我有一个在我的应用程序中的不同线程中运行的类。我可以一次运行多个线程,线程是守护进程。经过一段时间后,这些线程的某些需要接收和处理消息。我该怎么做?
我的代码示例如下:
import threading
import time
class MyThread(threading.Thread):
def __init__(self, args=(), kwargs=None):
threading.Thread.__init__(self, args=(), kwargs=None)
self.daemon = True
self.receive_messages = args[0]
def run(self):
print threading.currentThread().getName(), self.receive_messages
def do_thing_with_message(self, message):
if self.receive_messages:
print threading.currentThread().getName(), "Received %s".format(message)
if __name__ == '__main__':
threads = []
for t in range(10):
threads.append( MyThread(args=(t % 2 == 0,)))
threads[t].start()
time.sleep(0.1)
for t in threads:
t.do_thing_with_message("Print this!")
输出:
Thread-1 True
Thread-2 False
Thread-3 True
Thread-4 False
Thread-5 True
Thread-6 False
Thread-7 True
Thread-8 False
Thread-9 True
Thread-10 False
MainThread Received %s
MainThread Received %s
MainThread Received %s
MainThread Received %s
MainThread Received %s
但是,我希望最后五行与MainThread
无关,而不是%s
,我希望它Print this!
,就像这样:
Thread-1 True
Thread-2 False
Thread-3 True
Thread-4 False
Thread-5 True
Thread-6 False
Thread-7 True
Thread-8 False
Thread-9 True
Thread-10 False
Thread-1 Received Print this!
Thread-3 Received Print this!
Thread-5 Received Print this!
Thread-7 Received Print this!
Thread-9 Received Print this!
如何正确地将这样的消息发送到正在运行的线程?
附录:
如果我在Print this!
阻止之后有这个阻止,并使用@ dano的代码来解决上述问题,它似乎不会响应这些新消息。
for t in threads:
t.queue.put("Print this again!")
time.sleep(0.1)
在这种情况下,我希望输出的结尾看起来像这样
Thread-1 Received Print this!
Thread-3 Received Print this!
Thread-5 Received Print this!
Thread-7 Received Print this!
Thread-9 Received Print this!
Thread-1 Received Print this again!
Thread-3 Received Print this again!
Thread-5 Received Print this again!
Thread-7 Received Print this again!
Thread-9 Received Print this again!
答案 0 :(得分:13)
您可以使用Queue.Queue
(或Python 3中的queue.Queue
):
import threading
import time
from Queue import Queue
print_lock = threading.Lock()
class MyThread(threading.Thread):
def __init__(self, queue, args=(), kwargs=None):
threading.Thread.__init__(self, args=(), kwargs=None)
self.queue = queue
self.daemon = True
self.receive_messages = args[0]
def run(self):
print threading.currentThread().getName(), self.receive_messages
val = self.queue.get()
self.do_thing_with_message(val)
def do_thing_with_message(self, message):
if self.receive_messages:
with print_lock:
print threading.currentThread().getName(), "Received {}".format(message)
if __name__ == '__main__':
threads = []
for t in range(10):
q = Queue()
threads.append(MyThread(q, args=(t % 2 == 0,)))
threads[t].start()
time.sleep(0.1)
for t in threads:
t.queue.put("Print this!")
for t in threads:
t.join()
我们将Queue
个实例传递给每个帖子,并使用Thread
将邮件发送到queue.put
。我们等待消息到达run
方法,该方法是Thread
对象的一部分,它实际上是在一个单独的执行线程中运行的。收到消息后,我们调用do_thing_with_message
,它将在同一个后台线程中运行。
我还在代码中添加了threading.Lock
,因此stdout的打印件不会混淆。
修改强>
如果您希望能够向线程传递多条消息,只需使用循环:
def run(self):
print threading.currentThread().getName(), self.receive_messages
while True:
val = self.queue.get()
if val is None: # If you send `None`, the thread will exit.
return
self.do_thing_with_message(val)