我正在尝试为作业调度程序编写提交程序。由于我不知道作业何时到来以及作业将运行多长时间,我使用多处理为每个作业生成一个进程,并使用子进程和分离来处理下一个作业。同时,这也很有效,但是我想在作业完成后获得返回代码,这可能吗?我尝试了几个子流程变体,但是那些返回RC的变体在作业运行时阻塞了流程
#!/usr/bin/python3
# coding=utf-8
import time
import multiprocessing
import subprocess
JobsList = []
def SubmitJob(jobname):
""" Submit the next requested job """
print(f"Starting job {jobname}...")
JobDir ="/home/xxxxx/Jobs/"
JobMem = "{}{}.sh".format(JobDir, jobname)
SysoutFile = "./Sysout/{}.out".format(jobname)
fh = open(SysoutFile, 'w')
kwargs = {}
kwargs.update(start_new_session=True)
p = subprocess.Popen(JobMem, shell = False, stdout = fh, **kwargs)
pid = p.pid
print(f"Job {jobname} pid {pid} submitted...")
def PrepareSubmit():
""" Create and start one process per job """
jobs = []
for Job in JobsList:
process = multiprocessing.Process(target=SubmitJob,
args=(Job,))
jobs.append(process)
JobsList.remove(Job)
for j in jobs:
j.start()
for j in jobs:
j.join()
print("All jobs submitted...")
def main():
""" Check queue for new job requests """
number_of_lines = 0
jobs_list = []
while 1:
job_queue = open("/home/xxxxx/Development/Python/#Projects/Scheduler/jobs.que", 'r')
lines = job_queue.readlines()
if len(lines) > number_of_lines:
jobs_list.append(lines[len(lines)-1])
NewJob = lines[len(lines)-1][:-1]
JobsList.append(NewJob)
PrepareSubmit()
number_of_lines = number_of_lines+1
time.sleep(1)
if __name__ == "__main__":
main()
main()中的while循环仅用于测试目的
有谁能告诉我这是否可能,以及如何做到?提前谢谢
这是一个代码,它给了我一个返回代码,但在上一个作业完成之前不会发送作业。因此,如果我有一个长期运行的作业,它会延迟运行作业的过程,我称之为阻塞
def Submit(job):
""" Submit the next requested job """
print(f"Starting job {job}...")
JobDir ="/home/uwe/Jobs/"
JobMem = "{}{}.sh".format(JobDir, job)
SysoutFile = "./Sysout/{}.out".format(job)
fh = open(SysoutFile, 'w')
kwargs = {}
kwargs.update(start_new_session=True)
p = subprocess.Popen(JobMem, shell = False, stdout = fh, **kwargs)
pid = p.pid
while p.poll() == None:
a = p.poll()
print(a)
time.sleep(1)
else:
rc = p.returncode
print(f"PID: {pid} rc: {rc}")
def main():
JobsList = ['JOB90501','JOB00001','JOB00002','JOB00003']
for Job in JobsList:
Submit(Job)
Roy,这是我在你最后提示后的当前代码:
def SubmitJob(jobname):
""" Submit the next requested job """
JobDir ="/home/uwe/Jobs/"
JobMem = "{}{}.sh".format(JobDir, jobname)
SysoutFile = "./Sysout/{}.out".format(jobname)
fh = open(SysoutFile, 'w')
kwargs = {}
kwargs.update(start_new_session=True)
p = subprocess.Popen(JobMem, shell = False, stdout = fh, **kwargs)
ProcessList[p] = p.pid
print(f"Started job {jobname} - PID: {p.pid}")
def main():
c_JobsList = ['JOB00001','JOB00002','JOB00003']
for Job in c_JobsList:
SubmitJob(Job)
for p, pid in ProcessList.items():
RcFile = "./Sysout/{}.rc".format(pid)
f = open(RcFile, 'w')
while p.poll() == None:
a = p.poll()
time.sleep(1)
else:
rc = p.returncode
f.writelines(str(rc))
print(f"PID: {pid} rc: {rc}")
f.close()
以及输出:
Started job JOB00001 - PID: 5426
Started job JOB00002 - PID: 5427
Started job JOB00003 - PID: 5429
PID: 5426 rc: 0
PID: 5427 rc: 0
PID: 5429 rc: 8
编辑(以下原始答案供将来参考)
用于此目的的natuaram的意思是
Popen.poll
,但显然它在某些情况下不起作用(参见https://lists.gt.net/python/bugs/633489)。我想提出的解决方案是使用Popen.wait
和非常短的超时,如以下代码示例所示:我得到的结果是:
原创创意-Popen.poll
您应该使用的方法是
Popen.poll
(documentation)。它返回进程的退出状态,如果进程仍在运行,则返回None
要使用它,您必须保留在调用
subprocess.Popen
时得到的“popen”对象,以及稍后在这些对象上调用poll
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