我该如何制作它，以便每个循环每次出现只打印一次？

Question

I want this to print Start and stop times so I can verify lights are going On and OFF when they should. Example: Light schedule Lights ON 7:00 AM Lights OFF 1:00 AM Lights ON 7:00 AM...

What I'm getting instead is printing lights ON from 7am-1am printing lights OFF from 1am-7am...

try:
    while True:
        now = datetime.datetime.now().time()
        GPIO.output(R1,GPIO.LOW)
        if now.hour == 7:
            GPIO.output(R1, GPIO.HIGH)
            print("Lights ON")
            print(time.strftime("%-I:%M %p"))
        elif now.hour == 1:
            GPIO.output(R1, GPIO.LOW)
            print("Lights OFF") 
            print(time.strftime("%-I:%M %p"))       
finally:
    GPIO.cleanup()

What am I missing?

Answer 1

EDIT: Updated my answer based on more clarity.

1. Assuming that GPIO.output() is like an on-off switch, you need to make sure that the first default setting is outside the while loop.

2. Since you want the loop to run indefinitely, you cant use break . If you do that however, the loop will keep updating the GPIO indefinitely, which you dont want either. So, I would recommending adding a flag toggle .

3. If toggle==1 , the if condition is allowed to update the GPIO values else if can't. By default it is set to 1, so the second it encounters now.hour==7 or 1 , it will update it. But once its done implementing the code, it sets it to 0 again. This prevents it from continuously updating the GPIO values.

4. Now, when the hour changes, you want the toggle to go back to 1, so that it can check the hour and update if needed. This is done by storing the value of the previous hour in lasthour and checking it as part of a elif condition.

5. Lastly, you use time.strftime . Instead, you can simply use now.strftime to get the time printed in the right format.

import datetime

testtime = datetime.datetime(2020,1,6,7)

try:
    #GPIO.output(R1, GPIO.LOW)
    toggle = 1
    lasthour = 0
    
    while True:  
        now = testtime
        
        if now.hour == 7 and toggle==1:
            #GPIO.output(R1, GPIO.HIGH)
            print("Lights ON")
            print(now.strftime("%-I:%M %p"))
            lasthour = now.hour
            toggle=0
            
        elif now.hour == 1 and toggle==1:
            #GPIO.output(R1, GPIO.LOW)
            print("Lights OFF") 
            print(now.strftime("%-I:%M %p"))
            lasthour = now.hour              #<- Save last hour of updatre
            toggle=0                         #<- Set toggle to 0
        
        elif lasthour != now.hour:           #<- If change in hour
            toggle=1                         #<- Set toggle to 1
            
        print('_________running_________; toggle ->', toggle) #Comment this
finally:
    print("out of while loop")
    #GPIO.cleanup()

Lights ON
7:00 AM
_________running_________; toggle -> 0
_________running_________; toggle -> 0
_________running_________; toggle -> 0
_________running_________; toggle -> 0
_________running_________; toggle -> 0
_________running_________; toggle -> 0
_________running_________; toggle -> 0

Answer 2

    try:
    while True:
        now = datetime.datetime.now().time()
        GPIO.output(R1,GPIO.LOW)
        if now.hour == 7 and now.minute == 0:
            GPIO.output(R1, GPIO.HIGH)
            print("Lights ON")
            print(time.strftime("%-I:%M %p"))
        elif now.hour == 1 and now.minute == 0:
            GPIO.output(R1, GPIO.LOW)
            print("Lights OFF") 
            print(time.strftime("%-I:%M %p"))       
finally:
    GPIO.cleanup()

You have to check the minutes too, otherwise it will print Lights ON from 7:00 till 7:59 and the same with lights off.

Also, be aware of the while loop. The way you coded it, it will run forever unless you stop it manually.