Python SyntaxError: ("'return' with argument inside generator",)
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Python SyntaxError: ("'return' with argument inside generator",)
提问by sharkbait
I have this function in my Python program:
我的 Python 程序中有这个函数：
@tornado.gen.engine
def check_status_changes(netid, sensid):
como_url = "".join(['http://131.114.52:44444/ztc?netid=', str(netid), '&sensid=', str(sensid), '&start=-5s&end=-1s'])
http_client = AsyncHTTPClient()
response = yield tornado.gen.Task(http_client.fetch, como_url)
if response.error:
self.error("Error while retrieving the status")
self.finish()
return error
for line in response.body.split("\n"):
if line != "":
#net = int(line.split(" ")[1])
#sens = int(line.split(" ")[2])
#stype = int(line.split(" ")[3])
value = int(line.split(" ")[4])
print value
return value
I know that
我知道
for line in response.body.split
is a generator. But I would return the value variable to the handler that called the function. It's this possible? How can I do?
是一个发电机。但是我会将 value 变量返回给调用该函数的处理程序。这可能吗？我能怎么做？
采纳答案by Martijn Pieters
You cannot use returnwith a value to exit a generator in Python 2, or Python 3.0 - 3.2. You need to use yieldplus a returnwithoutan expression:
return在 Python 2 或 Python 3.0 - 3.2 中，您不能使用带值退出生成器。您需要使用不带表达式的yieldplus ：return
if response.error:
self.error("Error while retrieving the status")
self.finish()
yield error
return
In the loop itself, use yieldagain:
在循环本身中，yield再次使用：
for line in response.body.split("\n"):
if line != "":
#net = int(line.split(" ")[1])
#sens = int(line.split(" ")[2])
#stype = int(line.split(" ")[3])
value = int(line.split(" ")[4])
print value
yield value
return
Alternatives are to raise an exception or to use tornado callbacks instead.
替代方法是引发异常或使用龙卷风回调。
In Python 3.3 and newer, returnwith a value in a generator function results in the value being attached to the StopIteratorexception. For async defasynchronous generators (Python 3.6 and up), returnmust still be value-less.
在 Python 3.3 和更新版本中，return生成器函数中的值会导致该值附加到StopIterator异常。对于async def异步生成器（Python 3.6 及更高版本），return必须仍然是无值的。
