我使用“眼睛”作为主管,并且在模板更改上必须运行以下内容:
eye load service.rb
eye restart service.rb
我想将其定义为所有应用程序的单个处理程序,并将其命名为
eye reload appname
并在处理程序中这样操作:
- name: reload eye service
command: eye load /path/{{ service }}.rb && eye restart {{ service }}
但是我找不到将变量传递给处理程序的方法。是否可以?
最佳答案
处理程序/main.yml:
- name: restart my service
shell: eye load /path/{{ service }}.rb && eye restart {{ service }}
因此您可以通过默认设置变量
defaults/main.yml:
service : "service"
或者您可以通过命令行定义{{service}}:
ansible-playbook -i xxx path/to/playbook -e "service=service"
http://docs.ansible.com/ansible/playbooks_variables.html
PS:http://docs.ansible.com/ansible/playbooks_intro.html#playbook-language-
example
---
- hosts: webservers
vars:
http_port: 80
max_clients: 200
remote_user: root
tasks:
- name: ensure apache is at the latest version
yum: name=httpd state=latest
- name: write the apache config file
template: src=/srv/httpd.j2 dest=/etc/httpd.conf
notify:
- restart apache
- name: ensure apache is running (and enable it at boot)
service: name=httpd state=started enabled=yes
handlers:
- name: restart apache
service: name=httpd state=restarted
http://docs.ansible.com/ansible/playbooks_intro.html#handlers-running-operations-on-change
但是,如果要在1.2及更高版本中立即刷新所有处理程序命令,则可以:
tasks:
- shell: some tasks go here
- meta: flush_handlers
- shell: some other tasks
关于ansible:将变量传递给处理程序,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/26475761/