open a file with default program in node-webkit

问题

I want to give the user any option he want to edit a file, how can I open a file with the default program of the specific file type? I need it to work with Windows and Linux but Mac option would be great too.

回答1:

as PSkocik said, first detect the platform and get the command line :

function getCommandLine() {
   switch (process.platform) { 
      case 'darwin' : return 'open';
      case 'win32' : return 'start';
      case 'win64' : return 'start';
      default : return 'xdg-open';
   }
}

second , execute the command line followed by the path

var sys = require('sys');
var exec = require('child_process').exec;

exec(getCommandLine() + ' ' + filePath);

回答2:

For file on a disk:

var nwGui = require('nw.gui');
nwGui.Shell.openItem("/path/to/my/file");

For remote files (eg web page):

var nwGui = require('nw.gui');
nwGui.Shell.openExternal("http://google.com/");

回答3:

Detect the platform and use:

'start' on Windows
'open' on Macs
'xdg-open' on Linux

回答4:

I am not sure if start used to work as is on earlier windows versions, however on windows 10 it doesn't work as indicated in the answer. It's first argument is the title of the window.

Furthermore the behavior between windows and linux is different. Windows "start" will exec and exit, under linux, xdg-open will wait.

This was the function that eventually worked for me on both platforms in a similar manner:

  function getCommandLine() {
     switch(process.platform) {
       case 'darwin' :
         return 'open';
       default:
         return 'xdg-open';
     }
  }

  function openFileWithDefaultApp(file) {
       /^win/.test(process.platform) ? 
           require("child_process").exec('start "" "' + file + '"') :
           require("child_process").spawn(getCommandLine(), [file],
                {detached: true, stdio: 'ignore'}).unref(); 
  }

来源：https://stackoverflow.com/questions/29902347/open-a-file-with-default-program-in-node-webkit

标签

javascript

Linux

windows

node.js

node-webkit