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

Question

I have this function in my Python program:

@tornado.gen.engine
def check_status_changes(netid, sensid):        
    como_url = \"\".join([\'http://131.114.52         


        

Answer 1

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.