Python:线程处理中的TypeError.函数接受x位置参数,但给定y [英] Python: TypeError in Threading. function takes x positional argument but y were given

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问题描述

目前,我正在使用Python.我有一个启动功能,可以从消息中获取字符串.我想为每条消息启动线程.

I am working with Python at the moment. I have a start-function, that gets a string from a message. I want to start threads for every message.

此线程现在应该只打印出我的消息,如下所示:

The thread at the moment should just print out my message like this:

def startSuggestworker(message):
    print(message)

def start():
    while True:
        response = queue.receive_messages()
        try:
            message = response.pop()
            start_keyword = message.body
            t = threading.Thread(target=startSuggestworker, args = (start_keyword))
            t.start()
            message.delete()
        except IndexError:
            print("Messages empty")
            sleep(150)

start()

此刻我得到一个TypeError,却不明白为什么.异常消息是这样的:

At the moment I get a TypeError and don't understand why. The Exception message is this one:

Exception in thread Thread-1:
Traceback (most recent call last):
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/threading.py", line 914, in _bootstrap_inner
    self.run()
  File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/threading.py", line 862, in run
    self._target(*self._args, **self._kwargs)
TypeError: startSuggestworker() takes 1 positional argument but y were given

* y =我的字符串的长度

*y = length of my String

我在做什么错了?

推荐答案

threading.Threadargs kwarg需要一个可迭代的对象,并且该可迭代对象中的每个元素都将传递给目标函数.

The args kwarg of threading.Thread expects an iterable, and each element in that iterable is being passed to the target function.

由于您要提供args的字符串:
t = threading.Thread(target=startSuggestworker, args=(start_keyword))

Since you are providing a string for args:
t = threading.Thread(target=startSuggestworker, args=(start_keyword))

每个字符都作为单独的参数传递给startSuggestworker.

each character is being passed as a separate argument to startSuggestworker.

相反,您应该为args提供一个元组:

Instead, you should provide args a tuple:

t = threading.Thread(target=startSuggestworker, args=(start_keyword,))
#                                                                  ^ note the comma

这篇关于Python:线程处理中的TypeError.函数接受x位置参数,但给定y的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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