C# sets custom file icon to achieve double-click to start (modify the registry)

The custom file generated by the program, such as the suffix is.test

How to start and open the program directly and open this file?

1. Double-click to open

2. Customized file with icon display

3. Right-click to have corresponding properties for custom files

Background code: (How to modify information in the registry)

  //Tool startup path  string toolPath =  + "\\Mail Widget.exe";

  string extension = ;

  string fileType = "Email File";

  string fileContent = "text/plain";
  //Get information  Microsoft. registryKey = Microsoft.(extension);

  if (registryKey != null &amp;&amp; ("shell") != null &amp;&amp; ("shell").OpenSubKey("open") != null &amp;&amp;
    ("shell").OpenSubKey("open").OpenSubKey("command") != null)
  {
    var varSub = ("shell").OpenSubKey("open").OpenSubKey("command");
    var varValue = ("");

    if ((varValue, toolPath + " %1"))
    {
      return;
    }
  }
  //delete  Microsoft.(extension, false);
  //File Registration  registryKey = Microsoft.(extension);
  ("File Type", fileType);
  ("Content Type", fileContent);
  //Set the default icon  Microsoft. iconKey = ("DefaultIcon");
  ("",  + "\\");
  //Set the default open program path  registryKey = ("shell\\open\\command");
  ("", toolPath + " %1");
  //closure  ();

After modifying the registry information, double-clicking the file will start the software. How to operate it in the code?

//Double click to start opening //If there are spaces in the original path, it will be decomposed into multiple elementsif ( &gt; 0)
{
   string filePath = (" ", ());
   FileInfo file = new FileInfo(filePath);
   if ()
    {
       = ;
    }
 }

Then you can do it in the main programloadedIn the method, judgeDoubleClickSptdFilePath Whether there is a value, if so, get the file under the path and continue the operation.