9

I would like to make a Cakefile task to watch some CoffeeScript files just like if I had run coffee -c -w js/*.coffee.

Its watching and recompiling them successfully, but it doesn't log the usual output to the terminal when there's a compile error like it would if I just ran the script from the terminal. Any idea how to make this happen?

exec = require('child_process').exec

task 'watch','watch all files and compile them as needed', (options) ->
    exec 'coffee -c -w js/*.coffee', (err,stdout, stderr) ->
        console.log stdout

Also, if there's a better way to invoke a coffeescript command from a cakefile than running 'exec' please post that too.

4

3 に答える 3

6

spawn instead of exec?

{spawn} = require 'child_process'

task 'watch', -> spawn 'coffee', ['-cw', 'js'], customFds: [0..2]
于 2011-01-28T02:02:34.207 に答える
4

これを解決するためにspawnを使用しました。これは、ケーキファイルの例です。

{spawn, exec} = require 'child_process'

option '-p', '--prefix [DIR]', 'set the installation prefix for `cake install`'

task 'build', 'continually build with --watch', ->
    coffee = spawn 'coffee', ['-cw', '-o', 'lib', 'src']
    coffee.stdout.on 'data', (data) -> console.log data.toString().trim()

doccoプロジェクトで実際に動作していることを確認できます： https ：//github.com/jashkenas/docco/blob/master/Cakefile

于 2011-09-11T03:17:33.050 に答える
2

The problem with your original code was that exec only calls its callback once—after the child process has terminated. (The Node docs aren't so clear on this.) So instead of defining that callback, you should instead try

child = exec 'coffee -c -w js/*.coffee'
child.stdout.on 'data', (data) -> sys.print data

Let me know if that works for you.

于 2011-02-03T19:31:17.463 に答える