Load from text file at program startup - c#

Sorry for the likely noobish question, just starting to learn c#, and couldn't find anything that worked.
I'm making a text editor in c#, and so far it can open and save text files from inside the program with dialogs, but how can I make it load the text from a file that I open in windows explorer, outside of the editor, with the editor
Basically, I can already read from text files opened inside the editor, but how can i make it so that if I open a text file (and have the default program for opening text files set to my editor), it'll read it?
I saw something about getting the filename somehow and passing it as an argument, if that helps.

If I understood you correctly, you want to pass the filename/names as command line arguments ?
If you look at the Main, which starts the program you can see that it will store parameters in a string[] (string array) so if you pass arguments you can just check the args[] inside the program to get the parameters you sent in. Please ask more if you need more help !
UPDATED
As per your request if you open a file from windows explorer it will send the path of the file it to the Main method. So lets say you right click a file and choose to open it with your text editor. You have to use the path as I do below, and read the file's content. Then you can do whatever you want with the content.
class TestClass {
static void Main(string[] args) {
// Now you have all arguments in the string array
if (args.Length != 0) {
string pathToTextfile = args[0];
}
StreamReader textFile = new StreamReader(pathToTextfile);
string fileContents = textFile.ReadToEnd();
textFile.Close();
}
}

So you have a text editor coded in C#, and you want to be able to open a text file through double clicking on the file in Windows explorer. If so, basically 2 steps:
1. Your editor program must accept one argument as the file name. Carl had already given an example.
2. You need to associate *.txt files with your text editor. This could be done through editing Windows registry. please check What registry keys are responsible for file extension association

You can use the OpenFileDialog class to select a file to show in your program.

Related

Using Process.Start to open explorer

Edit: solution found, will note under image at the end of the question
After a bunch of research here on SO, I found that the way to open explorer with a selected file was:
Process.Start("explorer.exe", "/select, " + path);
However when I do this with controlled input, Explorer opens just its main window, however when I harcode the function call to the same value that's in the path variable (In my control test its a text file in C:\Temp) it works. So if I do the above when path is "C:\Temp\test.txt" It does't open explorer in the temp folder, however when I do:
Process.Start("explorer.exe", "/select, C:\\Temp\\test.txt");
It works perfectly, opening explorer and highlighting the file. What is happening here? Is there something wrong with the internal formatting on my string variable or something?
(Additionally, I ran into the same issue using the path variable to open a FileInfo. Hardcoded to the same value would work, but using the variable gave me a "the given path's format is not supported" exception")
Image showing that path and the harcoded value are the same:
The 2 explorer windows (cropped for Security) are the results of the 2 respective calls. The one with the variable shows te basic explorer main page. The one that's hardcoded shows the file selected as expected.
Edit: There was a hidden Left-To-Right Format character hidden in the front of the string.
public static class Program
{
static void Main()
{
Explore("C:\\Users\\art_g\\Desktop\\Sample.txt");
}
static void Explore(string path) =>
Process.Start("explorer.exe", "/select, " + path);
}
Works like a charm. Check your path string.

C# App which uses a file in the source dir, makes error about finding the file if it's started using other apps

I'm making a C# application which uses a text file located in the same directory as the application. when I start the app by double click, it runs without any problem. I want to start it using 4th mouse button but when I try, app makes error about finding the text file. My app is a simple launcher and i want it to run as easy as clicking a mouse button.
I defined text file path as below and I used public static to let other forms use the path and find the file too.
public static string list = System.IO.Directory.GetCurrentDirectory() + #"\list.txt";
Also tested:
public static string list = Environment.CurrentDirectory+ #"\list.txt";
The app and the text file are located in>> bin\Debug\
Error:
enter image description here
I use "Volume 2" Application (by Alexsandr Irza) to define functionality of 4th and 5th buttons of my mouse.I think it's so weird because running my app using double click makes no error and it can read the text file and write to it.
Please help me to fix it.
Don't use GetCurrentDirectory; that can change as your program runs.
If you want to find the directory of your executable, use
var path = System.IO.Path.GetDirectoryName(
System.Reflection.Assembly.GetExecutingAssembly().Location
);
To get the path of a file in the same directory as your executable:
var filePath = System.IO.Path.Combine(path, "list.txt");
How about something like:
string path = Application.ExecutablePath.Substring(0, Application.ExecutablePath.LastIndexOf('\'));?

Open a file without dialog

with openFileDialog you would select a file and after pressing "OPEN" it would paste the filepath of the selected file (c:\blob\template) into an textbox.
I would like to do automatically select the file c:\blob\template en then put the filepath in the textbox. basically the exact same thing as openfiledialog without a dialog. i have been trying to do this for some time now. can somebody help me out with this? i have no clue how to realise this.
i`m only able to get the filepath and paste the string in the textbox but this but only fills the box with a string. i need to load the file/template in there.
private void txt()
{
string fileName = "template";
string fullPath;
fullPath = Path.GetFullPath(fileName);
lblFirstTemplate.Text = fullPath;
}
Thank you in advance!
The code you now have will only get the file path. What you need to add is code that will actually open the file and read its content.
Let's say your file contains some text. You can use the following line to read the complete file as text:
System.IO.File.ReadAllText(fullPath);
If your file contains some other data like binary data, you can use:
System.IO.File.ReadAllBytes(fullPath);
And instead of reading all data at once, you can read it one line or a couple of bytes at a time. A good place in the documentation to start is: Common I/O Tasks

C# Saving RichTextBox text with formatting to a specific location without Dialog box

I had recently posted about this but didn't find the appropriate solution, probably due to not giving the whole context of my program.
I have two Form(s) and in one of the forms, I have a RichTextBox, in the same form, I have a method (public) which is used to save the rich text in that RichTextBox` automatically to a specific location which it takes as a parameter.
From the other form, I get the location and pass it to the method and the file is create (in the format of .rtf) in that location but the problem is that the file is empty i.e. there is no text at all in that file.
Can you please help me sort out this issue?
Code:
public void SaveIt(string parser)
{
MessageBox.Show(parser);
System.IO.StreamWriter file = new System.IO.StreamWriter(parser);
file.WriteLine(this.NotePad.Rtf);
file.Close();
}
I pass parser this : E:\Profiles\Muhammad Waqas\Data\BE\note1.Rtf
If you want to use StreamWriter use it like this:-
using (StreamWriter SW = new StreamWriter(parser))
{
Sw.WriteLine(this.NotePad.Rtf) ;
}

Unique File Renaming Issue

I am developing an app right now that reads in data from a windows from and generates an XML file based on the input.
I am tasked with creating a new file each time the form is updated (User presses "Submit"). (so far so good)
Here is the catch: The file has to be named after a prominent field input. (If the user types '993388CX' in the text box, the app would rename the pending file 993388CX.xml).
I understand how to actually rename a file in C#, but not how to rename it based on a form's input. Do any classes/methods exist that will dynamically rename the file based on the form input?
Code:
//Reads info1 from user input on the app UI and generates XML statement
XTemp = XDoc.CreateElement("New_Info");
XTemp.InnerText = info1.Text;
Xsource.AppendChild(XTemp);
XDoc.Save(#"C:\oldfile.xml");
I need the new file to be renamed after the string in info1.Text
If the user input was "John5", the file needs renamed to john5.xml
Thank you
Either directly save it with the correct name:
XDoc.Save(String.Format("C:\\{0}.xml",info1.Text));
OR
Rename it afterwards
File.Move("c:\\oldfile.xml", String.Format("C:\\{0}.xml",info1.Text));
XDoc.Save(#"C:\" + info1.Text + ".xml");
File.Move should do what you want.

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