 'IndexOutOfRangeException Unhandled' in program.cs Application.Run(new Form1(args[0]));
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It shows me 'IndexOutOfRangeException Unhandled' in program.cs Application.Run(new Form1(args[0]));
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I think you just COPY and PASTE the code. You must AWARE OF ERRORS. You MUST check if args.Length > 0 // you get the file directly args.Length == 0 // without file
Yes it works. Thanks a lot.
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Yes it works. Thanks a lot.
Will get u back if i get more doubts.