我从数据库中提取了一个格式如下的列表:
function confirmLogoff()
{
try {
var message = 'Do you want to proceed with logout?';
var confirmed = confirm(message);
if(confirmed == true){
closeWindow();
window.location = ""/logut/jsp/Signoff.jsp";
}
} catch (e) {
}
现在在for循环中,我试图结合日期和时间来设置作业以使用APscheduler进行调度,但是首先我想将每日作业和一次性作业分开:
task_list = [
("script_to_run.py", date(2019,8,12), time(10,20), "one time"),
("script2_to_run.py", date(2019,8,12), time(10,30), "daily"),
("script3_to_run.py", date(2019,8,12), time(10,40), "daily")]
现在,第一个任务是“一次”:
def send_jobs():
for i in task_list:
if i[3] =='one time':
one_time_schedule(i)
if i[3] =='daily':
daily_schedule(i)
然后在主要部分:
def one_time_schedule(row):
date_time = datetime.combine(row[1],row[2])
sched.add_jos(function, "date", run_date = date_time)
对我来说,问题是该程序进入send_jobs方法中,标识出第一个作业,然后将其发送到方法sched = BackgroundSchedule()
send_jobs()
sched.start()
中。在那里添加了作业,但是程序不会返回到下一个作业,即列表中的第二个元素,因此我仅添加了一个作业而不是许多作业来完成程序。
打印时的结果是:
one_time_scheduler()
答案 0 :(得分:0)
您能尝试一下吗?
from apscheduler.schedulers.background import BackgroundScheduler
from datetime import datetime, date, time
schedule = BackgroundScheduler()
def daily_job(): pass
def one_time_job(): pass
task_list = [
('script_to_run.py', date(2019, 8, 12), time(10, 20), 'one time'),
('script2_to_run.py', date(2019, 8, 12), time(10, 30), 'daily'),
('script3_to_run.py', date(2019, 8, 12), time(10, 40), 'daily')]
for script, task_date, task_time, frequency in task_list:
if frequency == 'daily':
run_date = datetime.combine(task_date, task_time)
schedule.add_job(func=daily_job,
trigger='date',
run_date=run_date)
else:
schedule.add_job(func=one_time_job,
trigger='cron',
hour=task_time.hour,
minute=task_time.minute)
schedule.start()
我得到的输出是:
>> print(schedule.get_jobs())
[<Job (id=13028449c2a44b169fee37dfbacb2742 name=one_time_job)>,
<Job (id=647e0e2c90ef4e35a3b03dbeec794197 name=daily_job)>,
<Job (id=65d352d400f64287a80f5cb06fcb5f03 name=daily_job)>]