选择在启动线程之前在线程内执行的方法

问题描述 投票:0回答:1

我创建了一个小烧瓶应用程序,用于从页面下载图像和文本,这可能需要很长时间,因此我想用Parell执行我的请求。我创建线程任务。我希望此任务能够从站点下载文本或图像。我将任务保存在工人清单中。但是我想选择一种方法,该方法将执行线程,然后启动整个线程。如何将我的方法传递给线程运行method()?这将是子守护程序线程吗?

import threading
import time

workers = []

class SavePage:
    def get_text(self):
        print("Getting text")

    def get_images(self):
        print("Getting images")


class Task(threading.Thread):
    def __init__(self):
        super().__init__()
        self.save_page = SavePage()

    def get_text_from_page(self):
        self.save_page.get_text()

    def get_images_from_page(self):
        self.save_page.get_images()


if __name__ == '__main__':
    task = Task()

    task.get_images_from_page()  # Why this executes, when I didn't put task.start() ?
    # Moreover, is this really threaded? or just uses a method from class Task?

    workers.append(task)  # I want this list to be empty, after job is finished
    print("".join(str(worker.is_alive()) for worker in workers))  #
    print(workers)

python multithreading flask
1个回答
0
投票
task.get_images_from_page()  # Why this executes, when I didn't put task.start() ?
# Moreover, is this really threaded? or just uses a method from class Task?

没有线程。这只是主线程中的普通方法调用。

Thread.start是将在另一个线程内启动Thread.run功能的方法。

您可以在__init__中设置一些状态来选择要执行的功能:

class Task(threading.Thread):
    def __init__(self, action):
        super().__init__()
        self.save_page = SavePage()
        self.action = action

    def get_text_from_page(self):
        self.save_page.get_text()

    def get_images_from_page(self):
        self.save_page.get_images()

    def run(self):
        if self.action == "text":
            self.get_text_from_page()
        elif self.action == "images":
            self.get_images_from_page()

请记住,可以通过传递目标函数来以更简单的方式运行线程:

def target_func():
    save_page = SavePage()
    save_page.get_images()

t = threading.Thread(target=target_func)
t.start()

# or in this simple case:

save_page = SavePage()

t = threading.Thread(target=save_page.get_images)
t.start()
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