当您按住按钮然后在释放该按钮时启动时,是否可以暂停python脚本? (我的按钮连接到我的Raspberry Pi上的GPIO引脚)
答案 0 :(得分:1)
我假设您使用的按钮位于GPIO18中,因此您可以使用此代码。
<DataGrid ItemsSource="{Binding Path= Shares}" HorizontalAlignment="Left" Margin="89,201,0,0" CanUserAddRows="False" AutoGenerateColumns="False" VerticalAlignment="Top" Height="280" Width="500">
<DataGrid.Columns>
<DataGridTextColumn Header="Company" Binding="{Binding CompanyName}" Width="250" />
<DataGridTextColumn Header="Share Price" Binding="{Binding Price}" />
</DataGrid.Columns>
</DataGrid>
或者您也可以尝试:
import RPi.GPIO as GPIO
import time
GPIO.setmode(GPIO.BCM)
GPIO.setup(18, GPIO.IN, pull_up_down=GPIO.PUD_UP)
while True:
input_state = GPIO.input(18)
while not input_state:
# as soon as your button is pressed you
# will be inside this loop
print('Button is being pressed')
我认为第二个更准确地针对您的要求。
答案 1 :(得分:0)
你看过gpiozero了吗?它使与GPIO的交互更加简单。
from gpiozero import Button
button = Button(2)
button.wait_for_press()
button.wait_for_release()
print("Button was pressed and released")
以下是Button类的链接:https://gpiozero.readthedocs.io/en/v1.3.1/api_input.html#gpiozero.Button.wait_for_release
以及如何使用它的示例: https://gpiozero.readthedocs.io/en/v1.3.1/recipes.html#button