Question

这是我的代码。我想返回str1并在20-30秒后返回str2。我知道我不能两次返回str1并在str2下面。我怎么能设置计时器，所以它返回，我试过这样，但没有任何想法。

import os
import sys
import time
from weather import *

greeting=(["Change", "change", "CHANGE"]) question_1=(["stop",
"STOP"])

str1 = """Hey [First Name]! 

Are your ready for a free overview of my favorite ways to find new
income opportunities?

This is going to CHANGE the way you look at your options!

It’s kinda weird, but Facebook requires that I confirm you really want
it…so please reply and type  CHANGE"""

str2 = """ Hey [First Name]! Thanks for subscribing! As promised here
is your requested link  https://thebusinessofnursing.com/earn-more-now


P.S. If you ever want to unsubscribe, just type “stop”."""

def classify(msg):
    msg=msg.strip()
    if(msg in greeting):
        return str1

def classify2(msg):
    time.sleep(5)
    return str2

if __name__ == '__main__':
    while(1):
        msg=raw_input("Write Something: ")
        print(classify(msg), classify2(msg))

Answer 1

你可以屈服于它

def gen():
    yield str1
    time.sleep(sec)
    yield str2

或

def gen(str_list, delay):
    yield str_list[0]
    for str in str_list[1:]:
        time.sleep(delay)
        yield str

其中一种可能的用法可能是

for str in gen(str_list, 20):
      do_some_stuff()

Answer 2

延迟打印的最简单方法

import time

print("first thing")
time.sleep(20)
print("second thing")

小延迟后返回两个字符串？

2 个答案: