TypeError:start()只取1个参数(给定0)？

Question

我有这样的事情:

class thread1(threading.Thread):
    def __init__(self):
        file = open("/home/antoni4040/file.txt", "r")
        data = file.read()
        num = 1
        while True:
           if str(num) in data:
               clas = ExportToGIMP
               clas.update()
           num += 1     
thread = thread1
        thread.start()
        thread.join()

我收到这个错误:

TypeError: start() takes exactly 1 argument (0 given)

为什么？

Answer 1

thread = thread1需要thread = thread1().否则,您尝试在类上调用方法,而不是类的实际实例.

另外,不要覆盖__init__Thread对象来完成工作 - 覆盖run.

(虽然您可以覆盖__init__以进行设置,但实际上并不在线程中运行,并且还需要调用super().)

以下是代码的外观:

class thread1(threading.Thread):
    def run(self):
        file = open("/home/antoni4040/file.txt", "r")
        data = file.read()
        num = 1
        while True:
           if str(num) in data:
               clas = ExportToGIMP
               clas.update()
           num += 1     

thread = thread1()
        thread.start()
        thread.join()