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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