Question

def crawler(id):
    print id
    crawer.getCourseFromUrl("http://www.imooc.com/view/"+id)
    time.sleep(3)
def main():
    print '*** Starting crawler ***'
    try:
        for id in xrange(100):
            threads = []
            for i in range(10):
                t = threading.Thread(target = crawler,args = str(i+1))
                threads.append(t)
            for t in threads:
                t.start()
            for t in threads:
                t.join()
            for t in threads:
                t.close()
    except:
        pass
    print '*** crawler End ***'

以上是我的代码，当args为1到9时，它运行良好，但是当它达到10和更大时，它会出现错误：

Exception in thread Thread-10:
Traceback (most recent call last):
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 810, in __bootstrap_inner
    self.run()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 763, in run
    self.__target(*self.__args, **self.__kwargs)
TypeError: crawler() takes exactly 1 argument (2 given)

我无法弄清楚它是什么错误。

Answer 1

尝试将args作为元组传递：

t = threading.Thread(target = crawler, args = (str(i+1),))

如此处https://docs.python.org/2/library/threading.html#threading.Thread

所述

Answer 2

undefined获取Thread并以这种方式使用函数：

args

当crawler(args[0], args[1], ...)只有一个字符时，它会给出

str(i+1)

当crawler(args[0])有两个字符时，它会给出

str(i+1)

但是你的函数只需要一个参数。

您必须在crawler(args[0], args[1])

中使用列表或元组

Thread

然后Thread(target = crawler,args = [ str(i+1) ] ) Thread(target = crawler,args = ( str(i+1), ) )会将所有args[0]作为一个元素。

python multithread和strangeTypeError：正好采用1个参数（给定2个）

2 个答案: