python时间范围验证器 [英] python time range validator
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问题描述
我在DB中有2个参数：开始和停止。它们的值可以是07：00-23：00或23：00-07：00
（从07开始，在23之后停止，或者在23之后，在07之后停止）
I have 2 parameters in DB: start and stop. value for them can be eg 07:00-23:00 or 23:00-07:00 (start after 07, stop after 23 or start after 23, stop after 07)
那时，状态必须为0或1，假设它是LED
In that time, a status must be 0 or 1, let's say it's LED
如何创建不会弄乱的统一逻辑控制器午夜之后/之前？
How to create unified logic controller that won't mess up after/before midnight?
我的执行不力（无法正常工作）在下面。实际上，我已经尝试了许多变体，但仍然以当前的状态结束。.
My poor implementation (wont work) is below. Actually, I've tried many-many variations and still ended up where I currently am..
if curtime >= vv_time_trig1 and curtime <= vv_time_trig2:
logger.info("turning socket on")
logger.debug("#1")
#check current status
#if current is 0
#turn socket on
if vvstatus == 0:
logger.debug("current status off, turning socket on")
GPIO.output(25, GPIO.HIGH)
#check current status
#if current is already 1
#do nothing
elif vvstatus == 1:
logger.info("skiping. already on")
#unhandeled current status
else:
logger.critical("unhandeled vvstatus!")
logger.critical("turning socket off")
GPIO.output(25, GPIO.LOW)
#if current time is before start
#turn off
elif curtime <= vv_time_trig1 and curtime >= vv_time_trig2:
logger.info("turning socket off")
logger.debug("#2")
#check current status
#if current is 1
#turn socket off
if vvstatus == 1:
logger.debug("current status on, turning socket off")
GPIO.output(25, GPIO.LOW)
#check current status
#if current is already 0
#do nothing
elif vvstatus == 0:
logger.info("skiping. already off")
#unhandeled current status
else:
logger.critical("unhandeled vvstatus!")
logger.critical("turning socket off")
GPIO.output(25, GPIO.LOW)
#if current time is after stop
#turn off
elif curtime >= vv_time_trig2:
logger.info("turning socket off")
logger.debug("#3")
#check current status
#if current is 1
#turn socket off
if vvstatus == 1:
logger.debug("current status: %s, turning socket off", vvstatus)
GPIO.output(25, GPIO.LOW)
#check current status
#if current is already 0
#do nothing
elif vvstatus == 0:
logger.info("skiping. already on")
#unhandeled current status
else:
logger.critical("unhandeled vvstatus!")
logger.critical("turning socket off")
GPIO.output(25, GPIO.LOW)
#if current time is before stop
#turn off
elif curtime <= vv_time_trig2 and curtime <= vv_time_trig1:
logger.info("turning socket on")
logger.debug("#4")
#check current status
#if current is 0
#turn socket on
if vvstatus == 0:
logger.debug("current status off, turning socket on")
GPIO.output(25, GPIO.HIGH)
#check current status
#if current is already 1
#do nothing
elif vvstatus == 1:
logger.info("skiping. already on")
#unhandeled current status
else:
logger.critical("unhandeled vvstatus!")
logger.critical("turning socket off")
GPIO.output(25, GPIO.LOW)
更新版本。确定相对于结束时间的当前位置，如果午夜过去，则将结束位置设置为明天
Updated version. Determine current position in time relative to the end. end is set to tomorrow if midnight passes
n1 = datetime.now()
startTrig = datetime(n1.year, n1.month, n1.day, 23, 00, 0)
logger.debug("start: %s",startTrig)
n = datetime.now()
endTrig = datetime(n.year, n.month, n.day, 07, 00, 0)
logger.debug("end: %s",endTrig)
if startTrig > endTrig:
logger.debug("start > stop")
endTrig += timedelta(days=1)
logger.debug("new stop trig: %s",endTrig)
if datetime.now() < endTrig:
if curStatus == 1:
logger.debug("socket %s already on. doing nothing.")
elif curStatus == 0:
logger.debug("socket %s sould be on. flipping switch")
flipSocketStatus(bcmNo,bcmDir)
else:
logger.critical("unhandeled socket %s current status %s",socName,curStatus)
if curStatus == 1:
logger.critical("shutting socket %s down",socName)
GPIO.output(bcmNo, GPIO.LOW)
elif curStatus == 0:
logger.warn("socket %s already off",socName)
else:
logger.critical("unhandeled current status for pin: %s",bcmNo)
logger.critical("forcing socket %s down",socName)
GPIO.output(bcmNo, GPIO.LOW)
else:
logger.critical("unhandeled start-stop rules")
推荐答案
有两种情况：当前时间在给定时间之间（顺时针）或在外部时间（想象时钟圈）之间：
There are two cases: the current time is between given times (clock-wise) or outside (imagine the clock circle):
#!/usr/bin/env python
from datetime import datetime
def in_between(now, start, end):
if start < end: # e.g., "07:00-23:00"
return start <= now < end
elif end < start: # e.g., "23:00-07:00"
return start <= now or now < end
else: # start == end
return True # consider it 24 hour interval
now = datetime.now().time()
for date_range in ["07:00-23:00", "23:00-07:00"]:
start, end = [datetime.strptime(s, "%H:%M").time()
for s in date_range.split("-")]
not_ = '' if in_between(now, start, end) else 'not '
print("{now:%H:%M} is {not_}in between {date_range}".format(**vars()))
输出
Output
02:26 is not in between 07:00-23:00
02:26 is in between 23:00-07:00
这篇关于python时间范围验证器的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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