Python-3.2协程: AttributeError：'generator‘对象没有'next’属性

Question

#!/usr/bin/python3.2
import sys

def match_text(pattern):
    line = (yield)
    if pattern in line:
        print(line)

x = match_text('apple')
x.next()

for line in input('>>>> '):
    if x.send(line):
        print(line)

x.close()

这是一个协程，但Python3.2将其视为生成器-为什么？这里发生什么事情？我指的是David Beazeley pg:20的Python Essential Reference。

引用相关章节：

Normally, functions operate on a single set of input arguments. However, a function can
also be written to operate as a task that processes a sequence of inputs sent to
it.This type of function is known as a coroutine and is created by using the yield 
statement as an expression (yield) as shown in this example:
 def print_matches(matchtext):
   print "Looking for", matchtext
   while True:
     line = (yield)       # Get a line of text
     if matchtext in line:
       print line

To use this function, you first call it, advance it to the first (yield), and then 
start sending data to it using send(). For example:
>>> matcher = print_matches("python")
>>> matcher.next() # Advance to the first (yield)
Looking for python
>>> matcher.send("Hello World")
>>> matcher.send("python is cool")
python is cool
>>> matcher.send("yow!")
>>> matcher.close() # Done with the matcher function call

为什么我的代码不能工作--数据库不能工作..

deathstar> python3.2 xxx   
Traceback (most recent call last):
  File "xxx", line 9, in <module>
    matcher.next() # Advance to the first (yield)
AttributeError: 'generator' object has no attribute 'next'

Answer 1

您被错误消息搞糊涂了；就类型而言，Python不做任何区分--您可以对任何使用yield的内容执行.send操作，即使它不会在内部对发送的值做任何操作。

在3.x中，这些函数不再附加.next方法；取而代之的是使用内置的自由函数next

next(matcher)

Answer 2

对于Python3.2版，next()内置函数的语法应该是matcher.__next__()或next(matcher)。