如何让用户输入日期并从中减去 [英] How To Have User Input Date and Subtract from It
问题描述
我想要的是用户输入一个选定的日期,然后从当前日期中减去该日期,然后根据结果创建一个睡眠定时器.
What I want is a user input a selected date, and subtract that date from the current date, and then create a sleep timer according to the results.
from datetime import tzinfo, timedelta, datetime
def ObtainDate():
isValid=False
while not isValid:
userIn = raw_input("Type Date: mm/dd/yy: ")
try:
d1 = datetime.datetime.strptime(userIn, "%m/%d/%y")
isValid=True
except:
print "Invalid Format!\n"
return d1
t = (datetime.now() - d1).seconds
我当前的当前代码看起来像这样,但我不知道如何获取 d1 并从中减去当前日期.
My current current code looks like this, but I cannot figure out how to get d1 and subtract the current date from it.
推荐答案
您的代码有一些简单的错误.此版本有效(虽然我不确定您到底需要什么,但它应该能让您解决眼前的问题).
Your code has a few simple errors. This version works (though I'm not sure exactly what you need, it should get you past your immediate problem).
from datetime import datetime
def ObtainDate():
while True:
userIn = raw_input("Type Date: mm/dd/yy: ")
try:
return datetime.strptime(userIn, "%m/%d/%y")
except ValueError:
print "Invalid Format!\n"
t0 = datetime.now()
t1 = ObtainDate()
td = (t1 - t0)
print t0
print t1
print td
print td.total_seconds()
您的主要问题是您没有调用您的函数.我还将您的 while 循环简化为无限循环.除非引发错误,否则 return 语句将跳出循环.
Your main problem was that you were not calling your function. I have also simplified your while loop to an infinite loop. The return statement will break out of the loop unless it raises an error.
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