Question

所以我试图弄清楚为什么选择不适用于调度程序。目前它每次都没有发表新的说法。调度程序运行正常，我在其他环境中成功使用的选项。

那我在这里做错了什么？另外，我通过解释器通过说“import [project name]”来运行它，如果该信息有用。

谢谢！

from apscheduler.scheduler import Scheduler
from random import choice

#change this to a text file
cat_sayings = [
"I can haz?",
"I pooped in your shoe.",
"I ate the fish.",
"I want out.",
"Food? What...? Food?",
"When are you coming home? There's food that needs eating!",
"Lulz, I am sleeping in your laundry.",
"I didn't do it. Nope."]

sayings = choice(cat_sayings)

def cat_job(sayings):

    print sayings


s = Scheduler()
s.add_cron_job(cat_job, args=[sayings], second='*/30')
s.start()

Answer 1

您只需在模块的顶层调用choice(cat_sayings)一次，而不会再次调用def cat_job(sayings): print choice(sayings) # ... s.add_cron_job(cat_job, args=[cat_sayings], second='*/30')。因此，它将选择一个随机选择，而不是选择一个新选项。

要解决此问题，只需将代码移动到函数中：

{{1}}

如何在Python中使用调度进行随机选择

1 个答案: