How can I find the correct program to open a given file type in node.js?

Question

How can I find the correct program to open a given file type in node.js?

I'm trying to open files from node, If I try and open them using this code then it works

exec("excelfile.xls", {cwd: "c:\\"}, function(error, stdout, stderror){
    if(error)console.log("Error: \""+stderror+"\"");
});

However, I'm trying to adapt the program to run as a service, and the only way I can get my program to open a file when it running as a service is to use spawn instead of exec:

var child = spawn("explorer.exe",[], {env: process.env, cwd: "C:\\Windows"});
child.unref();

The problem is if I try and spawn something that isn't executable then I get am ENOENT error as node doesn't know what program to use to open the file:

var child = spawn("excelfile.xls",[], {env: process.env, cwd: "c:\\"});
child.unref();

What I'm looking for is a way to extract the correct program to open a given file type from windows (and the path to that program).

I've tried using winreg to get the information from the registry, but to do that I need to lookup computer\HKEY_CLASSES_ROOT."fileExtension"\ and get the value from the "(Default)" string value, then I could look up computer\HKEY_CLASSES_ROOT\"(Default) value"\shell\Open\command\ and get the string I need to open the file.

Unfortunately, winreg doesn't appear to give you the value of the (Default) string.

var Winreg = require('winreg');

var regKey = new Winreg({
      hive: Winreg.HKCR,
      key : "\\.xls"
})

regKey.values(function (err, items) {
      if (err)
        console.log('ERROR: '+err);
      else
          for (var i in items){
              console.log('ITEM: '+items[i].name+'\t'+items[i].type+'\t'+items[i].value);
          }
});

OUTPUT

ITEM: Content Type REG_SZ application/vnd.ms-excel

EXPECTED OUTPUT

ITEM: (Default) Type REG_SZ Excel.Sheet.8
ITEM: Content Type REG_SZ application/vnd.ms-excel

So if someone could give me a way to open a non-executable file using child_process.spawn() or give me a way to extract the path to the correct program to open a file based on it's extension I'd be every so grateful.

javascript
windows
node.js

Answer 1

You could try to use START command.

var child = spawn("cmd.exe", ["/c", "start", "excelfile.xls"], {env: process.env, cwd: "c:\\"});
child.unref();

Answer 2

My complete code:

var EventLogger = require('node-windows').EventLogger;
var log = new EventLogger('File Launcher starting');
var exec = require('child_process').exec;
var userName = "mike";
var password= "mikesPassword";
var fullPath= "C:\folder\file.xls";

exec("PSExec.exe -accepteula -h -d -u "+userName+" -p "+password+" -i 1   C:\\WINDOWS\\SYSTEM32\\CMD.EXE /c start \"\" \""+fullPath+"\"",{cwd: process.cwd},      function(error, stdout, stderror) {
    if(error){
        log.error(stderror.replace('\n','').replace('\r',''));
    }
    if(stdout){
        log.info(stdout.replace('\n','').replace('\r',''));
    }
});