我試圖使用由application_python食譜所提供的Django的資源:https://github.com/opscode-cookbooks/application_python使用python_application廚師的菜譜
如果按照 https://github.com/opscode-cookbooks/application_python/blob/master/examples/recipes-packaginator.rb 提供和已經宣佈MT元depends application_python
的例子,我有以下錯誤:
No resource found for django. Tried application_django, application_python_django, django
於是,我就照做這裏Including a Chef LWRP from another cookbook使用「application_python_django」資源,而不僅僅是「的Django」。和我:
No resource found for application_python_django. Tried application_application_python_django, application_python_application_python_django, application_python_django
有趣的是,如果我刪除應用程序資源,只留下application_python_django資源的抱怨是:
You must supply a name when declaring a application_python_django resource
!所以它似乎找到了資源。因此,如果聲明資源名稱,配方會被執行,但是沒有任何操作會執行django資源(事實上,應用程序資源中缺少很多參數)。
啊,如果我刪除應用程序,只留下資源作爲Django的,一個也得到:
Cannot find a resource for django on ubuntu version 12.04
嗯,我相當丟失。
最初嘗試爲我的祕方是:
application 'radar_parlamentar' do
path '$HOME/radar_parlamentar'
owner 'radar'
repository 'https://github.com/leonardofl/radar_parlamentar.git'
revision 'master'
django do
debug true
collectstatic 'build_static --noinput'
database do
database 'radarparlamentar'
adapter 'mysql'
username 'radarparlamentar'
password 'secret'
end
end
gunicorn do
only_if { node['roles'].include? 'packaginator_application_server' }
app_module :django
port 8080
end
end
TKS, 萊昂納多
請不要忘記,以紀念一個答案是正確的! :) – sethvargo
hummm ...我沒有找到如何做到這一點:( 是真實的,答案幫我,但我在這條巨蟒的應用菜譜的新問題stucked ...現在我打算寫一個較低級別的配方...... –