不能重新運行我的應用程序後衛或重新運行

Question

所以我有一個應用程序，該樹是這樣的：

- Gemfile 
- Guardfile 
- source/ 
- dist/ 
- app.rb 

啓動服務器是ruby app.rb命令（或require_relative './app.rb'，它做同樣的事情）

我想運行此命令並在任何文件更改時重新運行它。

唯一的例外是dist /文件夾 - 其中的任何文件更改應該被忽略。

這裏是我的嘗試，到目前爲止與guard和guard-shell（道歉代碼轉儲）：

require 'childprocess' 

# Global constant tracking whether the app has been started 
RunningProcess = {gen_rb: false} 

# Method to stop the app if it's been started 
def ensure_exited_server 
    begin 
    RunningProcess[:gen_rb] && RunningProcess[:gen_rb].poll_for_exit(10) 
    rescue ChildProcess::TimeoutError 
    RunningProcess[:gen_rb].stop # tries increasingly harsher methods to kill the process. 
    end 
    nil 
end 

# Start the app using 'child-process' 
def start_app 
    # prevent 'port in use' errors 
    ensure_exited_server 
    # The child-process gem starts a process and exposes its stdout 
    RunningProcess[:gen_rb] = ChildProcess.build("ruby", "gen.rb") 
    RunningProcess[:gen_rb].io.inherit! 
    RunningProcess[:gen_rb].start 
    nil 
end 

# Always start the app, not just when a file changes. 
start_app 

# The guard-shell gem runs a block whenever some set of files has changed. 
guard :shell do 
    # This regex matches anything except the dist/ folder 
    watch /^[^dist\/].+/ do |m| 
    start_app 
    # Print a little message when a file changes. 
    m[0] + " has changed." 
    end 
    nil 
end 

# Make sure the app does not run after guard exits 
at_exit { ensure_exited_server } 

這不會永遠重新啓動我的應用程序。

與重播的問題是我對他們的瞭解回購提出的一個問題：看到https://github.com/alexch/rerun/issues/107

Answer 1

如何像這樣爲你Guardfile？

guard :shell do 
    watch(%r{^source/.+\.(rb)}) do |m| 
    `ruby app.rb` 
    end 

    watch('app.rb') do |m| 
    `ruby app.rb` 
    end 
end 

而不是列出該目錄忽略的，這watch州哪些目錄/文件來使用。