Python SyntaxError: ("'return' with argument inside generator",)

generator, python, return, tornado

Solution

You cannot use `return` with a value to exit a generator in Python 2, or Python 3.0 - 3.2. You need to use `yield` plus a `return` without an expression:

if response.error:
    self.error("Error while retrieving the status")
    self.finish()
    yield error
    return

In the loop itself, use `yield` again:

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.

In Python 3.3 and newer, `return` with a value in a generator function results in the value being attached to the `StopIterator` exception. For `async def` asynchronous generators (Python 3.6 and up), `return` must still be value-less.

Problem

I have this function in my Python program: ``` @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 ``` 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?

Original source