如何清除scrapy工作列表？

Question

如何清除scrapy工作列表？当我启动任何蜘蛛时，我有很多特定蜘蛛的工作，我知道如何杀死所有蜘蛛？阅读文档后，我完成了下一个代码，我在循环中运行：

cd = os.system('curl http://localhost:6800/schedule.json -d project=default -d spider=google > kill_job.text')
file = open('kill_job.text', 'r')
a = ast.literal_eval(file.read())
kill='curl http://localhost:6800/cancel.json -d project=default -d job={}'.format(a['jobid'])
pprint(kill)

cd = os.system(kill)

Answer 1

So, I do not know what is wrong with my first example, but I fixed problem with this:

cd = os.system('curl http://localhost:6800/listjobs.json?project=projectname > kill_job.text')
file = open('kill_job.text', 'r')
a = ast.literal_eval(file.read())
b = a.values()
c = b[3]
for i in c:
    kill = 'curl http://localhost:6800/cancel.json -d project=projectname -d job={}'.format(i['id'])
    os.system(kill)

Answer 2

Took @kolas's script and updated it for python 3:

import json, os
PROJECT_NAME = "MY_PROJECT"

cd = os.system('curl http://localhost:6800/listjobs.json?project={} > kill_job.text'.format(PROJECT_NAME))
with open('kill_job.text', 'r') as f:
    a = json.loads(f.readlines()[0])

pending_jobs = list(a.values())[2]
for job in pending_jobs:
    job_id = job['id']
    kill = 'curl http://localhost:6800/cancel.json -d project={} -d job={}'.format(PROJECT_NAME, job_id)
    os.system(kill)