ESPHOME 接收433信号设置为点动开关

xinba8

请问怎样修改这个代码为按一下遥控器   传感器状态 变为开启后 1秒后自动变回关闭状态？
现在的状态是按一下遥控器 开启，再按一下遥控器 才关闭。这样子不好做自动化。

请求各位帮个忙修改下。谢谢

remote_receiver:
  pin: GPIO2
  dump:
    - rc_switch
  tolerance: 50%
  filter: 250us
  idle: 4ms
  buffer_size: 2kb


binary_sensor:


# 1  433遥控器


  - platform: template
    name: "433button"
    id: a433button
  - platform: remote_receiver
    name: "433button"
    internal: true
    rc_switch_raw:
      code: "000111011110010110001010"
      protocol: 1
    filters:
      delayed_off: 100ms  
    on_press:
      if:
        condition:
          binary_sensor.is_on: a433button  
        then:
          - binary_sensor.template.publish:
              id: a433button
              state: OFF
        else:
          - binary_sensor.template.publish:
              id: a433button
              state: ON
              

复制代码

xinba8

hunterfox 发表于 2022-1-4 12:14
delay: 1s 然后关闭

请问怎样修改？谢谢

xinba8

mildrabbit 发表于 2022-1-4 17:20
esphome:
  name: dswitch_computer
  platform: ESP8266

你这代码是开关代码自动复位是没问题，但传感器不适用这代码。