Binary file in npm package

I try to create an npm package, which can be started as a command from shell. I have package.json

{
  "name": "myapp",
  "version": "0.0.6",
  "dependencies": {
    "async": "",
    "watch": "",
    "node-promise": "",
    "rmdir": "",
    "should": "",
    "websocket": ""
  },
  "bin": "myapp"
}

and myapp

#!/bin/bash

path=`dirname "$0"`
file="/myapp.js"

node $path$file $1 &

But I get an error:

module.js:340
    throw err;
          ^
Error: Cannot find module '/usr/local/bin/myapp.js'
    at Function.Module._resolveFilename (module.js:338:15)
    at Function.Module._load (module.js:280:25)
    at Function.Module.runMain (module.js:497:10)
    at startup (node.js:119:16)
    at node.js:902:3

The problem is that myapp.js is in another directory. How can I get this directory name from my script? Or maybe there is better way to do this?

---

Actually, you can put your myapp.js file into bin .
So, the bin key in package.json file should be like this :

"bin": { "myapp" : "<relative_path_to_myapp.js>/lib/myapp.js" }

At the first line in myapp.js , you must add this shebang line :

#!/usr/bin/env node

It tells the system to use node to run myapp.js .

---

... Or if you don't want to call myapp.js directly, you can create a script like this to be your executable file :

#!/usr/bin/env node

var myapp = require('<relative_path_to_myapp.js>/myapp.js');
myapp.doSth();

and in package.json :

"bin" : { "myapp" : "<relative_path_to_the_script>/script.js" }

By doing this either way, you can avoid finding the path to your nodemodule.

---

But... if you insist to use your old myapp bash script, then you can find the path to the module with this :

myapp_path=$( npm explore -g myapp -- "pwd" )

Hope these help :D

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