TypeError:第一个参数必须是可调用的 [英] TypeError: the first argument must be callable
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问题描述
我正在使用python并计划lib来创建类似cron的作业
I am using python and schedule lib to create a cron-like job
class MyClass:
def local(self, command):
#return subprocess.call(command, shell=True)
print "local"
def sched_local(self, script_path, cron_definition):
import schedule
import time
#job = self.local(script_path)
schedule.every(1).minutes.do(self.local(script_path))
while True:
schedule.run_pending()
time.sleep(1)
在主体中调用时
cg = MyClass()
cg.sched_local(script_path, cron_definition)
我知道了:
local
Traceback (most recent call last):
File "MyClass.py", line 131, in <module>
cg.sched_local(script_path, cron_definition)
File "MyClass.py", line 71, in sched_local
schedule.every(1).minutes.do(self.local(script_path))
File "/usr/local/lib/python2.7/dist-packages/schedule/__init__.py", line 271, in do
self.job_func = functools.partial(job_func, *args, **kwargs)
TypeError: the first argument must be callable
在类中调用另一个方法而不是sched_local时,例如
When calling another method within the class instead of sched_local, like
def job(self):
print "I am working"
工作正常.
推荐答案
do
期望可调用对象以及它接受的所有参数.
do
expects a callable and any arguments it make take.
因此,您对do
的呼叫应类似于:
Therefore your call to do
should look like:
schedule.every(1).minutes.do(self.local, script_path)
do
实现可在此处找到.
def do(self, job_func, *args, **kwargs):
"""Specifies the job_func that should be called every time the
job runs.
Any additional arguments are passed on to job_func when
the job runs.
"""
self.job_func = functools.partial(job_func, *args, **kwargs)
functools.update_wrapper(self.job_func, job_func)
self._schedule_next_run()
return self
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