帮忙简化个自动化代码，简单的判断开关！

lylus

我有一个开关面板和一个夜灯。他们之间是没有物理连接的，只是都通过HA桥接，我写了一个自动化的脚本，当开关面板状态为开时 夜灯就打开，当开关面板状态为关时，就关闭夜灯，同时如果不通过开关面板开关夜灯时，如果在手机关夜灯，开关面板也会跟据夜灯的状态开或关。为此，我写了四段代码，1和2，判断开关状态，状态为开 则灯开，关则灯关，另外两条是判断夜灯状态，状态为开 则灯开，关则灯关。为此我写了以下代码，但是我感觉太长了，肯定有办法可以简单化。求指点。

#夜灯联动

#当面板开关状态为 on 时 设置夜灯状态为 on
- alias: 'onoff'
  trigger:
    platform: state
    entity_id: switch.wall_switch_right_158d00012f7c
    to: 'on'
  condition:

  action:
    - service: light.turn_on
      entity_id: light.gateway_light_34ce008bf8

      
#当夜灯状态为 on 时 设置开关状态为 on
- alias: 'lonoff'
  trigger:
    platform: state
    entity_id: light.gateway_light_34ce008bf8
    to: 'on'
  condition:

  action:
    - service: switch.turn_on
      entity_id: switch.wall_switch_right_158d00012f7c
      
 
 #当面板开关状态为 off 时 设置夜灯状态为 off
- alias: 'offon'
  trigger:
    platform: state
    entity_id: switch.wall_switch_right_158d00012f7c
    to: 'off'
  condition:

  action:
    - service: light.turn_off
      entity_id: light.gateway_light_34ce008bf8


#当夜灯状态为 off 时 设置开关状态为 off      
- alias: 'loffon'
  trigger:
    platform: state
    entity_id: light.gateway_light_34ce008bf8
    to: 'off'
  condition:

  action:
    - service: switch.turn_off
      entity_id: switch.wall_switch_right_158d00012f7c  
#夜灯联动

复制代码

ABC

- alias: 'onoff'
  trigger:
    platform: state
    entity_id: switch.wall_switch_right_158d00012f7c
        
  action:
    service_template: >
       {% if is_state("switch.wall_switch_right_158d00012f7c", "on") %}
          light.turn_on
       {% elif is_state("switch.wall_switch_right_158d00012f7c", "off") %}
          light.turn_off
       {% endif %}
    entity_id: light.gateway_light_34ce008bf8

复制代码

ABC

自己调整格式

lylus

万分感激回复，我现在去试试。