In his book Erik Brown writes the following code
and compiles it from the command-line:
csc MyForm.cs
[assembly: System.Reflection.AssemblyVersion("1.1")]
namespace MyNamespace
{
public class MyForm : System.Windows.Forms.Form
{
public MyForm()
{
this.Text = "Hello Form";
}
public static void Main()
{
System.Windows.Forms.Application.Run(new MyForm());
}
}
}
I want to add another form and call it from the first.
Do I need a project file? An assembly file? I don't understand the build process. Can you explain the very basics to me: how do I tell the compiler to build a two-forms application?
First form (form1.cs):
public class MyForm : System.Windows.Forms.Form
{
public MyForm()
{
this.Text = "Hello Form";
this.Click += Form_Click;
}
public static void Main()
{
System.Windows.Forms.Application.Run(new MyForm());
}
private void Form_Click(object sender, System.EventArgs e)
{
MyForm2 form2 = new MyForm2();
form2.ShowDialog();
}
}
Second form (form2.cs):
public class MyForm2 : System.Windows.Forms.Form
{
public MyForm2()
{
this.Text = "Hello Form 2";
}
}
Now from the command line, locate to the location where you saved these .cs files and then run:
csc form1.cs form2.cs
It will create an EXE file. Run it and click in the form to open form2.
You need to use Visual Studio (VS) command prompt and mention the C# code file names (*.cs) of all the Windows forms in your project. You must include each form's code-behind file as well as designer code-behind file. Otherise, the compilation will not succeed.
A sample project structure is as below:
WindowsFormsApplication1.csproj
--Program.cs
--Form1.cs
----Form1.Designer.cs
--Form2.cs
----Form2.Designer.cs
The compilation command for above project will look like:
csc /target:winexe Program.cs Form1.cs Form1.Designer.cs Form2.cs Form2.Designer.cs
Note: It is compulsory to include Program.cs file during compilation, else the compiler fails to obtain the project's entry point(the Main method). The /target switch helps you to launch the output EXE file as a GUI based Windows Forms application.
There is a shorter version of the above command which uses wildcard to include all the files names in one go:
csc /target:winexe *.cs
The easiest alternative is to use the msbuild command in place of csc on Visual Studio command prompt. msbuild command can be used to build the project file which contains the reference of all the *.cs files. Here is how the command looks like:
msbuild WindowsFormsApplication1.csproj
This relieves us from mentioning all the C# code file names individually (whether in project root directory or nested sub-direcotries). Build the project file and you are done.
Related
Suppose I have two files in my current working directory:
// file1.cs
Console.WriteLine("file1");
//file 2.cs
Console.WriteLine("file2");
In powershell, I do a dotnet new and delete the automatically generated Program.cs file. Then I do a dotnet build and get an error:
Only one compilation unit can have top level statements
I understand why this occurs, but I would like to be able to have full control of which .cs file is being targetted, while the other ones get ignored.
Is there any way to achieve this without having to create a whole new project for every file?
Doing this with .NET doesn't seem to be possible as of now. An issue on the dotnet/sdk GitHub has requested for this feature to be implemented.
However, you can use the C Sharp Compiler to compile a Windows executable and specify a .cs file with csc file1.cs
file1.cs:
using System;
Console.WriteLine("File 1");
https://learn.microsoft.com/en-us/dotnet/csharp/fundamentals/program-structure/top-level-statements
These files both use top-level statements. It implies that they both contain the Main method where program execution starts. You can only have one entry point. Generally, C# code is going to be contained within classes. Define a class in one (or both) files and put your methods within.
// Program.cs
public class Program
{
static void Main()
{
Console.WriteLine("Program.cs");
}
}
// Util.cs
public class Util
{
public static void Display()
{
Console.WriteLine("Util.cs");
}
}
I have created a boolean algebraic simplifier. It simplifies expressions and I am content with it. However, I am trying to add a feature that allows users to check if two expressions are equivalent. For this I have created a new form that allows the user to input two expression by clicking buttons. To do this, I thought it best to simplify both expressions and then compare the two for equivalency. As I have got lots of subroutines and code that works for simplification in another form, I thought making the form a child form of the form with the code in would allow me to call the subroutines instead of copying them onto the form. I have made these protected in the parent form. I have inherited like so:
public partial class Expression_Equivalency_Form : Expression_Simplifier
However, when I click onto the form designer, this error appears and I cannot view the graphical interface of the form:
"Could not find file File Path"
The file is in the debug folder which is within the bin folder within the folder containing the program and is recongised in the parent class. The file is read from and appeneded by the parent form without issue. I have tried to research this but have been unable to find a solution. Does anyone know one?
I have read to the file and appended to it. I have also used the following code to remove any blank lines from my text file:
File.WriteAllLines("PreviousExpressionInputs.txt",
File.ReadAllLines("PreviousExpressionInputs.txt").Where(l => !string.IsNullOrWhiteSpace(l)));
Code that writes to the file:
using (BinaryWriter Writer = new BinaryWriter(File.Open("PreviousExpressionInputs.txt",
FileMode.Append)))
{
Writer.Write(expressionandanswertowritetotextfile);
}
Code that reads from the file:
foreach (string line in File.ReadLines("PreviousExpressionInputs.txt"))
{
try
{
LinesInFile.Add(line);
}
catch (Exception)
{
}
}
Consider following facts:
When you open a form in design mode, the constructor of its base class will run.
When you look for a relative file name, the path will be resolved relative to the current working directory of the application.
When the form is in design mode, the current application is Visual Studio and its working directory is where the devenv.exe is located.
It describes why you cannot find your text files. Because you have some code in the constructor of your base form(or fir example load event handler of the base form) which looks for the file and since the filename is relative, its looking for the file in the Visual Studio working directory and could not find file.
How to prevent the problem? Check DesignMode property to prevent running the code:
public partial class MyBaseForm : Form
{
public MyBaseForm()
{
InitializeComponent();
}
private void MyBaseForm_Load(object sender, EventArgs e)
{
MessageBox.Show("This will show both in run-time and design time.");
if (!DesignMode)
MessageBox.Show("This will show just in run-time");
}
}
Create the derived form and open it in designer to see what happens:
public partial class Form1 : MyBaseForm
{
public Form1()
{
InitializeComponent();
}
}
To learn more about how designer works take a look at this post.
I want to have a Windows Form Application use a menustrip with three options to launch a console application. The console application is a .exe file built in C# in Visual Studio with some basic code for as school project. The console application does not need to return any values, it only needs to run and allow the user to use it. This is what the form will look like: Menu Application
I have tried importing the System.Diagnostics.Process.Start namespace with Process.Start#("Path of file") in my menu item click event method to launch my C# console application but have not been successful. I am getting a "Win32Exception was unhandled: An unhandled exception of type 'System.ComponentModel.Win32Exception' occurred in System.dll. Additional information: The system cannot find the file specified"
Here is the code in the menu item click event:
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void Form1_Load(object sender, EventArgs e)
{
}
private void lesson13LabCToolStripMenuItem_Click(object sender, EventArgs e)
{
Process.Start(#"\C:\Users\Sam\Documents\Visual Studio 2015\Projects\LabMenu\LabMenu\Lesson13LabC.exe");
}
}
Any ideas on what I am doing wrong?
The error is clear The system cannot find the file specified. Check the path of file.
Also remove the starting \ in the path
Remove the backslash at the beginning of your path (before the drive letter).
I have a WPF project which I try to make it a single instance app using the recipe with Microsoft.VisualBasic dll described by Dale Ragan here at StackOverflow
Doing so in Visual Studio 2013 with Framework 4.5 give me 2x the same error while compiling: "... has more than one entry point defined..." for each entry point. Then I though that I would see both entry points in the comboBox choices of my "Startup Object" item of "Application" tab of my project properties. But it is empty. Why the "StartUp object" comboBox is empty and how to set the entry point? Could it be a Microsoft bug?
Additional information:
- The 2 files with entry points are "App.g.cs" (auto generated) and my newly defined class with entry point - main : "EntryPoint.cs"
I've solved the problem by hacking the csproj file:
<PropertyGroup>
<StartupObject>Test.Program</StartupObject>
</PropertyGroup>
Sorry folks,
The problem disappeared.
I restarted Visual Studio but I had same behavior.
I made a new project to send to Microsoft as a bug but it was working fine.
I then copied my startup class from my test project and the bug disappeared ?????????
I don't understand.
You can follow these steps (For an EntryPoint):
Right Click in your solution, Properties, Common Properties, Startup
Project, and select your Startup Project there.
Open your app.xaml and set the StartUpUri to your Main XAML file.
Unload your WPF project, and after that Edit it!
In App.xaml.cs file you can put these lines of code:
using System.Diagnostics;
...
Process[] activeProcess = Process.GetProcessByName(Process.GetCurrentProcess().ProcessName);
if (activeProcess.Length == 1)
{
Application.Run(new YOUR_MAIN_XAML_CLASS_HERE());
}
else
{
MessageBox.Show("You already have an instance of this program");
}
Hope it helps
It's hard to predict without reviewing the code. However, ensure following points are covered.
Create a class that derives from Microsoft.VisualBasic.ApplicationServices.
WindowsFormsApplicationBase, and use it to wrap your WPF System.Windows.Application. The
wrapper is initialized by supplying your own implementation of Main.
namespace SingleInstanceNamespace
{
using System;
using System.Windows;
using Microsoft.VisualBasic.ApplicationServices;
public class SingleInstanceManager : WindowsFormsApplicationBase
{
public SingleInstanceManager()
{
this.IsSingleInstance = true;
}
protected override bool OnStartup(
Microsoft.VisualBasic.ApplicationServices.StartupEventArgs eventArgs)
{
base.OnStartup(eventArgs);
App app = new App(); //Your application instance
app.Run();
return false;
}
protected override void OnStartupNextInstance(
StartupNextInstanceEventArgs eventArgs)
{
base.OnStartupNextInstance(eventArgs);
string args = Environment.NewLine;
foreach (string arg in eventArgs.CommandLine)
{
args += Environment.NewLine + arg;
}
string msg = string.Format("New instance started with {0} args.{1}",
eventArgs.CommandLine.Count,
args);
MessageBox.Show(msg);
}
}
}
The next code block details the content of the App.cs file where the application’s main
entry point is defined:
namespace SingleInstanceNamespace
{
public class MyApp
{
[STAThread]
public static void Main(string[] args)
{
//Create our new single-instance manager
SingleInstanceManager manager = new SingleInstanceManager();
manager.Run(args);
}
}
}
In my case, the forms were missing too from the Startup object dropdown list. So, I found another easey way to modify the startup form.
Open the Program.cs file on Code View
Go to the static void Main()
Modify the line that looks like the following:
Application.Run(new Form1());
To your desired Form, like this:
Application.Run(new BootloaderDialog());
Hope this helps!
In my case, I spelled entry name Main() to main(), then this entry not added into startup object list.When I changed it to Main(), the entry was added into startup object list.
So be careful case sensitive.
I wanted "MyMainWindow" to be the starting element for "MyProject" project.
In App.xaml had to set:
<Application x:Class="MyProject.App"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:local="clr-namespace:MyProject"
StartupUri="MyMainWindow.xaml">
<Application.Resources>
</Application.Resources>
I am using mcs version 2.10.8.1 in Ubuntu 12.04, I have the following code:
using System;
using System.Drawing;
using System.Windows.Forms;
namespace WindowsFormsApplication1
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
OpenFileDialog dlg = new OpenFileDialog();
dlg.ShowDialog();
if (dlg.ShowDialog() == DialogResult.OK)
{
string fileName;
fileName = dlg.FileName;
MessageBox.Show(fileName);
}
}
}
}
I compile using the command
$ mcs source_code.cs -r:System.Windows.Forms.dll -r:System.Drawing.dll
And I receive the error
source_code.cs(11,13): error CS0103: The name `InitializeComponent' does not exist in the current context
Compilation failed: 1 error(s), 0 warnings
I've seen many answers to this question in cases using Visual Basic; I would like to know what I should do to solve this problem. Thanks.
Was your C# code originally created in Visual Studio? If so, then you'll probably have a Form1.Designer.cs file as well as the file containing the code that you've written by hand. You need to include the file in the command line.
If this isn't C# code originally created in Visual Studo, you may not even have an InitializeComponent method... but in that case you'll need more code to do anything useful in your form (like creating the button and hooking up its Click event).