c#中如何将参数传递给另一个进程
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How to pass parameters to another process in c#
提问by Ldg
I just created an application that launches processes with the following code
我刚刚创建了一个使用以下代码启动进程的应用程序
string [] args = {"a", "b"};
???????????? Process.Start ("C: \ \ demo.exe" String.Join ("", args));
I would like to be able to pass the parameters from this application to the process I've launched.
我希望能够将参数从此应用程序传递到我启动的进程。
where I have to enter the parameters in the project of the process that I've launched? I tried to put them in
在我启动的流程项目中，我必须在哪里输入参数？我试着把它们放进去
static void Main (string [] args) {...
but they are not available in other forms. thanks for the help
但它们不能以其他形式提供。谢谢您的帮助
采纳答案by Robert
Process p= new Process();
p.StartInfo.FileName = "demo.exe";
p.StartInfo.Arguments = "a b";
p.Start();
or
或者
Process.Start("demo.exe", "a b");
in demo.exe
在演示.exe
static void Main (string [] args)
{
Console.WriteLine(args[0]);
Console.WriteLine(args[1]);
}
You asked how to save these params. You can create new class with static properties and save these params there.
你问如何保存这些参数。您可以创建具有静态属性的新类并将这些参数保存在那里。
class ParamHolder
{
public static string[] Params { get; set;}
}
and in main
并在主要
static void Main (string [] args)
{
ParamHolder.Params = args;
}
to get params in any place of your program use:
要在程序的任何位置获取参数，请使用：
Console.WriteLine(ConsoleParamHolder.Params[0]);
Console.WriteLine(ConsoleParamHolder.Params[1]);
etc.
等等。
回答by fhnaseer
Main Application code:
主要应用代码：
var process = new Process();
process.StartInfo.FileName = "demo.exe" // Path to your demo application.
process.StartInfo.Arguments = "a b c" // Your arguments
process.Start();
In your demo application (I am assuming that this is console application)
在您的演示应用程序中（我假设这是控制台应用程序）
static void Main (string [] args)
{
var agrumentOne = args[0];
var agrumentTwo = args[1];
var agrumentThree = args[3];
}
Here argumentOne, argumentTwo, argumentThree will contain "a", "b" and "c".
这里argumentOne、argumentTwo、argumentThree 将包含“a”、“b”和“c”。
