Question: The code below returns `none`. How can I fix it? I'm using Python 2.6.

```
import urllib

URL = "http://download.finance.yahoo.com/d/quotes.csv?s=%s&f=sl1t1v&e=.csv"
symbols = ('GGP', 'JPM', 'AIG', 'AMZN','GGP', 'JPM', 'AIG', 'AMZN')
#symbols = ('GGP')

def fetch_quote(symbols):
    url = URL % '+'.join(symbols)
    fp = urllib.urlopen(url)
    try:
        data = fp.read()
    finally:
        fp.close()

def main():
    data_fp = fetch_quote(symbols)
#    print data_fp
if __name__ =='__main__':
    main()

```

Answer: 
You have to explicitly `return` the `data` from `fetch_quote` function. Something like this:

```
def fetch_quote(symbols):
    url = URL % '+'.join(symbols)
    fp = urllib.urlopen(url)
    try:
        data = fp.read()
    finally:
        fp.close()
    return data # <======== Return

```

In the absence of an explicit return statement Python returns `None` which is what you are seeing.
Your method doesn't explicitly `return` anything, so it `returns` `None`